Physics (042) / XII / TERM-1 /2021-22

KENDRIYA VIDYALAYA SANGATHAN

AHMEDABAD REGION

STUDY MATERIAL

TERM-1

CLASS XII

PHYSICS (042)

Session 2021-22

Physics (042) / XII / TERM-1 /2021-22

INSPIRATION

Shri(Dr) Jaydeep Das

Deputy Commissioner

KVS RO Ahmedabad

Smt Shruti Bhargava

Assistant Commissioner

KVS RO Ahmedabad

MENTOR

Shri Avijit Panda

Principal

Kendriya Vidyalaya,Sabarmati

Physics (042) / XII / TERM-1 /2021-22

CONTENT DEVELOPEMENT TEAM S.NO. Name of Teacher Designation Name of KV 1 Mrs. Vandana Badola PGT Physics NO.1 SHAHIBAUG AHMEDABAD

2 Mr. R N Singh PGT Physics NO.1 SHAHIBAUG AHMEDABAD

3 Mr. Bhupesh Kumar Nagoda PGT Physics AHMEDABAD CANTT

4 Mr. R H Parmar PGT Physics SAC AHMEDABAD

5 Mr. M P Dabi PGT Physics ONGC CHANDKHEDA

6 Mr. T C Agrawal PGT Physics ONGC CHANDKHEDA

7 Mr. Gurmeet Singh PGT Physics HIMMATNAGAR

8 Mrs. Anju Kumari PGT Physics AFS WADSAR

9 Mr. Praveen Nogia PGT Physics BSF DANTIWADA

10 Mr. S K Devrani PGT Physics SEC 30 GANDHINAGAR

11 Mr. Sandeep Kumar Koli PGT Physics SEC 30 GANDHINAGAR

12 Mr. Bhoor Singh Meena PGT Physics CANTT GANDHINAGAR

13 Mr. Shabbir Hingorja PGT Physics CRPF GANDHINAGAR

14 Mrs. Anjana PGT Physics NO 1 AFS BHUJ

15 Mr. Akhilesh Suryavanshi PGT Physics NO 2 BHUJ CANTT

16 Mr. Deepak Kumar Singhal PGT Physics AFS NALIYA

17 Mr. Viswanath PGT Physics DHRANGADHRA

18 Mr. Yaksh Deep PGT Physics PORBANDAR

19 Mr. Sahil Aneja PGT Physics BHAVNAGAR PARA

20 Mr. Ranjay Kumar Sharma PGT Physics RAJKOT

21 Mr. Om Prakash Yadav PGT Physics RAJKOT

22 Mr. Mukesh Kumar PGT Physics NO 1 AFS JAMNAGAR

23 Ms. Suryakant Vats PGT Physics NO 1 AFS JAMNAGAR

24 Mr. Ram PGT Physics INF LINES JAMNAGAR

25 Dr Neeraj Singh PGT Physics AFS II JAMNAGAR

26 Mr. Umesh Kumar PGT Physics INS VALSURA

27 Mr. Pinkesh Rathod PGT Physics AFS SAMANA

28 Mr. Naveen Kumar PGT Physics DIU

29 Mr. Rakesh Kumar PGT Physics NO 1 BARODA

30 Mr. S V Singh PGT Physics EME BARODA

31 Mr. Sudhir Bhootda PGT Physics EME BARODA

32 Mr. Sanjeev Chaddha PGT Physics AFS BARODA

33 Mr. J K Jain PGT Physics AFS BARODA

34 Mrs. Sangeeta Arora PGT Physics ONGC BARODA

35 Mr. Chitresh Pandya PGT Physics DAHOD

36 Mr. C M Sharma PGT Physics ONGC ANKLESHWAR

37 Mr. V K Pathak PGT Physics NO 1 SURAT

38 Mr. Pawan Kumar PGT Physics KRIBHCO SURAT

39 Mr. R M Shukla PGT Physics ONGC SURAT

40 Mr. A P Singh PGT Physics SILVASSA

41 Mr. P K Sah PGT Physics VV NAGAR

42 Mrs. Anju PGT Physics ONGC MEHSANA

43 Mr. Kamlesh Kumar PGT Physics RAILWAY GANDHIDHAM

44 Mr. Balwant Kumar PGT Physics AHMEDABAD CANTT

45 Mr. Vikas Sharma PGT Physics SABARMATI

Physics (042) / XII / TERM-1 /2021-22

PHYSICS

Class-XII

INDEX

S.No Particulars 1 SYLLABUS (TERM-1)

2 Deleted Topics (TERM-1)

3 Question Paper Design (TERM-1)

4 Chapter-1; Electric Charge and Field

5 Chapter-2; Electric Potential and Capacitance

6 Chapter-3; Current Electricity

7 Chapter-4; Moving Charges and Magnetism

8 Chapter-5; Magnetism and Matter

9 Chapter-6; Electromagnetic Induction

10 Chapter-7; Alternating Current

11 CBSE Sample Paper(TERM-1) with Answer key 2021-22

*Every Chapter Contains: 1. Gist of Lesson

2. Multiple Choice Questions with Answer key

3. Assertion and Reason Question with Answer key

4. Case Study Based Questions with Answer key

Physics (042) / XII / TERM-1 /2021-22

SYLLABUS

TERM-1 Session 2021-22

PHYSICS, CLASS-XII

UNIT Chapters Marks

Unit 1 Chapter 1: Electric Charges and Fields

Chapter 2: Electric Potential and Capacitance

17

Unit 2 Chapter 3: Current Electricity

Unit 3 Chapter 4: Moving Charges and Magnetism

Chapter 5: Magnetism and Matter

18

Unit 4 Chapter 6: Electromagnetic Induction

Chapter 7: Alternating Current

Total 35

Physics (042) / XII / TERM-1 /2021-22

DELETED TOPICS (For Session 2021-22)

PHYSICS(042)

CLASS XII

TERM-1

S.No Name of the Chapter Deleted Topics

1 Electric Charges &

Fields

uniformly charged thin spherical shell (field inside and

outside).

2 Electric Potential &

Capacitance No topic is deleted

3 Current Electricity Carbon resistors, colour code for carbon resistors;

series and parallel combinations of resistors

4 Moving Charges and

Magnetism

Cyclotron

5 Magnetism and Matter magnetic field intensity due to a magnetic dipole (bar

magnet) along its axis and perpendicular to its axis,

torque on a magnetic dipole (bar magnet) in a

uniform magnetic field;

Para-, dia- and ferro - magnetic substances, with

examples. Electromagnets and factors affecting their strengths, permanent magnets.

6 Electromagnetic Induction

No topic is deleted

7 Alternating Current power factor, wattless current

Physics (042) / XII / TERM-1 /2021-22

QUESTION PAPER DESIGN TERM-1 EXAM

2021-22

CHAPTER 1

ELECTRIC CHARGE & FIELD

GIST OF LESSON:

Coulomb’s Law of Electrostatics

Electrostatic force of interaction acting between two stationary point charges is given by

𝐹 = 1

4𝜋𝜀0.𝑞1𝑞2

𝑟2

where q1, q2 are magnitude of point charges, r is the distance between them and εo is

permittivity of free space.

Here, 1 / 4πεo = 9 x 10 9 N-m2/C2

The value of εo is 8.85 X 10-12 C2 / N-mC2.

If there is another medium between the point charges except air or vacuum, then εo is

replaced by εoK or εoεr or ε.

where K or εr is called dielectric constant or relative permittivity of the medium.

K = εr = ε / εo where, ε = permittivity of the medium.

For air or vacuum, K = 1 For water, K = 81 For metals, K = ∞

Coulomb Law implies:

Force on q1 due to q2 = – Force on q2 due to q1

F12 = – F21

The forces due to two point charges are parallel to the line joining point charges; such forces

are called central forces and electrostatic forces are conservative forces.

Electric Field:

The space in the surrounding of any charge in which its influence can be experienced by

other charges is called electric field.

Electric Field Lines

“An electric field line is an imaginary line or curve drawn through a region of space so that

its tangent at any point is in the direction of the electric field vector at that point. The

relative closeness of the lines at some place give an idea about the intensity of electric field

at that point.”

Two lines can never intersect.

Electric field lines always begin on a positive charge and end on a negative charge and do

not start or stop in mid space.

Electric Field Intensity (E)

The electrostatic force acting per unit positive charge on a point in electric field is called

electric field intensity at that point.

Electric field intensity 𝐸 =𝐹

𝑞

Its SI unit is NC-1 or V/m and its dimension is [MLT-3 A-1].

It is a vector quantity and its direction is in the direction of electrostatic force acting on

positive charge.

Electric field intensity due to a point Charge:

due to a point charge q at a distance r is given by

Magnitude - 𝐸 = 1

4𝜋𝜀0.

𝑞

𝑟2

Properties of Electric Field Lines:

Electric lines of force are the imaginary lines drawn in electric field at Which a positive test

charge will move if it is free to do so.

Electric lines of force start from positive charge and terminate on negative charge.

A tangent drawn at any point on electric field represents the direction of electric field at

that point.

Two electric lines of force never intersect each other.

Electric lines of force are always perpendicular to an equipotential surface.

Electric Flux (φE)

Electric flux over an area is equal to the total number of electric field lines crossing this area.

Electric flux through a small area element dS is given by

φE = E. dS

where E= electric field intensity and dS = area vector.

Its SI unit is Nm2C-1.

Gauss’s Theorem:

The electric flux over any closed surface is 1 / εo times the total charge enclosed by that

surface, i.e.,

∅𝐸 = ∮ 𝐸. 𝑑𝑠 = 𝑄𝑖𝑛

𝜖0 where Qin = Net Charge enclosed in the surface.

Note: If a charge q is placed at the centre of a cube, then total electric flux linked with the

whole cube = q / εo, electric flux linked with one face of the cube = q / 6 εo.

Electric Field Intensity due to an Infinite Line Charge

𝐸 = 𝜆

2𝜋𝜖0𝑟 ( Direction –Radially outwards for q > 0 and inwards or q<0)

where λ is linear charge density and r is distance from the line charge.

Electric Field Near an Infinite Plane Sheet of Charge:

E = σ / 2 εo

where σ = surface charge density.

If infinite plane sheet has uniform thickness, then

E = σ / εo

Electric Dipole:

An electric dipole consists of two point charges of equal magnitude and opposite sign

separated by a very small distance. e.g., a molecule of HCL, a molecule of water etc.

Electric Dipole Moment p = q (2a)

Its SI unit is ‘coulomb-metre’ and its dimension is [LTA).

It is a vector quantity and its direction is from negative charge towards positive charge.

Electric Field Intensity and Potential due to an Electric Dipole

(i) On Axial Line

Electric field intensity 𝐸 = 1

4 𝜋 𝜀𝑜

2 𝑝𝑟

(𝑟2 – 𝑎2)2

If r > > 2a, then 𝐸 = 1

4 𝜋 𝜀𝑜

2 𝑝

𝑟3 ( Short Dipole)

Electric potential 𝑉 = 1

4 𝜋 𝜀𝑜

𝑝

𝑟2 – 𝑎2

If r > > 2a, then 𝑉 = 1

4 𝜋 𝜀𝑜

𝑝

𝑟 ( Short Dipole)

(ii) On Equatorial Line

Electric field intensity 𝐸 = 1

4 𝜋 𝜀𝑜

𝑝

(𝑟2 + 𝑎2)3/2

If r > > 2a, then 𝐸 = 1

4 𝜋 𝜀𝑜

𝑝

𝑟3 ( Short Dipole)

Electric potential V = 0

(iii) At any Point along a Line Making θ Angle with Axis

Electric field intensity 𝐸 = 1

4 𝜋 𝜀𝑜 𝑝 √1+3𝑐𝑜𝑠2𝜃

𝑟3

Electric potential 𝑉 = 1

4 𝜋 𝜀𝑜

𝑝𝑐𝑜𝑠𝑐𝑜𝑠 𝜃

𝑟2− 𝑎2𝑐𝑜𝑠2 𝜃

If r > > 2a, then 𝑉 = 1

4 𝜋 𝜀𝑜 𝑝𝑐𝑜𝑠𝑐𝑜𝑠 𝜃

𝑟2 ( Short Dipole)

Torque:

Torque acting on an electric dipole placed in uniform electric field is given by

τ = pEsin θ or τ = p x E

When θ = 90°, then ‘τmax = pE (Maximum)

When electric dipole is parallel to electric field, it is in stable equilibrium and when it is anti-

parallel to electric field, it is in unstable equilibrium. ( In this Case Torque = 0)

Dipole in Non-Uniform Electric Field:

When an electric dipole is placed in a non-uniform electric field, then a resultant force as

well as a torque act on it.

Net force on electric dipole = (qE1 – qE2), along the direction of greater electric field

intensity.

Therefore electric dipole undergo rotational as well as linear motion.

CHAPTER 1

ELECTRIC CHARGE AND FIELD

MULTIPLE CHOICE QUESTIONS:

1. Four charges +8Q, -3Q, +5Q and -10 Q are kept inside a closed surface. What will be

the outgoing flux through the surface

(a) 26 V-m (b) 0 V-m (c) 10 V-m (d) 8 V-m

2. The value of electric field inside a conducting sphere having radius R and charge Q

(a) KQ/R2 (b) KQ/R (c) Zero (d) KQ2/R2

3. Which is vector Quantity among the following–

(a) Electric flux (b) Electric charge

(c) Electric field (d) Electric potential

4. The SI unit of electric field is

(a) N/C2 (b) (N/C) x m2 (c) (N/m2) x C (d) N/C

5. The electric field intensity due to infinite charged sheet

(a) σ/Ɛ0 (b) σ/2Ɛ0 (c) σ/4Ɛ0 (d) σ/6Ɛ0

6. The direction of electric dipole moment

(a) Negative charge to positive charge.

(b) Positive charge to negative charge.

(c) Perpendicular to the line joining two charges.

(d) Positive Charge to positive charge.

7. The direction of electric field due to electric dipole at a point on its equatorial line

(a) Parallel to the direction of dipole moment.

(b) Perpendicular to the direction of dipole moment.

(c) Opposite to the direction of dipole moment.

(d) From Negative to Positive charge.

8. The electric field due to a uniformly charged linear conductor of linear charge

density λ at a distance r is -

(a) λ/4πƐ0r (b) 2λ/πƐ0r (c) λ/Ɛ0r (d) λ/2π Ɛ0r

9. The net torque on a dipole in an uniform electric field is Zero, When the angle

between the dipole movement and magnetic field is -

(a) 300 (b) 600 (c) 900 (d) 00

10. Gauss theorem gives the relation between

(a) Charge and flux (b) Charge and field

(c) Force and flux (d) Dipole movement and electric field.

11. What amount of charge we place at corner point A of the cube of sides a so that

the total flux through the faces of the cube is

(a) q (b) q/2 (c) 2q (d) q/8

12 Four equal charges q are placed at the four comers A, B, C, D of a square of length

a. The magnitude of the force on the charge at B will be

13. In Fig. (i) two negative charges q2 and q3 fixed along the y-axis, exert a net electric

force in the negative-x direction on a charge q1 fixed along the x-axis. If a positive

charge Q is added at (x, 0) in figure (ii), the force on q1

(a) shall increase along the positive x-axis.

(b) shall decrease along the positive x-axis.

(c) shall point along the negative x-axis.

(d) shall increase but the direction changes because of the intersection of Q with q2

and q3

14. The figure here shows electric field lines. The electric field strength at P1 is E1 and

that at P2 is E2 If distance between P1, P2 is r, then which of the following statement

is true?

(a) E1 > E2 (b) E1 < E2 (c) E2 = rE1 (d) E2 = E1/r²

15. Four charges are arranged at the corners of a square ABCD, as shown in the adjoining

figure. The force on the charge kept at the centre O is (a) Zero (b) Along the diagonal AC (c)Along the diagonal BD (d) Perpendicular to side AB

16. Consider a system of three charges q/3, q/3 and –2q/3 placed at points A, B and C respectively as shown in the figure. Take O to be the centre of the circle of radius R and

angle CAB = 60º

(a) The electric field at point O is q/8πε0R2

(b) The magnitude of the force between the charges at C and B is q2/54πε0R2 (c) The potential energy of the system is zero

(d) The potential at point O is q/12πε0R

17. A hemispherical surface of radius R is kept in a uniform electric field E as shown in

the figure. The flux th rough the curved surface is:

(a) EπR2 (b) E2πR2 (c) E4πR2 (d) Zero

18. A thin conducting ring of radius R is given a charge +Q. The electric field at the

centre O of the ring due to the charge on the part AKB of the ring is E. The electric

field at the centre due to the charge on the part ACDB of the ring is

(a) E along KO (b) 3E along OK (c) 3E along KO (d) E along OK

19. Three charges + 3q + q and Q are placed on a straight line with equal separation.

In order to make the net force on q to be zero, the value of Q will be : (a) +3q (b) +2q (c) -3q (d) -4q

20. Which of the following figures represent the electric field lines due to a single

negative charge?

21. The SI unit of electric flux is

(a) N C-1 m-2 (b) N C m-2 (c) N C-2 m2 (d) N C-1 m2

22. Each of the two point charges are doubled and their distance is halved. Force of

interaction becomes p times, where p is :

(a) 1 (b) 4 (c) 1/16 (d) 16

23. A hollow cylinder has a charge q coulomb within it. if Φ is the electric flux in units of

volt meter associated with the curved surface B, the flux linked with the plane

surface A in units of volt meter will be (charge is symmetrically placed within it)

24. Three charges, each +q, are placed at the corners of an isosceles triangle ABC of

sides BC and AC, 2a. D and E are the mid points of BC and CA. The work done in

taking a charge Q from D to E is

(a) (b) (c) zero (d)

25. Figure shows the part of an infinite plane sheet of charge

Which of the following graphs correctly shows the behaviour of electric field

intensity as we move from point O to A.

(a) (b) (c) (d)

26 Three charges + 3q + q and Q are placed on a straight line with equal separation. In

order to make the net force on q to be zero, the value of Q will be :

(a) +3q (b) +2q (c) -3q (d) -4

27 For a point charge, the graph between electric field versus distance is given by :

28 Which one of the following is the unit of electric field?

(a) Coulomb (b) Newton (c) Volt (d) N/C

29 Two charges 3 × 10-5 C and 5 × 104 C are placed at a distance 10 cm from each

other. Find the value of electrostatic force acting between them.

(a) 13.5 × 1011 C (b) 40 × 10-5 C

(c) 180 × 109 C (d) 3.05 × 10-11 C

30 Four charges + 8Q, - 3Q +5Q and -10Q are kept inside a closed surface. What

will be the outgoing flux through the surface.

(a) 26 V-m (b) 0 V-m (c) 10 V-m (d) 8 V-m

31

Electric field lines contracts lengthwise, It shows

(a) repulsion between same charges

(b) Attraction between apposite charges

(c) No relation between force & contraction.

(d) Electric field lines does not moves on straight path.

32 What is the S. I. unit of electric flux?

(a) N/C × m2

(b) N ×m2

(c) N2/C × m2

(d) N2/C2 × m2

33 What is the value of minimum force acting between two charges placed at 1 m

apart from each other

(a) Ke2 (b) Ke (c) Ke/4 (d) 2Ke2

34 A glass rod acquires charge by rubbing it with silk cloth. The charge on glass rod

is due to

(a) Friction (b) Conduction (c) Induction (d) Radiation

35

The number of electron-taken out from a body to produce 1 coulomb of charge will

be

(a) 6.25 × 1018 (b) 625 × 1018 (c) 6.023 × 1023 (d) None

36 If sphere of bad conductor is given charge then it is distributed on:

(a) surface (b) inside the surface (c) only inside the surface (d)None

37 Electric field in a cavity of metal:

(a) depends upon the surroundings (b) depends upon the size of cavity

(c) is always zero (d) is not necessarily zero

38 The dielectric constant of a metal is:

(a) 0 (b) 1 (c) ∞ (d) -1

39 About electric flux :

(a) SI unit is NC -1m 2 (b) it’s a scalar physical quantity.

(c) SI unit is Vm (d) All(a), (b) and (c) are correct.

40 Each of the two point charges are doubled and distance between the two point

charges is also dobled. Force of interaction becomes m times, where m is :

(a) 1 (b) 4 (c) 116 (d) 16

41 Coulomb’s law in vector form can be written as

F = (1/4πϵ0) (Q1 Q 2 /r 2)r

where ε0 is the permitivity of free space. The SI units of ϵ0 will be:

(a) N m2 / C2 (b) N-1 m -2 C2 (c) N m3 / C (d) 2 N m2 / C2

42 The dimensional formula for Charge is :

(a) [MLT 4A 2] (b) [M -1L -3 T 4A 2] (c)[MLT 1 A1 ] (d) None of these

43 If ∮ 𝐄 . 𝐝𝐬 = 0 , inside a surface, that means :-

(a) there is no net charge present inside the surface

(b) Uniform electric field inside the surface

(c) Discontinues field lines inside the surface

(d) Charge present inside the surface

44 A parrot comes and sits on a bare high power line. It will:

(a) experience a mild shock (b) experience a strong shock.

(c) get killed instantaneously (d) not be affected practically

45 If two conducting sphere are connected after charging

separately then:

(a) Electrostatic energy sphere energy will remain conserved

(b) Electrostatic energy charges remains conserved

(c) Electrostatic energy decreases and charge remains conserved

(d) None

46 An electron is sent in an electric field of magnitude 9.1 ×

10 6NC -1. The acceleration produced in it is :

(a) 1.6 ms-2 (b) 1.6 × 10 18ms -2

(c) 3.2 × 10 18ms-2 (d) 0.8 × 1018 ms-2

47 The value of electric field inside a conducting sphere having radius R and charge

Q will be

(a) KQ/ R 2 (b) KQ/ R (c) Zero (d) 2 KQ 2 / R

48 Charge on a body is integral multiple of +e . It is given by the law of -

(a) Conservation of charge (b) Conservation of mass

(c) Conservation of energy (d) Quantisation of charge

49 Charge Q is kept in a sphere of 5 cm first than it is kept in a cube of side 5 cm.

the outgoing flux will be-

(a) More in case of sphere (b) More in case of cube

(c) Same in both case (d) Information Incomplete

50 On charging a neutral Balloon its size -

(a) Increases (b) Decreases

(c) Remains same (d) No relation between charge & size

ANSWER KEY( MCQs):

Q. 1 2 3 4 5 6 7 8 9 10

ANS B C C D B A C D A A

Q. 11 12 13 14 15 16 17 18 19 20

ANS A C A A C B A D A B

Q. 21 22 23 24 25 26 27 28 29 30

ANS D D D b c C B D A B

Q. 31 32 33 34 35 36 37 38 39 40

ANS B A A A A D C C D A

Q. 41 42 43 44 45 46 47 48 49 50

ANS B C A D C B C D C A

ASSERTION AND REASON QUESTIONS:

For question numbers 1 to 24, two statements are given-one labelled Assertion (A) and the

other labelled Reason (R). Select the correct answer to these questions from the codes (a),

(b), (c) and (d) as given below.

a) Both A and R are true and R is the correct explanation of A

b) Both A and R are true but R is NOT the correct explanation of A

c) A is true but R is false

d) A is false and R is also false

1 Assertion : The electrostatics force increases with decrease the distance between the

charges.

Reason : The electrostatic force of attraction or repulsion between any two stationary point

charges is inversely proportional to the square of the distance between them

2 Assertion: When a body is charged, its mass changes.

Reason : Charge is quantized.

3 Assertion: The total amount of charge on a body equal to 4 x10-19 C is not possible.

Reason : Experimentally it is established that all free charges are integral multiples of a basic

unit of charge denoted by e. Thus, charge q on a body is always given by q = ne.

4 Assertion: The electrostatics force increases with decrease the distance between the

charges.

Reason : The electrostatic force of attraction or repulsion between any two stationary point

charges is inversely proportional to the square of the distance between them.

5 Assertion: The Coulomb force between two points charges depend upon the dielectric

constant of the intervening medium.

Reason: Coulomb’s force varies inversely with the dielectric constant of medium.

6 Assertion: The charge given to a metallic sphere does not depend on whether it is hollow or

solid.

Reason: The charge resides only at the surface of conductor.

7 Assertion: A comb run through one’s dry hair attracts small bits of paper.

Reason: Molecules in the paper gets polarized by the charged comb resulting in net force of

attraction

8 Assertion: A proton is placed in a uniform electric field, it tends to move along the direction

of electric field.

Reason: A proton placed in a uniform electric field experiences a force.

9 Assertion: Electric field at the surface of a charged conductor is always normal to the surface

at every point.

Reason: Electric field gives the magnitude & direction of electric force F experienced by any

charge placed at any point.

10 Assertion: Electric filed lines do not form closed loops.

Reason: Electric filed lines are always normal to the surface of a conductor.

11 Assertion: Electrostatic forces are conservative in nature.

Reason: Work done by electrostatic force is path dependent.

12 Assertion: A negatively charged body means that the body has gained electrons while a

positively charged body means the body has lost some of its electrons.

R: Charging process involves transfer of electrons.

13 Assertion: If dipole moment of water molecules were zero, then microwave cooking would

not be possible.

Reason: In a microwave oven the water molecules vibrate due to oscillating electric field in

microwave and heat the food.

14 Assertion: Electric field lines are continuous curves in free space.

Reason: Electric field lines start from negative charge and terminate at positive charge.

15 Assertion: A point charge cannot exert force on itself.

Reason: Coulomb force is a central force.

16 Assertion: An electron has negative charge by definition.

R: Charge of a body depends on its velocity.

17 Assertion: Since matter cannot be concentrated at a point, therefore point charge is not

possible.

Reason: An electron is a point charge.

18 Assertion: A finite size charged body may behave like a point charge if it produces an

inverse square electric field.

Reason: Two charged bodies may be considered as point charges if their distance of

separation is very large compared to their dimensions.

19 Assertion: The path traced by a positive charge is a field line.

Reason: A field line can intersect itself.

20 Assertion: If electric flux over a closed surface is negative then the surface encloses net

negative charge.

Reason: Electric flux is independent of the charge distribution inside the surface.

21 Assertion: Particles such as photon or neutrino which have no rest mass are uncharged.

Reason: Charge cannot exist without mass.

22 Assertion: The equilibrium of a charged particle under the action of electrostatic force

alone can never be stable.

Reason: Coulomb force is an action-reaction pair.

23 Assertion: The field in a cavity inside a conductor is zero which causes electrostatic

shielding.

Reason: Dielectric constant of conductors in electrostatics is infinite.

24 Assertion: Though quark particles have fractional electronic charges, the quantum of charge

is still e.

Reason: Quark particles do not exist in Free State.

25 Assertion. In a non-uniform electric field, a dipole will have translatory as well as rotatory

motion.

Reason. In a non-uniform electric field, a dipole experiences a force as well as a torque.

26 Assertion. Net electric field inside a conductor is zero .

Reason. Total positive charge equals to total negative charge in a charged conductor.

27 Assertion: Three equal charges are situated on a circle of radius r such that they form on equilateral triangle, then the electric field intensity at the centre is zero.

Reason:The force on unit positive charge at the centre, due to the three equal charges are represented by the three sides of a triangle taken in the same order. Therefore, electric field intensity at centre is zero.

28 Assertion: The surface densities of two spherical conductors of different radii are equal.

Then the electric field intensities near their surface are also equal.

Reason:Surface density is equal to charge per unit area.

29 Assertion: When a conductor is placed in an external electrostatic field, the net electric field

inside the conductor becomes zero after a small instant of time.

Reason: It is not possible to set up an electric field inside a conductor.

30 Assertion: The electric flux of the electric field ∮ E.dA is zero. The electric field is zero

everywhere on the surface.

Reason:The charge inside the surface is zero.

ANSWERS(Assertion & Reason):

Q.No. ANS Q.No. ANS Q.No ANS

1 A 11 C 21 A

2 B 12 A 22 B

3 A 13 A 23 B

4 A 14 C 24 A

5 A 15 B 25 A

6 A 16 C 26 C

7 A 17 C 27 A

8 B 18 B 28 B

9 B 19 D 29 C

10 B 20 B 30 D

CASE STUDY QUESTIONS:

1 Gold Leaf Electroscope: Gold leaf electroscope has two gold leafs suspended

from a metal(usually brass) stem in a vacuumed glass jar and connected to a

metal cap. The glass is grounded with the help of a metal foil to make it

uncharged. It can be used to: Detect charge: Body under test is touched with the

metal cap.

(i) Neutral atoms contain equal numbers of positive and negative .

(a) Electrons and Protons

(b) Protons and Electrons

(c) Neutrons and Electrons

(d) Protons and Neutrons

(ii) greater the charge on a given body, the divergence in the leaves will be…

(a) smaller (b) greater (c) same (d) first smaller then greater

(iii) Magnitude of force between a pair of proton and proton is F and between a

proton d electron is F’ then…

(a) F = F’ (b) F > F’ (c) F < F’ (d) F >> F’

(iv) Earthing means…

(a) to perform an experiment on Earth

(b) to burry the device in the Earth

(c) To put the device on Earth

(d) Process of sharing charges with Earth

(v) If a negatively charged rod touches a conductor, the conductor will be charged

by what method?

(a) Friction (b) Conduction (c) Convection (d) Induction

2 Concept of Electric Field: Electric field is an elegant way of characterizing the

electrical environment of a system of charges. Electric field at a point in the space

around a system of charges tells you the force a unit positive test charge would

experience if placed at that point (without disturbing the system). Electric field is

a characteristic of the system of charges and is independent of the test charge

that you place at a point to determine the field

(i) Which of the following statement is correct? The electric field at a point is

(a) always continuous.

(b) continuous if there is a charge at that point.

(c) discontinuous only if there is a negative charge at that point.

(d) discontinuous if there is a charge at that point

(ii) The force per unit charge is known as…

(a) Electric Field (b) Electric Flux

(c) Electric Current (d) Electric potential.

(iii) The SI unit of electric field is…

(a) N/C (b) V/m (c)V m (d) both (a) and (b)

(iv) A proton of mass ‘m’placed in electric field region remains stationary in air

then magnitude of electric field is…

(a) mge (b) mg/e (c) e/mg (d) e²g/m²

(v) Electric field is …

(a) Vector (b)Scalar (c) tensor (d) none of the above

3 Electric Charge: When a glass rod is rubbed with silk, the rod acquires one kind

of charge and the silk acquires the second kind of charge. This is true for any pair

of objects that are rubbed to be electrified. Now if the electrified glass rod is

brought in contact with silk, with which it was rubbed, they no longer attract each

other. They also do not attract or repel other light objects as they did on being

electrified. Thus, the charges acquired after rubbing are lost when the charged

bodies are brought in contact. What can you conclude from these observations?

It just tells us that unlike charges acquired by the objects neutralise or nullify each

other’s effect. Therefore, the charges were named as positive and negative by

the American scientist Benjamin Franklin. We know that when we add a positive

number to a negative number of the same magnitude, the sum is zero. This might

have been the philosophy in naming the charges as positive and negative. By

convention, the charge on glass rod or cat’s fur is called positive and that on

plastic rod or silk is termed negative. If an object possesses an electric charge, it

is said to be electrified or charged. When it has no charge it is said to be

electrically neutral.

(i) When you charge a balloon by rubbing it on your hair this is an example of what

method of charging?

(a)Friction (b)Conduction (c)Grounding (d)Induction

(ii) Neutral atoms contain equal numbers of positive __ and negative __.

(a)Electrons and Protons (b)Protons and Electrons

(c)Neutrons and Electrons (d)Protons and Neutrons

(iii) Which particle in an atom can you physically manipulate?

(a)protons (b)electrons

(c)neutrons (d)you can't manipulate any particle in an atom

(iv) If a negatively charged rod touches a conductor, the conductor will be

charged by what method?

(a) Friction (b)Conduction (c)Induction (d)Convection

(v) A negatively charged rod is touched to the top of an electroscope, which on is

correct in the given figure

(a) A (b) B (c) C (d) D

4 Electrostatic Induction: Figure (a) shows an uncharged metallic sphere on an

insulating metal stand. If we Bring a negatively charged rod close to the metallic

sphere, as shown in Fig. (b). As the rod is brought close to the sphere, the free

electrons in the sphere move away due to repulsion and start piling up at the

farther end. The near end becomes positively charged due to deficit of electrons.

This process of charge distribution stops when the net force on the free

electrons inside the metal is zero. Now if we Connect the sphere to the ground

by a conducting wire. The electrons will flow to the ground while the positive

charges at the near end will remain held there due to the attractive force of the

negative charges on the rod, as shown in Fig. (c). Disconnect the sphere from the

ground. The positive charge continues to be held at the near end Fig.(d). if we

remove the electrified rod. The positive charge will spread uniformly over the

sphere as shown in Fig. (e). In this experiment, the metal sphere gets charged by

the process of induction and the rod does not lose any of its charge.

(i) What do you call the process of charging a conductor by bringing it near

another charged object?

(a) Induction (b) Polarisation (c) neutralization (d) conduction

(ii)The negatively charged balloon is brought near the two cans. What happens?

(a)The negative charges in Can B move towards the balloon

(b)The negative charges in Can A move away from the balloon

(c)The positive charges in Can B move towards the balloon

(d)The positive charges in Can A move away from the balloon

(iii) Transferring a charge without touching is ___

(a)Conduction (b)Induction (c)Grounding (d)Newtons 3rd law

(iv) Due to electrostatic induction in aluminum rod due to charged plastic rod,

the total charge on the aluminum rod is

(a)Zero (b)Positive (c) Negative (d) Dual

(v) If we bring charged plastic rod near-neutral aluminum rod, then rods will

(a)Repel each other (b)Attract each other

(c)Remain their position (d)Exchange charges

5

Electric Dipole: The electric field due to a

charge configuration with total charge

zero is not zero, but for distances large

compared to the size of the

configuration, its filed falls off faster than

1/r2, typical of the field due to a single

charge. An electric dipole is the simplest

example of this fact. An electric dipole is

a pair of equal and opposite charges +q

and -q separated by some distance 2a. Its

dipole moment vector p has magnitude

2qa and is in the direction of the dipole axis from -q to +q. The electric field of the

pair of charges can be found out from Coulomb’s law and the superposition

principle. The magnitude and the direction of the dipole field depend not only on

the distance r but also on the angle between the position vector r and the dipole

moment p. In some molecules, like H2O, the centers of -ve charges and of +ve

charges do not coincide. So they have permanent dipole moment. Such molecules

are called polar molecules.

(i) What will be the value of electric field at the center of the electric dipole?

(a) Zero

(b) Equal to the electric field due to one charge at Centre

(c) Twice the electric field due to one charge at Centre

(d) Half the value of electric due to one charge at Centre

(ii) If r is the distance of a point from the Centre of a short dipole, then the electric field intensity due to the short dipole remains proportional to

(a) r2 (b) r3 (c) r-2 (d) r-3

(iii) An electric dipole coincides on Z-axis and its midpoint is on origin of the

coordinate system. The electric field at an axial point at a distance z from origin

is Ez and electric field at an equatorial point at a distance y from origin is Ey. Here

z=y>>a, so |Ez|/|Ey| is equal to…

(a) 1 (b)4 (c) 3 (d) 2

(iv) An electric dipole of moment p is placed in a uniform electric field E. The

maximum torque experienced by the dipole is…

(a) pE (b) p/E (c) E/p (d) p . E

(v) The frequency of oscillation of electric dipole having dipole moment p and

rotational inertia I, oscillating in a uniform electric filed E, is given by…

(a) (1/2π) √𝐼/𝑝𝐸 (b) (1/2π) √𝑝𝐸/𝐼

(c) (2π) √𝑝𝐸/𝐼 (d) (2π) √𝐼/𝑝𝐸

6 Electric Flux & Gauss’s Theorem: The term electric field flux implies some kind of

flow.

Flux is the property of any vector field. Electric flux is

a

property of Electric field. It is equal to the product of

the given area

and the normal component of the electric filed

through it.

Gauss’s theorem gives a relationship between the

total flux passing

through any closed surface S and the charge q enclosed with in the surface. Its

states that the total flux surface S and the charge q enclosed with in the surface.

It states that the total flux through a closed surface is 1/Ɛ0 times the net charge

enclose by the surface.

Mathematically, ɸE = ∫ 𝐸. 𝑑𝑆 = 𝑞/𝑠

Ɛ0

Gauss’s theorem is quite useful in calculating the electric field in problems where

it is possible to choose a closed surface that the electric field E has a normal

component which is either zero

or has a single fixed value at every point on the surface. Symmetry considerations

in many problems make the application of Gauss’s theorem much easier. The

closed surface we choose

(having symmetry consideration in view) to solve a given problem is called

Gaussian surface. Gauss’s theorem is based on inverse square dependence on the

distance contained in Coulomb’s law. Any violation of Gauss’s theorem will reflect

a deviation from the inverse square law.

(i) What is the SI unit of electric flux?

(a) (N/C) * m2 (b) N * m2

(c) N/m2 (d)(N2/m2) * C2

(ii)∫ 𝐸. 𝑑𝑆 = 0, inside a surface, that means there is..

(a) Charge present inside the surface.

(b)uniform electric field inside the surface

(c) discontinuous field lines inside the surface

(d) some the charge present inside the surface

(iii) For a given surface the Gauss’s law is stated as∫ 𝐸. 𝑑𝑆 = 0. From this we

can conclude that..

(a) E is necessarily zero on the surface.

(b) E is perpendicular to the surface at every point

(c) the total flux through the surface is zero

(d) the flux is only going out of the surface.

(iv) The electric flux for Gaussian

surface A that encloses the charged

particles in free space is…

(Given q1 = -14nC, q2 = 78.85nC, q3 =

-56nC)

(a)103 Nm2C-1

(b)103 CN-1 m-2

(c) 6.32 x 103 Nm2C-1

(d) 6.32 x 103 CN-1 m-2

(v) Charge q is first kept in a sphere of radius 5 cm and then it is kept in a cube of

side 5 cm.

The outgoing flux….

(a) will be more in case of sphere

(b) will be more in case of cube

(c) will be same in both cases

(d) cannot be determined

7 Electric Field Lines: Electric field strength is proportional to the density of lines of

force i.e., electric field strength at a point is proportional to the number of lines

of force cutting a unit area element placed normal to the field at that point. As

illustrated in given figure, the electric field at P is stronger than at Q.

(i) Electric lines of force about a positive point charge are (a) radially outwards (b) circular clockwise (c) radially inwards (d) parallel straight lines

(ii) Which of the following is false for electric lines of force? (a) They always start from positive charge and terminate on negative charges. (b) They are always perpendicular to the surface of a charged conductor. (c) They always form closed loops. (d) They are parallel and equally spaced in a region of uniform electric field.

(iii) Which one of the following patterns of electric line of force is not possible in

field due to stationary charges?

(iv) Electric field lines are curved

(a) in the field of a single positive or negative charge

(b) in the field of two equal and opposite charges.

(c) in the field of two like charges.

(d) both (b) and (c)

(v) The figure below shows the electric field lines due to two positive charges. The magnitudes EA, EB and EC of the electric fields at point A, B and C respectively are related as

(a) EA>EB>EC

(b) EB>EA>EC

(c) EA=EB>EC

(d) EA>EB=EC

8 Torque on Dipole in Uniform electric field: When electric dipole is placed in uniform electric field, its two charges experience equal and opposite forces, which cancel each other and hence net force on electric dipole in uniform electric field is zero. However, these forces are not collinear, so they give rise to some torque on the dipole. Since net force on electric dipole in uniform electric field is zero, so no work is done in moving the electric dipole in uniform electric field. However, some work is done in rotating the dipole against the torque acting on it.

(i) The dipole moment of a dipole in a uniform external field Ē is B. Then the torque τ acting on the dipole is (a) τ=p x E (b) τ = p. E (c) τ = 2(p + E) (d) τ = (p + E)

(ii) An electric dipole consists of two opposite charges, each of magnitude 1.0 μC separated by a distance of 2.0 cm. The dipole is placed in an external field of 105 NC-1. The maximum torque on the dipole is (a) 0.2 x 10-3 Nm (b) 1x 10-3 Nm (c) 2 x 10-3 Nm (d) 4x 10-3 Nm

(iii) Torque on a dipole in uniform electric field is minimum when θ is equal to (a) 0° (b) 90° (c) 180° (d) Both (a) and (c)

(iv) When an electric dipole is held at an angle in a uniform electric field, the net force F and torque τ on the dipole are (a) F= 0, τ = 0 (b) F≠0, τ≠0 (c) F=0, τ ≠ 0 (d) F≠0, τ=0

(v) An electric dipole of moment p is placed in an electric field of intensity E. The dipole acquires a position such that the axis of the dipole makes an angle with the direction of the field. Assuming that potential energy of the dipole to be zero when 0 = 90°, the torque and the potential energy of the dipole will respectively be (a) pEsinθ, -pEcosθ (b) pEsinθ, -2pEcosθ (c) pEsinθ, 2pEcosθ (d) pEcosθ, – pEsinθ

ANSWERS (CASE STUDY QUESTIONS):

ANS.(i) ANS.(ii) ANS.(iii) ANS.(iv) ANS.(v)

Q.1 b b a d b

Q.2 b a d b a

Q.3 a b b b C

Q.4 a b b a b

Q.5 c d d a b

Q.6 a a c a c

Q.7 a c c d a

Q.8 a c d c a

CHAPTER 2

ELECTRIC POTENTIAL & CAPACITANCE

GIST OF LESSON:

Electric Potential (V)

Electric potential at any point is equal to the work done per positive charge in carrying it from infinity to

that point in electric field.

Electric potential, V = W / q

Its SI unit is J / C or volt and its dimension is [ML2T-3A-1].

It is a scalar quantity.

Electric potential due to a point charge at a distance r is given by

𝑉 = 1

4𝜋𝜀0.𝑞

𝑟

Potential Gradient:

The rate of change of potential with distance in electric field is called potential gradient.

Potential gradient = 𝑑𝑉

𝑑𝑟

Its unit is V / m.

Relation between potential gradient and electric field intensity is given by

𝐸 = −𝑑𝑉

𝑑𝑟

Equipotential Surface

Equipotential surface is an imaginary surface joining the points of same potential in an electric field. So,

we can say that the potential difference between any two points on an equipotential surface is zero. The

electric lines of force at each point of an equipotential surface are normal to the surface.

(i) Equipotential surface may be planer, solid etc. But equipotential surface can never be point size.

(ii) Electric field is always perpendicular to equipotential surface.

(iii) Equipotential surface due to an isolated point charge is spherical.

(iv) Equipotential surface are planer in an uniform electric field.

(v) Equipotential surface due to a line charge is cylindrical.

Properties of Electric Field Lines:

Electric lines of force are the imaginary lines drawn in electric field at which a positive test charge will

move if it is free to do so.

Electric lines of force start from positive charge and terminate on negative charge.

A tangent drawn at any point on electric field represents the direction of electric field at that point.

Two electric lines of force never intersect each other.

Electric lines of force are always perpendicular to an equipotential surface.

Potential Energy:

Potential energy of an electric dipole in a uniform electric field is given by

U = – pE cosθ = p.E

Dipole in Non-uniform Electric Field

When an electric dipole is placed in a non-uniform electric field, then a resultant force as well as a torque

acts on it.

Net force on electric dipole = (qE1 – qE2), along the direction of greater electric field intensity.

Therefore electric dipole undergoes rotational as well as linear motion.

Electrostatic Potential Energy of Charge System:

(1) Two point charge system, contains charges q1 and q2 separated by a distance r is given by 𝑈 =

1

4 π εo 𝑞1𝑞2

𝑟

(2) Three point charge system

𝑈 = 1

4 π εo 𝑞1𝑞2

𝑟12+

1

4 π εo 𝑞2𝑞3

𝑟23+

1

4 π εo 𝑞1𝑞3

𝑟13

Important Points

When charge is given to a soap bubble its size gets increased.

In equilibrium for a charged soap bubble, pressure due to surface tension

= electric pressure due to charging

4T / r = σ2 / 2 εo

or 4T / r = 1 / 2 εo (q / 4 πr2)2

or q = 8 πr √2 εo rT

where, r is radius of soap bubble and T is surface tension of soap bubble.

Behaviour of a Conductor in an Electrostatic Field:

(i) Electric field at any point inside the conductor is zero.

(ii) Electric field at any point on the surface of charged conductor is directly proportional to the surface

density of charge at that point, but electric potential does not depend upon the surface density of charge.

(iii) Electric potential at any point inside the conductor is constant and equal to potential.

(iv) Direction of electric field at any point on its surface is always normal to the surface.

Electrostatic Shielding:

The process of protecting certain field from external electric field is called, electrostatic shielding.

Electrostatic shielding is achieved by enclosing that region in a closed metallic chamber.

Dielectric

Dielectrics are of two types Non-polar Dielectric the non-polar dielectrics (like N2, O2, benzene, methane)

etc. is made up of non-polar atoms/molecules, in which the centre of positive charge coincides with the

centre of negative charge of the atom/molecule.

Polar Dielectric

The polar dielectric (like H2O, CO2, NH3 etc) are made up of polar atoms/molecules, in which the centre

of positive charge does not coincide with the centre of negative charge of the atom.

Capacitor

A capacitor is a device which is used to store huge charge over it, without changing its dimensions.

When an earthed conductor is placed near a charged conductor, then it decreases its potential and therefore

more charge can be stored over it.

A capacitor is a pair of two conductors of any shape, close to each other and have equal and opposite

charges.

Capacitance of a conductor C = q / V

Its SI unit is coulomb/volt or farad (F)

Its other units are 1 μ F = 10-6 F

1 μμ F = 1 pF = 10-12 F

Capacitance of an Isolated Spherical Conductor

C = 4 π εo K R For air K = 1

∴ C = 4 π εo R = R / 9x 109

Parallel Plate Capacitor

The parallel plate capacitor consists of two metal plates parallel to each other and separated by a distance d.

Capacitance C = KA εo / d

For air capacitor Co = A εo / d

When a dielectric slab is inserted between the plates partially, then its capacitance.

𝐶 = A εo

(d – t + t / K)

If a conducting (metal) slab is inserted between the plates, then

𝐶 = A εo

(d –t)

Force Between the Plates of the Capacitor:

The plates of a parallel plate capacitor attract each other with a force

F = Q2 / 2 A εo

Charge Induced:

When a dielectric slab is placed between the plates of a capacitor than charge induced on its side due to

polarization of dielectric is

q’ = q (1 – 1 / K)

Capacitors Combination:

(i) In Series:

Resultant capacitance = 1 / C = 1 / C1 + 1 / C2 + 1 / C3 + ….

In series charge is same on each capacitor, which is equal to the charge supplied by the source.

If V1, V2, V3,…. are potential differences across the plates of the capacitors then total voltage applied by

the source

V = V1 + V2 + V3 + ….

(ii) In Parallel:

Resultant capacitance C = C1 + C2 + C3 + ….

In parallel potential differences across the plates of each capacitor is same.

If q1, q2, q3 are charges on the plate of capacitors connected in parallel then total charge given by the source

q = q1 + q2 + q3 + ….

Electrostatic Energy Stored in Capacitor:

Electric potential energy of a charged conductor or a capacitor is given by,

𝑈 = 1

2𝑄𝑉 =

1

2𝐶𝑉2 =

𝑄2

2𝐶 Where Q = Charge. V = Potential of Capactor, C = Capacitance

Redistribution of Charge:

When two isolated charged conductors are connected to each other then charge is redistributed in the ratio

of their capacitances.

Common potential 𝑉 = q1 + q2

C1 + C2 =

C1V1 + C2V2

C1 + C2

Energy loss ∆𝐸 = 1

2

𝐶1𝐶2(𝑉1−𝑉2)2

(𝐶1+𝐶2)

This energy is lost in the form of heat in connecting wires.

CHAPTER-2

ELECTROSTATIC POTENTIAL AND CAPACITANCE

MULTIPLE CHOICE QUESTIONS:

1. When charge is supplied to a conductor, its potential depends upon

(a) the amount of charge (b) Geometry & size of conductor

(c) both (a)&(b) (d) only on(a)

2. A parallel plate capacitor is charged by a battery. Once it is charged battery is removed. Now a dielectric material is inserted between the plates of the capacitor, which of the following does not change?

(a) electric field between the plates (b)potential difference across the plates

(c) charge on the plates (d) energy stored in the capacitor.

3. A dipole is placed parallel to electric field. If W is the work done in rotating the dipole

from 0° to 60°, then work done in rotating it from 0° to 180°is

(a) 2 W (b) 3W

(c) 4 W (d) W/2

4. The variation potential V with r & electric field E with r for a point charge is correctly

shown in the graphs

5. A charge Q is supplied to a metallic conductor. Which is true?

(a) Electric field inside it is same as on the surface.

(b) Electric potential inside is zero.

(c) Electric potential on the surface is zero

(d) Electric potential inside it is constant

6. A parallel plate capacitor C has a charge Q. The actual charges on the plates are

(a) Q, Q (b) Q / 2, Q /2

(c) Q,-Q (d) Q1/2 ,Q2/2

7. Three capacitors of capacitances 1µf, 2µF & 3µF are connected in series and a potential

difference of 11V is applied across the combination them the potential difference

across the plates of 1µF capacitor is

(a) 2V (b) 4V (c) 1V (d)6V

8. The electric field at the Centre of a cavity is-

(a) Zero (b) kq /a2 (c) kq/a (d) kq/2a2

9. Two conducting spheres A and B of radii a & b respectively are at the same

potential. The ratio of surface charge densities of A and B is

(a) b/a (b) a/b (c) a2/b2 (d) b2/a2

10. Work done to bring a unit positive charge un-accelerated from infinity to a point

inside electric field is called:

(a) Electric field (b) Electric potential

(c) Capacitance (d) Electric flux

11. Electric potential due to a point charge –q at distance x from it is given by:

(a) Kq/x2 (b) Kq/x

(c) -Kq/x2 (d) -Kq/x

12. Electric field is always:

(a) Parallel to equipotential surface

(b) Perpendicular to equipotential surface

(c) It can be perpendicular and parallel as well

(d) It does not depend on distribution of charge

13. Electric field and electric potential inside a charged

spherical shell:

(a) E = 0; V= 0 (b) E = 0; V ≠0

(c) E ≠ 0; V= 0 (d) E ≠ 0; V ≠0

14. Shape of equipotential surface in uniform electric field will be:

(a) Spherical normal to electric field

(b) Random

(c) circular normal to electric field

(d) Equidistant Planes normal to electric field

15. On reducing potential across a capacitor, its capacitance:

(a) Decreases (b) Increases

(c) Remains constant (d) First increases then decreases

16. Energy stored in a in a charged capacitor is given by:

(a) U= CV/2 (b) U = CV2/2

(c) 2CV2 (d) VC2/2

17. A spherical conductor has charge Q on its surface. Radius of the conductor is R. Which of the following is correct? (a) Electric Potential is a non-zero constant (b) Electric Potential is a constant (c) Electric Field is zero (d) Electric Field is a non-zero constant

18. The electrostatic potential on the surface of a charged conducting sphere is 200 V.

Two statements are made in this regard: S1 : At any point inside the sphere, electric intensity is zero. S2 : At any point inside the sphere, the electrostatic potential is 200V. Which of the following is a correct statement? (a) S1 is true but S2 is false. (b) Both S1 & S2 are false. (c) S1 is False but S2 is true. (d) both S1 and S2 are true..

19. In a region of constant potential (a) the electric field is uniform (b) the electric field is zero. (c) there can be no charge inside the region. (d) the electric field shall necessarily change if a charge is placed outside the region.

20. If I place an second uniformly charge shell (Q) of radius R'>R, will the value of potential change at r<R (a) increases (b) decreases (c) constant (d) none of the above

21. The electric charges are distributed in a small volume. The flux of the electric field through a spherical surface of radius 10 cm surrounding the total charge is 20 Vm. The flux over a concentric sphere of radius 20 cm will be

(a) 20 Vm. (b) 25 Vm. (c) 40 Vm. (d) 200 Vm.

22. A soap bubble is given negative charge, its radius will

(a) Increase (b) decrease (c) remains unchanged (d) fluctuate

23. Three charges Q, +q and +q is placed at the vertices of a right-angled triangle of side

‘a’. The net electrostatic energy of the configuration is zero, if Q is equal to-

(a) -q/(1+√2) (b) -2q/(2+√2) (c) -2q (d) +q

24 . The electrostatic force between the metal plates of an isolated parallel plate capacitor C having a charge Q and area A, is (a) proportional to the square root of the distance between the plates.

(b) linearly proportional to the distance between the plates. (c) independent of the distance between the plates. (d) inversely proportional to the distance between the plates.

25. A capacitor is charged by a battery. The battery is removed and another identical uncharged capacitor is connected in parallel. The total electrostatic energy of resulting system (a) increases by a factor of 4. (b) decreases by a factor of 2. (c) remains the same. (d) increases by a factor of 2.

26. 64 drops each having the capacity C and potential V are combined to form a big drop. If the charge on the small drop is q, then the charge on the big drop will be (a) 2q (b) 4q (c) 16q (d) 64q

27. A parallel plate condenser is connected with the terminals of a battery. The distance between the plates is 6mm. If a glass plate (dielectric constant K = 9) of 4.5 mm is introduced between them, then the capacity will become (a) 2 times. (b) the same. (c) 3 times. (d) 4 times.

28. Two metal plates form a parallel plate capacitor. The distance between the plates is d. A

metal sheet of thickness and of the same area is introduced between the plates. What is the ratio of the capacitance in the two cases? (a) 2 : 1 (b) 3 : 1 (c) 2 : 1 (d) 5 : 1 29. A capacitor of 4 μF is connected as shown in the circuit. The internal resistance and emf of the battery is 0.5 Ω and 2.5 V respectively. The amount of charge on the capacitor plates will be ---

(a) 0 (b) 4 (c) 16 μC (d) 8 μC

30. A capacitor is charged by using a battery which is then disconnected. A dielectric slab then slipped between the plates, which results in (a) reduction of charge on the plates and increase of potential difference across the plates (b) increase in the potential difference across the plate, reduction in stored energy, but no change in the charge on the plates (c) decrease in the potential difference across the plates, reduction in the stored energy, but no change in the charge on the plates. (d) none of these

31. In a parallel plate capacitor, the capacity increases if (a) area of the plate is decreased (b) distance between the plates increases. (c) area of the plate is increased. (d) dielectric constantly decreases

32. A parallel plate air capacitor is charged to a potential difference of V volts. After disconnecting the charging battery the distance between the plates of the capacitor is increased using an insulating handle. As a result the potential difference between the plates (a) increases (b) decreases (c) does not change (d) becomes zero

33. Two identical capacitors are joined in parallel, charged to a potential V, separated and then connected in series; the positive plate of one is connected to the negative of the other. Which of the following is true? (a) The charges on the free plated connected together are destroyed (b) The energy stored in the system increases (c) The potential difference between the free plates is 2V (d) The potential difference remains constant

34. A capacitor has some dielectric between its plates, and the capacitor is connected to a dc source. The battery is now disconnected and then the dielectric is removed, then (a) capacitance will increase (b) energy stored will decrease (c) electric field will increase (d) voltage will decrease

35. Two spherical conductors each of capacity C are charged to potential V and -V. These are then connected by means of a fine wire. The loss of energy is

(a) zero (b) CV2 (c) CV2 (d) 2 CV2

36 .In series combination of capacitors, the net capacitance of combination : (a) increases (b) decreases (c) remains same (d) zero

37 . A slab of material of dielectric constant K has the same area as the plates of a parallel plate capacitor but has a thickness (3/4)d, where d is the separation of the plates. The ratio of the capacitance C (in the presence of the dielectric) to the capacitance C0 (in the absence of the dielectric) is:

(a)

(b)

(c)

(d) 38 . A capacitor is charged by using a battery which remains connected. A dielectric slab is then slipped between the plates, which results in: (a) Reduction of charge on the plates and increase of potential across the plates (b) Increase in the potential difference across the plates, reduction in stored energy, but no change in the charge on the plates (c) Decrease in the potential difference across the plates, reduction in electric field, but no change in the charge on the plates (d) Increase in charge on the plates and no change in potential difference across the plates 39 . In the given figure, three capacitors C1, C2 and C3 are joined to a battery, with symbols having their usual meanings, the correct conditions will be:

(a) Q1 = Q2 = Q3 and V1 = V2 = V3 + V (b) Q1 = Q2 + Q3 and V = V1 + V2 + V3 (c) Q1 = Q2 + Q3 and V = V1 + V2 (d) Q3 = Q2 and V2 = V3 40. Two capacitors, 3 μF and 4 μF, are individually charged across a 6 volt battery. After being disconnected from the battery, they are connected together with the negative plate of one attached to the positive plate of the other. What is the common potential? (a) 6 volt (b) (6/7) volt (c) 2 volt (d) (3/2) volt

41 . Condenser A has a capacity of 15 μF when it is filled with a medium of dielectric constant 15. Another condenser B has a capacity of 1 μF with air between the plates. Both are charged separately by a battery of 100 V. After charging, both are connected in parallel without the battery and the dielectric material being removed. The common potential is now: (a) 400 V (b) 800 V (c) 1200 V (d) 1600 V 42 . In the circuit as shown in the figure:

The effective capacitance between A and B is: (a) 3 μF (b) 2 μF (c) 4 μF (d) 8 μF 43. A network of four capacitors of capacity equal to C1 = C, C2 = 2C, C3 = 3C and C4 = 4C are connected to a battery as shown in the figure. The ratio of the charges on C2 and C4 is:

(a) 4/7 (b) 3/22 (c) 7/4 (d) 22/3 44. In the circuit diagram, potential difference between point A and B is 200 volt, the potential difference between point a and b when the switch S is open, is:

(a) 100 volt (b) volt (c) volt (d) 50 volt

45 . Between the plates of a capacitor, a dielectric slab is slowly removed and then reinserted. The net work done by the system in this process is:

(a) Zero (b) 1

2(𝐾 − 1)𝐶𝑥2

(c) (d) (K – 1)CV2 ANSWERS MULTIPLE CHOICE QUESTIONS:

Q.NO. ANS Q.NO. ANS Q.NO. ANS Q.NO. ANS Q.NO. ANS

1 C 11 D 21 A 31 C 41 B

2 C 12 B 22 A 32 A 42 C

3 C 13 B 23 B 33 C 43 B

4 B 14 D 24 C 34 C 44 B

5 D 15 C 25 B 35 C 45 A

6 C 16 B 26 D 36 B

7 D 17 A 27 C 37 C

8 A 18 D 28 A 38 D

9 A 19 B & C 29 D 39 C

10 B 20 A 30 C 40 B

ASSERTION & REASONING QUESTIONS:

Note: -These questions consist of two statemcents, each printed as Assertion and Reason.

While answering these questions, you are rbequired to choose any one of the following

four responses.

(a) If both Assertion and Reason are correct and the Reason is a correct explanation of the

Assertion.

(b) If both Assertion and Reason are correct but Reason is not a correct explanation of the

Assertion.

(c) If the Assertion is correct but Reason is incorrect.

(d) If both the Assertion and Reason are incorrect.

1. Assertion: The equatorial plane of the dipole is an equipotential surface.

Reason: The electric potential at any point on equatorial plane is zero.

2. Assertion: The total charge stored in a capacitor is zero.

Reason: The electric field just outside the capacitor is σ/ε0. Where σ is surface charge

density.

3. Assertion: The electric potential at any point on the equatorial plane of a dipole is zero.

Reason: The work done in bringing a unit positive charge from infinity to a point in

equatorial plane is equal for the two charges of the dipole.

4. Assertion: Electric field inside a conductor is zero.

Reason: The electric potential at all points inside the conductor is the same.

5. Assertion: For a charged particle moving from point A to point B, the net work done by an

electrostatic field on the particle is independent of the path connecting point A and B.

Reason: The net work done by a conservative force on an object moving along a closed

loop is zero.

6. Assertion: If the distance between parallel plates of a capacitor is halved and dielectric

constant is three times, then the capacitance becomes six times.

Reason: capacitance of capacitor does not depend upon the nature of the material.

7. Assertion: A parallel plate capacitor is connected across battery through a key. A

dielectric constant K is introduced between the plates. The energy which is stored

becomes K times.

Reason: The surface density of charge on the plate remains constant.

8. Assertion: Two equipotential surfaces cannot intersect each other.

Reason: Two equipotential surfaces are parallel to each other.

9. Assertion: The potential difference between any two points in an electric field depends

on initial and final position.

Reason: Electric field is a conservative field so the work done per unit positive charge

does not depend on the path followed.

10. Assertion: Work done in moving a charge between any two points in an electric field is

independent of the path followed by the charge between these points.

Reason: Electrostatic force is a non-conservative force.

11. Assertion: Electric field is discontinuous across the surface of a spherical charged shell.

Reason: Electric potential is continuous across the surface of a spherical charged shell.

12. Assertion: Polar molecules have permanent dipole moment.

Reason: In polar molecules, the centre of positive and negative charges coincide even

when there is no external electric field.

13. Assertion: Electric energy resides out of the spherical isolated conductor.

Reason: The electric field at any point inside the conductor is zero.

14. Assertion: Dielectric polarisation means formation of positive and negative charges

inside the dielectric.

Reason: Free electrons are formed in this process.

15. Assertion: A dielectric is inserted between the plates of a battery connected to a

capacitor. The energy of the capacitor increases.

Reason: Energy of the capacitor is U = CV2/2.

16. Assertion: When a dielectric slab is gradually inserted between the plates of an isolated

parallel plate capacitor, the energy of the system decreases.

Reason: The electrostatic force between the plates decreases.

ANSWERS(Assertion & Reason):

1. b 2.c 3. d 4. b 5. a 6. c 7.c 8.c

9. a 10.c 11. b 12. c 13.a 14. c 15. a 16. c

CASE STUDY BASED QUESTIONS:

Q1. CAPACITOR: The capacity of a condenser increases both when a conducting slab or an

insulating slab is introduced between the plates of the condenser. In the former case,

electric field inside the conductor is zero and in the latter case, electric field inside the

insulator decreases. Thus potential difference decreases in both cases between the plates of

the condenser and hence capacity of the condenser increases.

Now answer the followings.

1. New capacitance of the capacitor becomes when a dielectric medium of dielectric

constant k is introduced between its plates

(a) K times (b) 1/K times (c) no change (d) K2 times

2. Charge on the capacitor is doubled. Its capacity becomes K times, where K is equal to

(a) 2 (b) 1 (c)1/2 (d) 4

3. Which are the dielectric medium

(a)Air or vacuum (b) mica, glass, ceramic

(c) Paper, Oil, Rubber (d) all the above

4. Energy of a capacitor of capacitance 10μ F having charge 10 μC is

(a) 5μJ (b) 10μJ (c) 15μJ (d) 2.5μJ

5. Capacitor stores which type of energy

(a) kinetic energy (b) potential energy

(c) vibrational energy (d) heat energy

6. Dielectric constant is equal to

(a) Cair / Cmed (b) Cmed / Cair (c) Cair x Cmed (d) none of these

Q2. CAPACITOR: A capacitor contains two oppositely charges metallic conductors at a finite

separation. It is a device by which capacity of storing charge may be varied simply by

changing separation and or medium between the conductors. Capacitance is numerically

equal to charge on the plates of capacitor divided by the potential. Now answer the

followings.

(1) Electric field between plates of capacitor is

(a) Uniform (b) Non –uniform

(c) depends on size of capacitor (d) None of these

(2) Capacitance of capacitor is increased by

(a) Inserting medium between its plates

(b) Decreasing area of its plates

(c) Increasing distance between its plates

(d) None of these

(3) The electric field intensity between plates of capacitor is

(a) σ / ε0 (b) σ / 2ε0 (c) 2σ / ε0 (d) σ / 3ε0

(4) Energy stored in a capacitor of capacitance 6 μ F having charge on its plate 6 μ C is

(a)3 μ J (b) 6 μ J (c)12 μ J (d)4 μ J

(5) When a number of capacitors are connected in series between two points, all the

capacitors possesses same

(a) capacity (b) Potential (c) charge (d)none of these

(6) On reducing potential across a capacitor, its capacitance

(a) decreases (b) increases (c) remains same (d) none of these

Q3. ELECTROSTATIC POTENTIAL: Electrostatic potential of a charged body represents the

degree of electrification of the body. It determines the flow of electric charge between two

charged bodies placed in contact with each other. The charge always flows from a body of

higher potential to another body at lower potential. The flow of charge stops as soon as the

potential of the two bodies become equal. Now answer the followings.

1. Nm/C is the unit of

(a) Electric potential (b) electric field intensity

(c) electric flux (d) None of these

2. A uniform electric field of 100 N/C exists vertically upward direction. The decrease in

electric potential as one goes up through a height of 5cm is

(a) 20V (b) 120 V (c) 5V (d) zero

Hint* V = E x d

3. Work done to bring a unit positive charge unaccelerated from infinity to a point inside

electric field is called

(a) electric field (b) electric potential (c) capacitance (d) electric flux

4. Two conducting spheres A and B of radii a and b respectively are at the same potential.

The ratio of surface charge densities of A and B is

(a) b/a (b)a/b (c)a2/ b2 (d) b2/ a2

Hint* 4. Q1/q2 =C1V/ C2V =C1/C2 = (4π ε0 a/4π ε0 b) = a/b

and σ1= q/4π a2 then σ1/σ2 = (q1/q2)x (b/a)2 = b/a

5. Electric potential at a distance of 27 cm from a point charge 9 nC, is given by

(a) 200 V (b) 100 V (c) 300 V (d) 50 V

Hint* V = kQ/R (R = radius of shell)

6. Two charges -6μC and 10μC are placed at distant 18 cm apart. Electric potential at the

midpoint joining these two will be

(a) 4x 105V (b) 4x 104V (c) 4x 103V (d) 4x 106V

Hint* V = k/r ( Q1 +Q2) where r = 9 cm

Q4. Electrostatic Shielding: The electric field inside the cavity is zero, whateverbe the size

and shape of the cavity and whatever be the charge on the conductor and the external

fields in which it might be placed. The electric field inside a charged spherical shell is zero.

But the vanishing of electric field in the (charge-free) cavity of a conductor is , as mentioned

above, a very general result. A related result is that even if the conductor is charged or

charges are induced on a neutral conductor by an external field, all charges reside only on

the outer surface of a conductor with cavity. The proofs of the results noted in Fig.are

omitted here, but we note their important implication. Whatever be the charge and field

configuration outside, any cavity in a conductor remains shielded from outside electric

influence: the field inside the cavity is always zero. This is known as electrostatic shielding.

The effect can be made use of in protecting sensitive instruments from outside electrical

influence.

1. A metallic shell having inner radius R1and outer radii R2 has a point

charge Q kept inside cavity. Electric field in the region R1< r < R2where r is

the distance from the centre is given by a) Depends on the value of r b) Zero c) Constantand nonzeroeverywhere d) Noneoftheabove

2. The electric field inside the cavity is depend on a) Size of the cavity b) Shape of the cavity c) Charge on the conductor d) None of the above

3. Electrostatic shielding is based (a) Electric field inside the cavity of a conductor is less than zero (b) Electric field inside the cavity of a conductor is zero (c) electric field inside the cavity of a conductor is greater than zero (d) electric field inside the cavity of a plastic is zero

4. During the lightning thunderstorm, it is advised to stay a) inside the car b) under trees

c) in the open ground d) on the car

5. Which of the following material can be used to make a Faraday cage (based on electrostatic shielding) a) Plastic b) Glass c) Copper d) Wood

Q5.Electrostatic Potential & Field: Read the passage given below and answer the following questions: The potential at any observation point P of a static electric field is defined as the work done by the external agent (or negative of work done by electrostatic field)in slowly bringing a unit positive point charge from infinity to the observation point. Figure shows the potential variation along the line of charges. Two point charges Q1 and Q2 lie along a line at a distance from each other.

The following questions are multiple choice questions. Choose the most appropriate answer: 1. At which of the points 1, 2 and 3 is the electric field is zero? (a)1 (b) 2 (c) 3 (d) Both (a) and (b)

Hint: As−𝑑𝑉

𝑑𝑟 =Er, the negative of the slope of V versus r curve represents the component of

electric field along r. Slope of curve is zero only at point 3. Therefore, the electric field vector is zero at point 3.

2. The signs of charges Q1 and Q2 respectively are (a) positive and negative (b) negative and positive (c) positive and positive (d) negative and negative Hint: Near positive charge, net potential is positive and near a negative charge, net potential is negative. Thus, charge Q1 is positive and Q2, is negative.

3. Which of the two charges Q1, and Q2 is greater in magnitude? (a) Q2 (b) Q1 (c) Same (d) can’t determined Hint: From the figure, it can be seen that net potential due to two charges is positive everywhere in the region left to charge Q1. Therefore the magnitude of potential due to charge Q1 is greater than due to Q2.

4. Which of the following statement is not true? (a) Electrostatic force is a conservative force. (b) Potential energy of charge q at a point is the work done per unit charge in bringing a charge from any point to infinity. (c) When two like charges lie infinite distance apart, their potential energy is zero. (d) Both (a) and (c).

5. Positive and negative point charges of equal magnitude are kept at (0,0,𝑎

2) and(0,0,

−𝑎

2)

respectively. The work done by the electric fieldwhen another positive point charge is moved from(-a, 0, 0) to (0, a, 0) is (a) positive (b) negative (c) zero (d) depends on the path connecting the initial and final position Hint: It can be seen that potential at the points both A and Bare zero. When the charge is moved from A to B, work done by the electric field on the charge will be zero.

6. Work done in moving a positive charge on a equipotential surface is: (a) negative (b) zero (c) positive (d) infinte

ANSWERS(Case Study Questions):

ANS. 1 ANS. 2 ANS. 3 ANS. 4 ANS. 5 ANS. 6

Q.1 a b d a b b

Q.2 a a a a c c

Q.3 a c b a c a

Q.4 b d b a c ……

Q.5 c a b b c b

CHAPTER-3 CURRENT ELECTRICITY

GIST OF LESSON: • Current carriers – The charge particles which flow in a definite direction constitutes the electric

current are called current carriers. E.g.: Electrons in conductors, Ions in electrolytes, Electrons and holes in semi-conductors.

• Electric current is defined as the amount of charge flowing through any cross section of the conductor in unit time. I = Q/t.

• Current density J = I/A.

• Ohm’s law: Current through a conductor is proportional to the potential difference across the ends of the conductor provided the physical conditions such as temperature, pressure etc. Remain constant. V α I i.e. V = IR, Where R is the resistance of the conductor. Resistance R is the ratio of V & I

• Resistance is the opposition offered by the conductor to the flow of current.

• Resistance R = ρl/A where ρ is the resistivity of the material of the conductor- length and A area of cross section of the conductor. If l is increased n times, new resistance becomes n2R. If A is increased

n times, new resistance becomes Rn2

1

• Resistivity ρ = m/ne2τ, Where m, n, e are mass, number density and charge of electron respectively, τ-relaxation time of electrons. ρ is independent of geometric dimensions.

• Relaxation time is the average time interval between two successive collisions

• Conductance of the material G =1/R and conductivity σ=1/ρ

• Drift velocity is the average velocity of all electrons in the conductor under the influence of applied electric field. Drift velocity Vd = (eE/m)τ also I = neAvd

• Mobility (μ) of a current carrier is the ratio of its drift velocity to the applied field dV

E =

• Effect of temperature on resistance: Resistance of a conductor increase with the increase of

temperature of conductor (1 )T oR R T= + , where α is the temperature coefficient of resistance of

the conductor. α is slightly positive for metal and conductor, negative for semiconductors and insulators and highly positive for alloys.

• Cells: E.M.F of a cell is defined as the potential difference between its terminals in an open circuit. Terminal potential difference of a cell is defined as the p.d between its ends in a closed circuit.

• Internal resistance r of a cell is defined as the opposition offered by the cell to the flow of current.

r = RV

E

−1 where R is external resistances.

• Grouping of cells :

i) In series grouping circuit current is given by s

nEI

R nr=

+,

ii) In parallel grouping circuit current is given by p

mEI

r mR=

+ where n, m are number of cells in series

and parallel connection respectively.

• Kirchhoff’s Rule:

i) Junction Rule:-The algebraic sum of currents meeting at a point is zero. 0I =

ii) Loop rule:-The algebraic sum of potential difference around a closed loop is zero V o=

• Wheatstone bridge is an arrangement of four resistors arranged in four arms of the bridge and is used to determine the unknown resistance in terms of other three resistances. For balanced

Wheatstone Bridge,P R

Q S=

• Slide Wire Bridge or Metre Bridge is based on Wheatstone bridge and is used to measure unknown resistance. If unknown resistance S is in the right gap, R

l

ls

−=

100

• Potentiometer is considered as an ideal voltmeter of infinite resistance.

• Principle of potentiometer: The potential drop across any portion of the uniform wire is proportional to the length of that portion of the wire provided steady current is maintained in it i.e. v α l

• Potentiometer is used to (i) compare the e.m.f.s of two cells (ii) determine the internal resistance of a cell and (iii) measure small potential differences.

• Expression for comparison of e.m.f of two cells by using potentiometer,2

1

2

1

l

l=

where 21, ll are the

balancing lengths of potentiometer wire for e.m.fs 1 and 2 of two cells.

• Expression for the determination of internal resistance of a cell I is given by Rl

ll

−

2

21

Where 1l is the balancing length of potentiometer wire corresponding to e.m.f of the cell, 2l that of

terminal potential difference of the cell when a resistance R is connected in series with the cell whose internal resistance is to be determined

• Expression for determination of potential differenceL

rl

rRV

+=

. where L is the length of the

potentiometer wire, l is balancing length, r is the resistance of potentiometer wire, R is the resistance included in the primary circuit.

• Joule’s law of heating states that the amount of heat produced in a conductor is proportional to (i) square of the current flowing through the conductor,(ii) resistance of the conductor and (iii) time for which the current is passed. Heat produced is given by the relation H=I2Rt

• Electric power: It is defined `as the rate at which work is done in maintaining the current in electric circuit. P =VI = I2R =V2/R. Power P is the product of V & I

• Electrical energy: The electrical energy consumed in a circuit is defined as the total work done in maintaining the current in an electrical circuit for a given time. Electrical energy = VIt = I2Rt =(V2/R)t = Pt

• Commercial unit of energy 1KWh= 3.6×106J

CONCEPT MAP

Flow of Charges

MULTIPLE CHOICE QUESTIONS:

1 Consider a current carrying wire current I in the shape of a circle. Note that as the current

progresses along the wire, the direction of j (current density) changes in an exact manner,

while the current/remain unaffected. The agent that is essentially responsible for is-

(a) source of emf.

(b) electric field produced by charges accumulated on the surface of wire.

(c) the charges just behind a given segment of wire which push them just the right way by

repulsion.

(d) the charges ahead.

2 Which of the following is wrong? Resistivity of a conductor is

(a) independent of temperature.

(b) inversely proportional to temperature.

(c) independent of dimensions of conductor.

(d) less than resistivity of a semiconductor

3 Drift velocity vd varies with the intensity of electric field as per the relation

(a) vd ∝ E

(b) vd ∝ 1/E

(c) vd = constant

(d) vd ∝ E²

4 For a cell, the graph between the potential difference (V) across the terminals of the cell and

the current (I) drawn from the cell is shown in the figure.

The e.m.f. and the internal resistance of the cell are

(a) 2V, 0.5 Ω

(b) 2V, 0.4 Ω

(c) Greater than 2V, 0.5 Ω

(d)less than 2V, 0.4 Ω

5 When there is an electric current through a conducting wire along its length, then an electric

field must exist

(a) outside the wire but normal to it.

(b) outside the wire but parallel to it.

(c) inside the wire but parallel to it.

(d) inside the wire but normal to it.

6 From the graph between current I and voltage V shown below, identify the portion

corresponding to negative resistance

(a) AB (b) BC (c) CD (d) DE

7 Which of the following I-V graph represents ohmic conductors?

8 The I-V characteristics shown in figure represents

(a) ohmic conductors (b) non-ohmic conductors (c) insulators (d) superconductors

9 . Which of the following is correct for V-I graph of a good conductor?

10 A cell having an emf E and internal resistance r is connected across a variable external

resistance R. As the resistance R is increased, the plot of potential difference V across R is

given by

11 In parallel combination of n cells, we obtain

(a) more voltage (b) more current (c) less voltage (d) less current

12 For measurement of potential difference, a potentiometer is preferred over voltmeter

because

(a) potentiometer is more sensitive than voltmeter.

(b) the resistance of potentiometer is less than voltmeter.

(c) potentiometer is cheaper than voltmeter.

(d) potentiometer does not take current from the circuit.

13 Resistivity of conductor depend on

(a)Temperature of conductor

(b)Length of conductor

(c)Area of conductor

(d)Current in conductor

14 We use alloy for making of resistors, because they have:

Temp. coefficient Resistivity

(a) Low Low

(b) High High

(c) High Low

(d) Low High

15 Resistances R1 and R2 are connected respectively in the left gap and right gap of a metre

bridge. If the balance point is located at 55 cm, the ratio R2/R1 is

(a) 4/5 (b) 5/4 (c) 9/11 (d) 11/9

16 In a potentiometer arrangement; a cell of emf 1.25 V gives a balance point at 35.0 cm length

of the wire. If the cell is replaced by another cell and the balance point shifts to 63.0 cm,

what is the emf of the second cell?

(a) 1.00 V (b)1.25 V (c)2.00 V (d)2.25 V

17 Massesofthreewiresofcopperareintheratioof1:3:5andtheirlengthsareintheratioof 5 : 3 : 1.

The ratio of their electrical resistances are

(a) 1 : 3 : 5 (b) 5 : 3 : 1 (c)1 : 15 : 125 (d)125 : 15 : 1

18 S I unit of resistivity is

(a) Ohm (b) mho (c) ohm-m (d) mho-m-1

19 Resistances 3 Ω and 6 Ω are connected across the left gap and right gap respectively of a

metre bridge. When a 6 Ω resistance is connected in series with the 3 Ω resistance in the

left gap, the shift in the balance point is

(a) 25 cm (b) 22 cm (c) 26.67 cm (d) 29.67 cm

20 Two cells when connected in series are balanced on 8m on a potentiometer. If the cells

are connected with polarities of one of the cell is reversed, they balance on 2m. The

ratio of e.m.f.'s of the two cellsis

(a)3:5 (b)5 :3 (c)3: 4 (d)4 :3

21 A wire of 10 Ω resistance is stretched to thrice its original length. Its new

resistance will be

(a) 90 Ω (b) 80 Ω (c) 70 Ω (d) 60 Ω

22 A cell of internal resistance 3 ohm and emf 10 volt is connected to a uniform wire of length

500 cm and resistance 3 ohm. The potential gradient in the wire is

(a) 30mV/cm (b) 10 mV/cm (c) 20mV/cm (d)4 mV/cm

23 Which of the following characteristics of electrons determines the current in a

conductor?

(a) Drift velocity alone

(b)Thermal velocity alone.

(c)Both drift velocity and drift velocity

(d)Neither drift nor thermal velocity

24 Which of the following statements are true with regard to resistance?

(a)Resistance is directly proportional to a length of the wire

(b)Resistance is directly proportional to an area of cross section of the wire

(c)Resistance is inversely proportional to the length of the wire

(d)Resistance is inversely proportional to the resistivity of the wire

25 What is the value of current if a 50C charge flows in a conductor over a period of 5

seconds?

(a)5A (b)10A (c)15A (d) 20A

26 For a cell of e.m.f. 2 V, a balance is obtained for 50 cm of the potentiometer wire. If the

cell is shunted by a 2 Ω resistor and the balance is obtained across 40 cm of the wire,

then the internal resistance of the cell is

(a) 1 Ω (b) 0.5 Ω

(c) 1.2 Ω (d) 2.5 Ω

27 With rise in temperature, relaxation time in metals

(a)Decreases (b)Increases

(c)Remains same (d)First increase then decrease

28 Two bulbs of 500 W and 200 W rated at 250 V will have resistance ratio as

(a) 4:25 (b) 25:4 (c) 2:5 (d) 5:2

29 In the circuit shown in figure E1 = 7V, E2 = 7V R1 = R2 = 1 Ω and R3 = 3 Ω respectively. The

current through the resistance R3 is

(a) 2 A

(b) 3.5 A

(c) 1.75 A

(d) 2.5 A

30 In a Whetstone’s bridge, all the four arms have equal resistance R. If resistance of the

galvanometer arm is also R, then equivalent resistance of the combination is

(a)4 R (b) 3 R (c) 2 R (d) 1 R

31 The current in the given circuit will be ________ .

(a) 1A (b) 0.1A (c) 0.2 A (d) 10A

32 A resistance R is to be measured using a meter bridge, a student chooses the standard

resistance S to be 100 Ω He finds the null point at l1 = 2.9 cm. He is told to attempt to

improve the accuracy. Which of the following is a useful way?

(a) He should measure l1, more accurately

(b) He should change S to 1000 Ω and repeat the experiment

(c) He should change S to 3 Ω and repeat the experiment

(d) He should have given up hope of a more accurate measurement with a meter bridge

33 Two cells of emfs approximately 5 V and 10 V are to be accurately compared using a

potentiometer of length 400 cm.

(a) The battery that runs the potentiometer should have voltage of 8 V.

(b) The battery of potentiometer can have a voltage of 15 V and R adjusted so that the

potential drop across the wire slightly exceeds 10 V.

(c) The first portion of 50 cm of wire itself should have a potential drop of 10 V.

(d) Potentiometer is usually used for comparing resistances and not voltages.

34 A metal rod of length 10 cm and a rectangular cross-section of 1 cm x 1/2 cm is connected to

a battery across opposite faces. The resistance will be

(a) maximum when the battery is connected across 1 cm x 1/2 cm faces

(b) maximum when the battery is connected across 10 cm x 1 cm faces

(c) maximum when the battery is connected across 10 cm x 1/2 cm faces

(d) same irrespective of the three faces

35 Kirchhoff’s junction rule is a reflection of

(a) conservation of current density vector.

(b) conservation of charge.

(c) the fact that the momentum with which a charged particle approaches a junction is

unchanged (as a vector) as the charged particle leaves the junction.

(d) the fact that there is no accumulation of charges at a junction.

36 Temperature dependence of resistivity ρ(T) of semiconductors, insulators and metals is

significantly based on the following factors:

(a) number of charge carriers can change with temperature T.

(b) time interval between two successive collisions can depend on T.

(c) length of material can be a function of T.

(d) mass of carriers is a function of T.

37 In a meter bridge, the point D is a neutral point (figure).

(a) The meter bridge can have no other neutral. A point for this set of resistances.

(b) When the jockey contacts a point on meter wire left of D, current flows to B from the

wire

(c) When the jockey contacts a point on the meter wire to the right of D, current flows from

B to the wire through galvanometer

(d) When R is increased, the neutral point shifts to left

38 For measurement of potential difference, a potentiometer is preferred over voltmeter

because

(a) potentiometer is more sensitive than voltmeter.

(b) the resistance of potentiometer is less than voltmeter.

(c) potentiometer is cheaper than voltmeter.

(d) potentiometer does not take current from the circuit.

39 When there is an electric current through a conducting wire along its length, then an electric

field must exist

(a) outside the wire but normal to it.

(b) outside the wire but parallel to it.

(c) inside the wire but parallel to it.

(d) inside the wire but normal to it.

40 An electric heater is connected to the voltage supply. After few seconds, current gets its

steady value then its initial current will be

(a) equal to its steady current

(b) slightly higher than its steady current

(c) slightly less than its steady current

(d) zero

41 In a Wheatstone bridge if the battery and galvanometer are interchanged then the deflection

in galvanometer will

(a) change in previous direction

(b) not change

(c) change in opposite direction

(d) none of these.

42 In a potentiometer of 10 wires, the balance point is obtained on the 7th wire. To shift the

balance point to 9th wire, we should

(a) decrease resistance in the main circuit.

(b) increase resistance in the main circuit.

(c) decrease resistance in series with the cell whose emf is to be measured.

(d) increase resistance in series with the cell whose emf is to be determined.

43 AB is a wire of potentiometer with the increase in the value of resistance R, the shift in the

balance point J will be

(a) towards B

(b) towards A

(c) remains constant

(d) first towards B then back towards A

44 Consider a simple circuit shown in figure stands for a variable resistance R’. R’ can vary from

R0 to infinity, r is internal resistance of the battery (r << R << R0). [NCERT Exemplar]

(a) Potential drop across AB is not constant as R0 is varied.

(b Current through R0 is nearly a constant as R0 is varied.

(c) Current I depends sensitively on R0.

(d) I≥Vr+R always.

45 n the circuit shown, potential difference between X and Y is ________ and across 40 Ω is

________ .

(a) 120V, 0V

(b) 60V, 2V

(c) 30V, 4V

(d) None of these

ANSWERS(Multiple Choice Questions):

Q.No. Answer Q.No. Answer

1 b 26 b

2 a 27 a

3 a 28 c

4 b 29 a

5 c 30 d

6 c 31 c

7 a 32 c

8 b 33 b

9 b 34 a

10 b 35 b

11 b 36 a,b

12 d 37 a,c

13 a 38 d

14 a 39 c

15 d 40 b

16 d 41 b

17 d 42 d

18 d 43 a

19 c 44 d

20 b 45 a

21 a

22 b

23 c

24 a

25 b

ASSERTION AND REASON BASED QUESTIONS: Two statements are given-one labeled Assertion (A) and the other labeled Reason (R). Select the correct answer to these questions from the codes (a), (b), (c) and (d) as given below. a) Both A and R are true and R is the correct explanation of A b) Both A and R are true but R is NOT the correct explanation of A c) A is true but R is false d) A is false and R is also false

1. Assertion (A): The dimensional formula for product of resistance and conductance is same as

for dielectric constant.

Reason(R): Both have dimension for time constant.

2. Assertion (A): A wire carrying an electric current has no electric field around it.

Reason(R): rate of flow of electron is one direction is equal to the rate of flow of protons in opposite direction.

3. Assertion(A): The binding of an insulated wire increase the resistance of wire

Reason(R): Drift velocity of electron in bent wire decrease.

4. Assertion (A): The EMF of the driver cell in the potentiometer experiment should be greater

than the EMF of the cell to be determined.

Reason(R): The fall of potential across the potentiometer wire should not be less than the EMF of the cell to be determined.

5. Assertion(A).The value of temperature coefficient of resistance of positive for metals

Reason(R): The temperature coefficient of resistance for insulator is always positive

6. Assertion (A).In a meter bridge experiment null point for an unknown resistance is put

inside an enclosure maintained at a higher temperature. The null point can be obtained at

the same point as before by discharging the value of standard resistance.

Reason(R): Resistance of metal increase with increase in temperature.

7. Assertion (A).There is no current in metals in the absence of electric field.

Reason(R): Motion of free electron is randomly.

8. Assertion (A).In a simple battery circuit the point of lowest potential is positive terminal of

the battery.

Reason(R): The current flows towards the point of the higher potential as it flows in such a circuit from the negative to positive terminal.

9. Assertion (A). If a wire is stretched to increase its length X times then its resistance also

increases.

Reason(R): Resistance of conductor is directly depend upon the length of conductor.

10. Assertion (A). A potentiometer of longer length is used for accurate measurement.

Reason(R): The potential gradient for a potentiometer of longer length with a given source of EMF becomes small.

11. Assertion (A). When the Cell is in the open circuit there is no force on a test charge inside

the electrolyte of the cell.

Reason(R): There is no field inside the cell when the cell is in the open circuit. 12. Assertion (A). if the potential difference between two point is zero and resistance between

those point is zero current may flow between the points

Reason(R): Kirchhoff 1st law is based on conservation of charge

13. Assertion (A). The constant potential difference is applied across a conductor if temperature

of conductor is increase, drift speed of electron will decrease

Reason(R): resistivity increase with increase in temperature 14. Assertion (A). A potentiometer is preferred over a volt meter of the measurement of emf of

a cell

Reason(R): A potentiometer is preferred as it does not draw any current from the cell

ANSWERS(Assertion and Reason): 1 C 5 C 9 D 13 B

2 C 6 D 10 A 14 A

3 D 7 A 11 C

4 A 8 D 12 B

Explanation of Answers(A & R): 1. c) Resistance x conductance = R * 1/R = [M0L0T0] 2. c)There is no net charge on the wire 3. d) both are false 4. a) 5. c) α is positive for metal and negative for insulator 6. d) same null point can be obtained by increase in standard resistance 7.a) 8.d) current flow from higher potential to lower potential 9.d) resistance of a wire becomes x2 times 10.a) 11.c) 12.b) 13. b) 14. a)

CASE STUDY BASED QUESTIONS:

Read the source given below and answer any four out of the following questions:

1. Potentiometer: The potentiometer consists of a long resistive wire (L) and a battery of known EMF,

'V' whose voltage is known as driver cell voltage. Assume a primary circuit arrangement by connecting

the two ends of L to the battery terminals. One end of the primary circuit is connected to the cell

whose EMF 'E' is to be measured and the other end is connected to galvanometer G. This circuit is

assumed to be a secondary circuit.

(i) How can we increase the sensitivity of a potentiometer?

(a) Increasing the potential gradient

(b) Decreasing the potential gradient

(c) Decreasing the length of potentiometer wire

(d) Increasing resistance put in parallel

(ii) If l1 and l2 are the balancing lengths of the potentiometer wire for the cells of EMFs ε1 and ε2 , then

(a)ε1+ ε2= l1 + l2

(b) ε1/ ε2= l1/l2

(c) ε1* ε2 =l1*l2

(d) None of these

iii) Example of a potentiometer is

(a) Mobile

(b)Modem

(c) Joystick

(d) All of these

(iv) The emf of a cell is always greater than its terminal voltage. Why?

(a) Because there is some potential drop across the cell due to its small internal resistance

(b) Because there is some potential drop across the cell due to its large internal resistance

(c) Because there is some potential drop across the cell due to its low current

(d) Because there is some potential drop across the cell due to its high current

(v) Why is a ten wire potentiometer more sensitive than a four-wire one?

(a) Small potential gradient

(b) Large potential gradient

(c) Large length

(d) None of these

2. Factors Affecting Resistance: According to Ohm's law, the current flowing through a conductor is

directly proportional to the potential difference across the ends of the conductor i .e., I α V ,so V/I=

R, where R is resistance of the conductor. Electrical resistance of a conductor is the obstruction posed

by the conductor to the flow of electric current through it. It depends upon length, area of cross-

section, nature of material and temperature of the conductor. We can write, R or R = p =, where p is

electrical resistivity of the material of the conductor.

(i) Dimensions of electric resistance is

(a) [ML2T-2A-2] (b) [ML2T-3A-2] (c) [M-1L-2T-1A] (d) [M-1L2T2A-1]

(ii) If 1 µA current flows through a conductor when potential difference of 2 volt is applied across its

ends, then the resistance of the conductor is

(a) 2x 106 Ω (b) 3 x 105 Ω (c) 1.5 x 105 Ω (d) 5 x 107 Ω

(iii) Specific resistance of a wire depends upon

(a) Length (b) cross-sectional area (c) mass (d) none of these

(iv) The slope of the graph between potential difference and current through a conductor is

(a) A straight line (b) curve

(c) First curve then straight line (d) first straight line then curve

(v) The resistivity of the material of a wire 1.0 m long, 0.4 mm in diameter and having a resistance of

2.0 ohm is

(a) 1.57 x 10-6 Ω m (b) 5.25 x 10-7 Ω m (c) 7.12 x 10-5 Ω m (d) 2.55 x 10-7 Ω m

3.Kirchhoff's Rules:

In 1942, a German physicist Kirchhoff extended Ohm's law to complicated circuits and gave two laws,

which enable us to determine current in any part of such a circuit. According to

Kirchhoff's first rule, the algebraic sum of the currents meeting at a junction x

in a closed electric circuit is zero. The current flowing in a conductor towards

the junction is taken as positive and the current flowing away from the junction is taken as negative.

According to Kirchhoff's second rule, in a closed loop, the algebraic sum of the emf's and algebraic

sum of the products of current and resistance in the various arms of the loop is zero. While traversing

a loop, if negative pole of the cell is encountered first, then its emf is negative, otherwise positive.

(i) Kirchhoff's Ist law follows

(a) Law of conservation of energy (b) Law of conservation of charge

(c)Law of conservation of momentum (d) Newton's third law of motion

(ii) The value of current I in the given circuit is

(a) 4.5A

(b) 3.7 A

(c) 2.0 A

(d) 2.5A

(iii) Kirchhoff's loop law is based on

(a) Law of conservation of momentum of electron

(b)Law of conservation of charge and energy

(c) Law of conservation of energy

(d) None of these.

(iv) Point out the right statements about the validity of Kirchhoff's Junction rule.

(a) The current flowing towards the junction is taken as positive.

(b) The currents flowing away from the junction are taken as negative.

(c) Bending or reorienting the wire does not change the validity of Kirchhoff's Junction rule.

(d) All of the above

(v) Potential difference between A and B in the circuit shown here is B

(a) 4V

(b) 5.6 V

(c) 2.8 V

(d) 6 V

4. Wheatstone Bridge and its Applications:

WHEATSTONE bridge is an arrangement of four resistances P, Q, R and S connected as shown in the

figure.

Their values are so adjusted that the galvanometer G shows no deflection. The bridge is then said to

be balanced when this condition is achieved happens. In the setup shown here, the points B and D

are at the same potential and it can be shown that P/Q=R/S. This is called the balancing condition. If

any three resistances are known, the fourth can be found .The practical form of Wheatstone bridge is

slide Wire Bridge or Meter Bridge. Using this the unknown resistance S= (100-I)/l*R, where l is the

balancing length of the Meter bridge.

(i) In a Wheatstone bridge circuit, P = 5Ω, Q = 6Ω, R = 10Ω and S= 5Ω. What is the value of additional

resistance to be used in series with S, so that the bridge is balanced?

(a) 9Ω (b) 7Ω (c) 10Ω (d) 5 Ω

(ii) A Wheatstone bridge consisting of four arms of resistances P, Q, R, S is most sensitive when

(a) All the resistances are equal

(b) All the resistances are unequal

(c) The resistances P and Q are equal but R >> P and S >>Q

(d) The resistances P and Q are equal but R<< P and S << Q

(iii) When a metal conductor connected to left gap of a meter bridge is heated, the balancing point

(a) Shifts towards right (b) shifts towards left

(c) Remains unchanged (d) remains at zero.

(iv) The percentage error in measuring resistance with a meter bridge can be minimized by adjusting

the Balancing Point close to

(a) 0 (b) 20 cm (c) 50 cm (d) 80 cm

(v) In a meter bridge experiment, the ratio of left gap resistance to right gap resistance is 2: 3. The

balance point from left i

(a) 20 cm (b) 50 cm (c) 40 cm (d) 60 cm

5. Potentiometer: An Ideal Voltmeter:

Potentiometer is an apparatus used for measuring the emf of a cell or potential difference between

two points in an electrical circuit accurately. It is also used to determine the internal resistance of a

primary cell. The potentiometer is L based on the principle that, if V is the potential difference across

any portion of the wire of length ‘l’ and resistance R, then V directly prop. to l or V = kl where k is the

potential gradient. Thus, potential difference across any portion of potentiometer wire is directly

proportional to length of the wire of that portion. The potentiometer wire must be uniform. The

resistance of potentiometer wire should be high.

(i) Which one of the following is true about potentiometer?

(a) Its sensitivity is low.

(b) It measures the emf of a cell very accurately.

(c) It is based on deflection method.

(d) It takes current from the circuit.

(ii) A current of 1.0 mA is flowing through a potentiometer wire of length 4 cm and of resistance 4Ω

The potential gradient of the potentiometer wire is

(a) 10-3 Vm -1 (b) 10-5 Vm-2 (c) 2x 10-3 Vm-1 (d) 4x10-3 Vm-1

(iii) Sensitivity of a potentiometer can be increased by

(a) Decreasing potential gradient along the wire

(b)Increasing potential gradient along the wire

(c) Decreasing current through the wire

(d)Iincreasing current through the wire

(iv) A potentiometer is an accurate and versatile device to make electrical measurements of EMF

because the method involves

(a) potential gradients

(b) a condition of no current flow through the galvanometer

(c) a combination of cells, galvanometer and resistances

(d) cells

(v) In a potentiometer experiment, the balancing length is 8 m, when the two cells E, and E, are joined

in series.When the two cells are connected in opposition the balancing length is 4 m. The ratio of the

e. m. f. of two cells (E/ E) is

(a) 1: 2 (b) 2: 1 (c) 1: 3 (d) 3: 1

ANSWERS(Case Study Questions):

Q.1 Q.2 Q.3 Q.4 Q.5

ANS.i b b a b b

ANS.ii d a c a a

ANS.iii c d c a a

ANS.iv a a d c b

ANS v a d b c d

CHAPTER 4

MOVING CHARGES & MAGNETISM

GIST OF LESSON:

1. Biot-Savart Law The magnetic field B due to a current element dl carrying a steady current I at a point, at distance r from the current element is

𝑑𝐵 =𝜇0𝐼𝑑𝑙𝑆𝑖𝑛𝜃

4𝜋𝑟2 μ0=4π x 10

-7 Tm/A

*Direction of dB can be found by using Maxwell’s Right hand thumb rule.]

2. Applications : (i) Magnetic field at a centre of a current carrying circular coil B= μ0I/2a (ii) Magnetic field at a point on the axis of current carrying coil.

𝐵 =𝜇0 𝑁𝐼𝑎2

2(𝑎2+𝑥2)32

N=no. of turns in the coil

3. Ampere’s circuital law:The line integral of magnetic field B around any closed path in vacuum is equal to 𝜇0 time the total current enclosed by the path ∮ B.dl= μ0I

4. Applications i) Magnetic field due to straight infinitely long current carrying straight conductor. B= μ0I/2πr ii) Magnetic field due to a straight solenoid carrying current B= μ0nI

n= no. of turns per unit length iii) Magnetic field due to toroidal solenoid carrying current. B= μ0NI/2πr

N= Total no. of turns.

5. Force on a moving charge (i) In magnetic field F=q(V x B) (ii) In magnetic and electric field(Lorentz force) F=qxE+(ν x B) (ii) Cyclotron frequency or magnetic resonance frequency ν=qB/2πm, T=2πm/Bq (iii) Maximum velocity of charged particle. Vm=Bqr/m

6. Force on a current carrying conductor in uniform F= (Il x B) l=length of conductor Direction of force can be found out using Fleming’s left hand rule.

7. Force per unit length between parallel infinitely long current carrying straight conductors.

𝐹/𝑙 =𝜇0𝐼1𝐼2

2𝜋𝑑

(a) If currents are in same direction the wires will attract each other. (b) If currents are in opposite directions they will repel each other.

8. Torque experienced by a current loop in a uniform B. τ = NIBA Sinθ τ=m X B Where m=NIA

9. Motion of a charge in a magnetic field (a) Perpendicular magnetic field F=q(vxB),F=qvBSin90=qvB (circular path) (b) Parallel or antiparallel field F=qvBSin0 (or) qvBSin180=0(Straight-line path)

10. Moving coil galvanometer It is a sensitive instrument used for detecting small electric Currents. Principle: When a current carrying coil is placed in a magnetic field, it experiences a

torque. I α θ and I = K θ where, K= NAB / C Current sensitivity, I s= θ / I=NBA/K voltage sensitivity, Vs= θ /V=NBA/KR

Changing N -> Current sensitivity changes but Voltage Sensitivity does not change 11. (a) Conversion of galvanometer into ammeter

A small resistance S is connected in parallel to the galvanometer coil S=IgG/I-Ig RA=GS/G+S

(b) Conversion of galvanometer into a voltmeter. A high resistance R is connected in series with the galvanometer coil. R=V/Ig-G Rv=G+R

using

CHAPTER 4

MOVING CHARGES & MAGNETISM

MULTIPLE CHOICE QUESTIONS:

1. A wire of length L carries a current of I ampere on bending it into a circle, the value of its magnetic

moment will be:-

(a) IL2 (b) IL2/2 (c) IL2/ (d) IL2/4

2. Unit of magnetic flux density is :-

(a) Tesla (b) Weber/metre2

(c) Newton/ampere-metre (d) All of the above

3. A charged particle is moving with velocity v under the magnetic field B. The force acting on the

particle will be maximum if:-

(a) v and B are in same direction

(b) v and B are in opposite direction

(c) v and B are perpendicular

(d) None

4. The radius of circular loop is r and a current i is flowing in it. The equivalent magnetic moment will

be:-

(a) ir (b) 2πr

(c) iπr2 (d) i/r2

5. A magnetic field of 5.0 × 10–4 T just perpendicular to the electric field of 15 kV/m in their effect an

electron beam passes undeflected and perpendicular to both of them. The speed of the electrons

is:-

(a) 75 m/s (b) 3 × 104 m/s

(c) 7.5 × 104 m/s (d) 3 × 107 m/s

6. There are two straight long wires, insulated from each other, along x and y axis carrying equal

currents as shown in figure. AB and CD are lines in xy plane and at 45° with the axes. The magnetic

field of the system is zero at points on the line :-

(a) AB (b) OB but not on OA

(c) CD (d) OC but not on OD

7. An electron, moving in a circular orbit of radius 'R' with a period 'T'. The equivalent magnetic dipole

moment of circular orbit is:-

(a) 2eR/T (b) eR/T

(c) eR2T (d) R2e/T

8. A coil of one loop is made by a wire of length L and there after a coil of two loops is made by same

wire. The ratio of magnetic field at the centre of coils respectively:-

(a) 1 : 4 (b) 1 : 1

(c) 1 : 8 (d) 4 : 1

9. Two long parallel wires are at a distance of 1m. If both of them carry 1A of current in same direction.

The magnetic force of attraction on unit length of each wire will be:-

(a) 2 × 10–7 N/m (b) 4 × 10–7 N/m

(c) 8 × 10–7 N/m (d) 10–7 N/m

10. Two identically charged particles A and B initially at rest, are accelerated by a common potential

difference V. They enters into a uniform transverse magnetic field B and describe a circular path of

radii r1 and r2 respectively then their mass ratio is:-

(a) (r1/r2)2 (b) (r2/r1)2 (c) (r1/r2) (d) (r2/r1)

11. A long solenoid carrying a current produces a magnetic field B along its axis. If the current is doubled

and the number of turns per cm is halved, the new value of the magnetic field is:-

(a) B/2 (b) B (c) 2B (d) 4B

12. A charged particle moves through a magnetic field in a direction perpendicular to it. Then the:-

(a) Speed of the particle remains unchanged

(b) Direction of motion of particle remains unchanged

(c) Acceleration of particle remains unchanged

(d) Velocity of particle remains unchanged.

13. A closely wound flat circular coil of 25 turns of wire has diameter of 10 cm and carries a current of

4 ampere. Determine the magnetic flux density at the centre of the coil:-

(a) 1.679 × 10–5 T (b) 2.028 × 10–4 T

(c) 1.257 × 10–3 T (d) 1.512 × 10–6 T

14. A proton and an α-particle moving with the same velocity and enter into a uniform magnetic field

which is acting normal to the plane of their motion. The ratio of the radii of the circular paths

described by the proton and a-particle respectively:-

(a) 1 : 2 (b) 1 : 4 (c) 1 : 16 (d) 4 : 1

15. Two parallel beams of positrons moving in the same direction will:-

(a) Repels to each other

(b) Will not interact with each other

(c) Attracts to each other

(d) Be deflected normal to the plane containing the two beams.

16. An electron is moving around a proton in an orbit of radius 1Å and produces 16 Wb/m2 of magnetic

field at the centre, then find the angular velocity of electron :-

(a) 20 X 1016 rad/sec

(b) 10 X 1016 rad/sec

(c) 5/2 x 1016 rad/sec

(d) 5/4 X 1016 rad/sec

17. Current I is flowing in a conducting circular loop of radius R. It is kept in a magnetic field B which is

perpendicular to the plane of circular loop, the magnetic force acting on the loop is :-

(a) IRB (b) 2IRB

(c) Zero (d) IRB

18. In the shown figure magnetic field at point A will be :-

(a) (b) (c) (d) Zero

19. If the permeability of iron piece is 3 × 10–3 Wb/A-m and intensity of magnetizing field of iron piece

is 120 A/m, then what is the magnetic induction of iron piece:-

(a) 0.36 Wb/m2 (b) 5 × 10–3 Wb/m2

(c) 40 Wb/m2 (d) 2.5 × 10–4 Wb/m2

20. A long solenoid with 20 turns per cm is made to produce a magnetic field of 20 mT inside the

solenoid. The necessary current will nearly be:-

(a) 8.0 ampere (b) 4.0 ampere

(c) 2.0 ampere (d) 1.0 ampere

0

4

I 0

4

I

R

0

4

I

R

21. A current is flowing in a circular coil of radius R and the magnetic field at its centre is Bo. It what

distance from the centre on the axis of the coil, the magnetic field will be Bo/8:-

(a) 3 R (b) 3 R (c) 2R (d) 8R

22. Wire in the form of a right angle ABC, with AB=3cm and BC = 4cm, carries a current of 10A. There is

a uniform magnetic field of 5T perpendicular to the plane of the wire. The force on the wire will be:-

(a) 1.5N (b) 2.0N (c) 2.5N (d) 3.5N

23. A rectangular coil 20 cm × 20cm, has 100 turns and carries a current of 1A. It is placed in a uniform

magnetic field 0.5T with the direction of magnetic field parallel to the plane of the coil. The

magnitude of the torque required to hold this coil in this position is:-

(a) zero (b) 200N-m

(c) 2N-m (d) 10N-m

24. A straight wire carries a current vertically upwards. A point P lies to the east of it at a small distance

and another point Q lies to the west at the same distance. The magnetic field at P is (consider earth

field):-

(a) Greater than at Q

(b) Same as at Q

(c) Less than at Q

(d) Greater or less than at Q, depending upon the strength of current

25. A circular coil of radius 20 cm and 20 turns of wire is mounted vertically with its plane in magnetic

meridian. A small magnetic needle is placed at the centre of the coil and it is deflected through 45°,

when a current is passed through the coil. When horizontal component of earth's field is

0.34 × 10–4 T, the current in coil is:-

(a) 0.6 A (b) 6A (c) 6 × 10–3 A (d) 0.06 A

26. A current loop of area 0.01m2 and carrying a current of 10A is held perpendicular to a magnetic

field of 0.1T, the torque in N–m acting on the loop is :-

(a) 0 (b) 0.001 (c) 0.01 (d) 1.1

27. In a region constant uniform electric and magnetic field is present. Both field are parallel. In this

region a charge released from rest, then path of particle is:-

(a) Circle (b) Helical

(c) Straight line (d) Ellipse

28. A current carrying coil is placed in a constant uniform magnetic field B. Torque is maximum on this

coil when plane of coil is:-

(a) Perpendicular to B (b) parallel to B

(c) at 45° to B (d) at 60° to B

29. 2A current is flowing in a circular loop of radius 1m. Magnitude of magnetic field at the centre of

circular loop will be:-

(a) 0/2 (b) 20 (c) 0 (d) 0/4

30. A particle of mass m, charge Q and kinetic energy T enters a transverse uniform magnetic field of

induction. After 3 seconds the kinetic energy of the particle will be:-

(a) T (b) 4 T (c) 3T (d) 2T

31. A long straight wire of radius 'a' carries a steady current i. The current is uniformly distributed across

its cross section. The ratio of the magnetic field at a/2 and 2a distances from the axis of the wire is:

(a) 4 (b) 1 (c) 1/2 (d) 1/4

32. A horizontal overhead powerline is at a height of 4m from the ground and carries a current of 100

A from east to west. The magnetic field directly below it on the ground is:-

(a) 2.5 × 10-7 T, southward (b) 5 × 10-6 T, northward

(c) 5 ×10-6 T, southward (d) 2.5 ×10-7 T, northward

33. A current I flows along the length of an infinitely long, straight, thin walled pipe, then:-

(a) the magnetic field is different at different points inside the pipe

(b) the magnetic field at any point inside the pipe is zero

(c) the magnetic field at all points inside the pipe is the same, but not zero

(d) the magnetic field is zero only on the axis of the pipe

34. A and B are two concentric circular loops carrying current i1 and i2 as shown in

figure. If ratio of their radii is 1:2 and ratio of the flux densities at the centre O due

to A and B is 1:3 then the ratio of current will be :-

(a) 1/2 (b) 1/3

(c) 1/4 (d) 1/6

35. A magnetic field

(a) Always exerts a force on a charged particle

(b) Never exerts a force on a charged particle

(c) Exerts a force, if the charged particle is moving across the magnetic field lines

(d) Exerts a force, if the charged particle is moving along the magnetic field lines

36. A current of 10 A is flowing in a wire of length 1.5 m. A force of 15 N acts on it when it is placed

in a uniform magnetic field of 2 T. The angle between the magnetic field and the direction of the

current is

(a) 30°

(b) 45°

(c) 60°

(d) 90°

37. Biot-Savart law indicates that the moving electrons with velocity (v) produce a magnetic field B

such that

(a) B parallel to v

(b) B is perpendicular to v

(c) it obeys inverse cube law

(d) it is along the line joining electron and point of observation

38. If we double the radius of a coil keeping the current through it unchanged, then the magnetic

field at any point at a large distance from the centre becomes approximately

(a) double

(b) three times

(c) four times

(d) one-fourth

39. A current i ampere flows in a circular arc of wire whose radius is R, which subtend an angle 3/2

radian at its centre. The magnetic field B at the centre is

(a) 𝜇0 𝑖

𝑅

(b) 𝜇0 𝑖

2𝑅

(c) 2𝜇0 𝑖

𝑅

(d) 3𝜇0 𝑖

8𝑅

40. A positively charged particle moving due east enters a region of uniform magnetic field directed

vertically upward. The particle will

(a) get deflected in vertically upward direction

(b) move in circular path with an increased speed

(c) move in circular path with a decreased speed

(d) move in a circular path with uniform speed.

41. The radius of the circular path of an electron moving in magnetic field perpendicular to its path

is equal to

(a) 𝐵𝑒

𝑚𝑣

(b) 𝑚𝑒

𝐵

(c) 𝑚𝑣

𝑒𝐵

(d) 𝐵

𝑚𝑣

42. A uniform electric field and a uniform magnetic field are acting along the same direction in a

certain region. If an electron is projected along the direction of the fields with a certain velocity,

then

(a) its velocity will decrease

(b) its velocity will increase

(c) it will turn towards right of direction of motion

(d) it will turn towards left of direction of motion.

43. If a current is passed in a spring, it

(a) gets compressed

(b) gets expanded

(c) oscillates

(d) remains unchanged

44. A galvanometer acting as ammeter will have

(a) high resistance in parallel

(b) high resistance in series

(c) low resistance in parallel

(d) low resistance in series

45. If resistance of a galvanometer is 6 Ω and it can measure a maximum current of 2 A. Then

required shunt resistance to convert it into an ammeter reading up to 6 A, will be

(a) 2 Ω

(b) 3 Ω

(c) 4 Ω

(d) 5 Ω

46. A galvanometer can be changed into voltmeter by providing

(a) low resistance in series

(b) low resistance in parallel

(c) high resistance in series

(d) high resistance in parallel

47. If the number of turns in a moving coil galvanometer is increased, its current sensitivity

(a) increases

(b) remains same

(c) decreases

(d) may increase or decrease

48. The magnetic field due to a straight conductor of uniform cross-section of radius a and carrying a

steady current is represented by

49. A charged particle moves through a magnetic field perpendicular to its direction. Then

(a) kinetic energy changes but the momentum is constant

(b) the momentum changes but the kinetic energy is constant

(c) both momentum and kinetic energy of the particle are not constant

(d) both momentum and kinetic energy of the particle are constant.

ANSWERS(Multiple Choice Questions):

Q.NO. ANS Q.NO. ANS Q.NO. ANS Q.NO. ANS Q.NO. ANS

1 D 11 B 21 A 31 B 41 C

2 D 12 A 22 C 32 C 42 A

3 C 13 C 23 C 33 B 43 A

4 C 14 A 24 A 34 D 44 C

5 D 15 A 25 A 35 B 45 B

6 A 16 A 26 A 36 A 46 C

7 D 17 C 27 C 37 A 47 A

8 A 18 B 28 B 38 C 48 A

9 A 19 A 29 C 39 D 49 B

10 A 20 A 30 A 40 D ……. …….

ASSERTION AND REASONING QUESTIONS:

For these questions two statements are given-one labeled Assertion (A) and the other labeled

Reason (R). Select the correct answer to these questions from the codes (a), (b), (c) and (d) as

given below.

a. Both A and R are true and R is the correct explanation of A

b. Both A and R are true but R is NOT the correct explanation of A

c. A is true but R is false

d. A is false and R is also false.

1. Assertion: Free electron always keep on moving in a conductor even then no magnetic force act

on them in magnetic field unless a current is passed through it.

Reason: The average velocity of free electron is zero.

2. Assertion: A circular loop carrying current lies in XY plane with its center at origin

having a magnetic flux in Z-axis direction.

Reason: Magnetic flux direction is independent of the direction of current in the

conductor.

3. Assertion: The energy of charged particle moving in a uniform magnetic field does not

change.

Reason: Work done by magnetic field on the charge is zero.

4. Assertion: If an electron, while coming vertically from outer space, enters the earth's

magnetic field, it is deflected towards west.

Reason: Electron has negative charge.

5. Assertion: A direct current flows through a metallic rod, produced magnetic field

only outside the rod.

Reason: There is no flow of charge carriers inside the rod.

6. Assertion: Magnetic field interacts with a moving charge and not with a stationary

charge.

Reason: A moving charge produces a magnetic field.

7. Assertion: The magnetic field at the ends of a very long current carrying solenoid is half

of that at the center.

Reason: If the solenoid is sufficiently long, the field within it is uniform.

8. Assertion: If a charged particle is moving on a circular path in a perpendicular

magnetic field, the momentum of the particle is not changing.

Reason: Velocity of the particle in not changing in the magnetic field.

9. Assertion: If two long wires, hanging freely are connected to a battery in series, they

come closer to each other.

Reason: Force of attraction acts between the two wires carrying current.

10. Assertion: A current I flows along the length of an infinitely long straight and thin

walled pipe. Then the magnetic field at any point inside the pipe is zero.

Reason: ∮ . 𝑑𝑙 = 𝜇0𝐼.

ANSWERS(Assertion & Reason Question):

Q.No. 1 2 3 4 5 6 7 8 9 10

ANS a c a b d a b d d a

Explanation of Answers ( A and R) :

1. a. In the absence of the electric current, the free electrons in a conductor are in a state of

random motion, like molecules in a gas. Their average velocity is zero. i.e. they do not have any

net velocity in a direction. As a result, there is no net magnetic force on the free electrons in the

magnetic field. On passing the current, the free electrons

acquire drift velocity in a definite direction, hence magnetic force acts on them,

unless the field has no perpendicular component.

2. c. The direction of magnetic field due to current carrying conductor can be found by applying

right hand thumb rule or right hand palm rule. When electric current is passed through a circular

conductor, the magnetic field lines near the center of the conductor are almost straight lines.

Magnetic flux direction is determined only by the direction of current.

3. a. The force on a charged particle moving in a uniform magnetic field always acts in direction

perpendicular to the direction of motion of the charge. As work done by magnetic field on the

charge is zero, [𝑊 = 𝐹𝑆 cos 𝜃], so the energy of the charged particle does not change.

4. b. We know that the direction of the earth’s magnetic field is toward north and the velocity of

electron is vertically downward. Applying Fleming’s left hand rule, the direction of force is

towards west. Therefore, an electron coming from outer space will be deflected toward west.

5. d. In the case of metallic rod, the charge carriers flow through whole of the cross section.

Therefore, the magnetic field exists both inside as well as outside. However magnetic field inside

the rod will go on decreasing as we go towards the axis.

6. a. A moving charge experiences a force in magnetic field. It is because of interaction of

two magnetic fields, one which is produced due to the motion of charge and other in which

charge is moving.

7. b. For a solenoid . 𝐵𝑒𝑛𝑑 =1

2𝐵𝑖𝑛. Also for a long solenoid, magnetic field is uniform within it but

this reason is not explaining the assertion.

8. d. When a charged particle is moving on a circular path in a magnetic field, the magnitude of

velocity does not change but direction of velocity is changing every moment. Hence velocity is

changing, so momentum mv is also changing.

9. d. When two long parallel wires, are connected to a battery in series. They carry currents in

opposite directions, hence they repel each other.

10. a. Here, both Assertion and Reason are correct and reason is the correct explanation of

assertion.

CASE STUDY BASED QUESTIONS:

1. Motion of Charged Particle Inside Magnetic Field :A charged particle moving in a magnetic field

experiences a force that is proportional to the strength of the magnetic field, the component of the

velocity that is perpendicular to the magnetic field and the charge of the particle. This force is given by

= 𝒒( × ) , where q is the electric charge of the particle, is the instantaneous velocity ofthe

particle, and is the magnetic field (in tesla).

The direction of force is determined by the rules of cross product of two vectors.

Force is perpendicular to both velocity and magnetic field. Its direction is same as × if q is positive

and opposite of × if q is negative.

The force is always perpendicular to both the velocity of the particle and the magnetic field that created

it. Because the magnetic force is always perpendicular to the motion, the magnetic field can do no work

on an isolated charge. The direction of force on a moving charge particle can also be found using

Fleming’s Left Hand Rule as shown in figure below.

(i) When a magnetic field is applied on a stationary electron, it

(a) remains stationary

(b) spins about its own axis

(c) moves in the direction of the field

(d) moves perpendicular to the direction of the field.

(ii) A proton is projected with a uniform velocity along the axis of a current carrying solenoid,

then

(a) the proton will be accelerated along the axis

(b) the path of proton will be circular about the axis

(c) the proton moves along helical path

(d) the proton will continue to move with velocity v along the axis.

(iii) A charged particle experiences magnetic force in the presence of magnetic field. Which of the

following statement is correct?

(a) The particle is stationary and magnetic field is perpendicular.

(b) The particle is moving and magnetic field is perpendicular to the velocity.

(c) The particle is stationary and magnetic field is parallel.

(d) The particle is moving and magnetic field is parallel to velocity.

(iv) A charge q moves with a velocity 2 m/s along x-axis in a uniform magnetic field = (𝑖 + 2𝑗 +

3), then charge will experience a force

(a) in y-z plane

(b) along -y axis

(c) along +z axis

(d) along –z axis

(v) A beam of positively charged particles moving along +x axis, experience a force in +y direction

due to a magnetic field. The direction of magnetic field is

(a) In x-y plane

(b) Along +z axis

(c) Along –z axis

(d) Along +x axis

2. HELICAL MOTION OF CHARGED PARTICLES : A charged particle of mass m and charge q, entering

a region of magnetic field with an initial velocity making an angle θ with . Let this velocity

have a component 𝒗𝒑(= 𝒗 𝐜𝐨𝐬 𝛉) parallel to the magnetic field and a component𝒗𝒏 (=

𝒗 𝐬𝐢𝐧𝛉) normal (perpendicular) to it. There is no force on a charged particle in the direction of

the field. Hence the particle continues to travel with the velocity𝒗𝒑 parallel to the field. The

normal component 𝒗𝒏of the particle results in a Lorentz force𝒒(𝒗𝒏 × 𝑩 )which is perpendicular

to both 𝒗𝒏 and 𝑩 .The particle thus has a tendency to perform a circular motion in a plane

perpendicular to the magnetic field. When this is coupled with the velocity parallel to the field,

the resulting trajectory will be a helix along the magnetic field line, as shown in Figure below.

The radius of the helical path is given by 𝒓 =𝒎 𝒗𝒏

𝒒𝑩=

𝒎𝒗𝐬𝐢𝐧𝛉

𝒒𝑩 (By the fact that required centripetal

force is provided by Lorentz force).The distance moved along the magnetic field during the

period of one rotation is called pitch (𝑝 = 𝑣𝑝𝑇, where 𝑇 is time taken by particle to complete

one rotation). Since the Lorentz force is normal to the velocity of each point, the field does no

work on the particle and the magnitude of velocity remains the same.

(i) Two charged particles A and B having the same charge, mass and speed enter into a

magnetic field in such a way that the initial path of A makes an angle of 30° and that of B

makes an angle of 90° with the field. Then the trajectory of

(a) B will have smaller radius of curvature than that of A

(b) both will have the same curvature

(c) A will have smaller radius of curvature than that of B

(d) both will move along the direction of their original velocities.

(ii) An electron having momentum 3.2 x 10-23 kg m/s enters a region of uniform magnetic field

of 0.2 T. The magnetic field vector makes an angle of 30° with the initial velocity vector of

the electron. The radius of the helical path of the electron in the field will be

(a) 1 mm

(b) 2 mm

(c) 20 mm

(d) 0.5 mm

(iii) A neutron enters a region of uniform magnetic field 0.15 T with a velocity making an angle

30° with the magnetic field vector. The path of the neutron will be

(a) Circular

(b) Helical

(c) Straight line

(d) Parabolic

(iv) The magnetic field in a certain region of space is given by = 𝟐𝝅 × 𝟏𝟎−𝟐 𝑻 . A proton is

shot into the field with velocity 𝒗 = (𝟐 × 𝟏𝟎𝟓 + 𝟒 × 𝟏𝟎𝟓𝒋) 𝒎/𝒔. The proton follows a

helical path in the field. The distance moved by proton in the x-direction during the period

of one rotation in the yz-plane will be

(a) 0.312m

(b) 0.209m

(c) 0.450m

(d) 0.157m

(v) A negatively charged particle having velocity 𝒗 enters a region of uniform magnetic field at

an angle 30° with the direction of magnetic field. Then its kinetic energy

(a) Increase

(b) Decrease

(c) Remains same

(d) First increases and then decreases

3. A VELOCITY SELECTOR: A velocity selector is a region in which there is a uniform electric field and a

uniform magnetic field. The fields are perpendicular to one another, and perpendicular to the initial

velocity of the charged particles that are passing through the region. The force exerted on a charged

particle having charge q by the electric field is given by:𝐹 = 𝑞 . The magnitude of the force exerted

by the magnetic field is 𝐹 = 𝑞𝑣𝐵, as long as the velocity𝑣 is perpendicular to the magnetic field .

The idea is that, if the two forces are equal and opposite, the net force is zero, and the particle

passes through the region without changing direction. With the magnetic force being speed

dependent, however, any charges traveling faster or slower than the ones that go straight through

will be deflected one way or another out of the beam.

(i) Assuming the particle in the figure is positively charged, what are the directions of the forces due

to the electric field and to the magnetic field?

(a) The force due to the electric field is directed up (toward the top of the page); the force due to

the magnetic field is directed down (toward the bottom of the page).

(b) The force due to the electric field is directed down (toward the bottom of the page); the force

due to the magnetic field is directed up (toward the top of the page).

(c) The force due to the electric field is directed out of the plane of the paper; the force due to the

magnetic field is directed into the plane of the paper.

(d) The force due to the electric field is directed into the plane of the paper; the force due to the

magnetic field is directed out of the plane of the paper.

(ii) Suppose a particle with twice the velocity of the particle in the figure enters the velocity selector.

The path of this particle will curve

(a) upward (toward the top of the page)

(b) downward (toward the bottom of the page)

(c) out of the plane of the paper

(d) into the plane of the paper

(iii) If the electric field has a magnitude of 480 N/C and the magnetic field has a magnitude of 0.16T,

what speed must the particles have to pass through the selector undeflected?

(a) 2500 m/s

(b) 2700 m/s

(c) 3000 m/s

(d) 3200 m/s

(iv) A beam of neutrons is passed through the velocity selector, then

(a) All the neutrons will pass without any deflection

(b) Only the neutrons having a particular value of velocity will pass without any deflection

(c) Neutrons will be deflected in the direction of electric field

(d) All the neutrons will be deflected from their path

(v) A beam of electrons is moving with a constant velocity 40m/s in a region having uniform electric

field and uniform magnetic field perpendicular to each other and also perpendicular to the of

velocity of electrons. The magnitude of electric field is 20 N/C. The magnitude of magnetic field

is (in tesla)

(a) 2 T

(b) 1 T

(c) 0.5 T

(d) 1.5 T

4. Mass Spectrometer: The mass spectrometer is an instrument which can measure the masses and

relative concentrations of atoms and molecules. It makes use of the basic magnetic force on a moving

charged particle. The process begins with an ion source, a device like an electron gun. The ion source

gives ions their charge(q), accelerates them to some velocity 𝑣, and directs a beam of them into the

next stage of the spectrometer. This next region is a velocity selector that only allows particles with a

particular value of 𝑣 to get through undeflected. The velocity selector has both an electric field and a

magnetic field, perpendicular to one another and also to velocity of ions. . In the final region, there is

only a uniform magnetic field perpendicular to the velocity of ions and so the charged particles move in

circular arcs with radii proportional to particle mass.

𝒓 =𝒎𝒗

𝒒𝑩

(i) Velocity selector works on the principle

(a) Electric and magnetic forces are in same direction

(b) Electric and magnetic forces are equal and opposite to each other

(c) Electric force is more than magnetic force

(d) Electric force is less than magnetic force

(ii) Which of the following concept is used in mass spectroscopy?

(a) Charged particles are evenly deflected in a magnetic field

(b) Particles of lower mass are deflected in a magnetic field with a greater radius of curvature

(c) Particles of greater mass are deflected in a magnetic field with a greater radius of curvature

(d) None of the above

(iii) The ratio of radius of semi-circular path of two ions having same charge in mass spectrometer is

𝑟1

𝑟2=

3

4. What is the ratio of their masses?

(a) 𝑚1

𝑚2=

3

4

(b) 𝑚1

𝑚2=

1

2

(c) 𝑚1

𝑚2=

3

5

(d) 𝑚1

𝑚2=

4

3

(iv) Which of the following will trace a circular trajectory with largest radius?

(a) Proton

(b) Electron

(c) alpha-particle

(d) A particle with charge twice and mass thrice that of electron.

(v) The radius of curvature of the ion's path depends on the ratio of m/q. (m is the mass and q is the

charge of the ion). Which of the following ions will bend the most (have lesser radius of

curvature)?

(a) Ion A with m = 20 and q = +1

(b) Ion B with m = 10 and q = +1

(c) Ion C with m = 20 and q = +2

(d) Ion D with m = 10 and q = +2

5. Moving coil galvanometer: Moving coil galvanometer is an electromagnetic device that can

measure small values of current. It consists of permanent cylindrically cut magnets, coil, soft iron

core, pivoted spring, non-metallic frame, scale, and pointer. The magnetic field in moving coil

galvanometer is radial in nature it means the magnetic field always lies in the plane of coil which

makes the torque acting on the coil maximum. When the current is passed through the coil a

torque acts on the coil. Due to this torque the coil rotates and with the coil spring also rotates.

This rotation in the coil generates a restoring torque in the spring. The restoring torque is given

by 𝑘∅, where ∅ is the deflection in coil and 𝑘 is torsional constant of spring. In equilibrium

condition torque in the coil due to current is equal to the restoring torque in the spring.

(i) A moving coil galvanometer is an instrument which

(a) is used to measure emf

(b) is used to measure potential difference

(c) is used to measure resistance

(d) is a deflection instrument which gives a deflection when a current flows through its

coil

(ii) To make the field radial in a moving coil galvanometer

(a) number of turns of coil is kept small

(b) poles are cylindrically cut

(c) soft iron core is used

(d) both b and c

(iii) In a moving coil galvanometer, having a coil of N-turns of area A and carrying current I is

placed in a radial field of strength B. The torque acting on the coil is

(a) NA2BI

(b) NABI2

(c) NABI

(d) N2ABI

(iv) The deflection in a moving coil galvanometer is

(a) directly proportional to torsional constant of spring

(b) directly proportional to the number of turns in the coil

(c) inversely proportional to the area of the coil

(d) inversely proportional to the current in the coil

(v) Moving coil galvanometer can be used to measure alternating current

(a) True

(b) False

ANSWERS(Case Study Based Questions):

Q.1 Q.2 Q.3 Q.4 Q.5

ANS.i A C A B D

ANS.ii D D B C D

ANS.iii B C C A C

ANS.iv A B A C B

ANS v C C C D B

CHAPTER 5

MAGNETISM & MATTER

GIST OF LESSON:

Magnetic Line of Force

It is the path along which a unit north pole would, if it is tends to free to do so.

Representation of uniform magnetic field

-------------------------> o o o o X X X X

-------------------------> o o o o X X X X

-------------------------> o o o o X X X X

Current loop as a magnetic dipole

(i).Dipole moment of current loop m=nIA

(ii).Dipole moment of revolving electron m=evr/2

Gauss law for magnetism

The net magnetic flux through a closed surface is zero.

Dipole in a unifrom magnetic field

ζ = mBSin Ꝋ

I d2 Ꝋ/dt2 = - mBSin Ꝋ

d2 Ꝋ/dt2 = (mB Ꝋ)/I ( this equation represent simple harmonic motion)

ω2= (mB)/I

T=2𝛱√(𝑰/𝒎𝑩)

Magnetic element of Earth magnetic field:

1) Magnetic declination

The angle between the geographic meridian and magnetic meridian at a place.

2) Angle of dip

The angle between the direction of total intensity of earth’s magnetic field and its

horizontal component.

Angle of dip at equator δ = 00

Angle of dip at poles δ = 900

3) Horizontal component of earth’s magnetic field

Horizontal component of earth magnetic field BH = B Cosδ

Horizontal component of earth magnetic field BV = B Sinδ

It is the component of total intensity of earth’s magnetic field along the horizontal.

Tanδ= Bv/BH and B = √( Bv)2 +(BH)2

CHAPTER 5

MAGNETISM AND MATTER

MULTIPLE CHOICE QUESTIONS:

1. Magnetic moment (or magnetic dipole moment) of a current carrying coil is given by

(a) m = INA

(b) m = I/A

(c) m = NA

(d) m = NI/A

2. Magnetic field at the centre of a circular loop of area A is B. the magnetic moment of the

loop will be

(a) 𝐵𝐴2

𝝁𝟎𝝅

(b) 𝐵𝐴3/2

𝝁𝟎𝝅

(c) 𝐵𝐴3/2

𝝁𝟎 𝝅𝟏/𝟐

(d) 2𝐵𝐴3/2

𝝁𝟎 𝝅𝟏/𝟐

3. For an atom shown in the figure, the direction of magnetic moment related to revolving

electron is,

(a) Vertically upwards

(b) Vertically downwards

(c) along the velocity of electron

(d) none of these

4. In a hydrogen atom, an electron revolves around the nucleus in an orbit of radius r with

velocity v, the current corresponding to the revolving electron is

(a) I = 𝑒𝑟

2 𝝅𝒗 (b) I =

𝑒𝑟

2 𝝅

(c) I = 𝑒𝑣

2 𝝅𝒓 (d) I =

𝑒𝑣

𝝅𝒓

5. Bohr magneton(µB) is a unit of

(a) Magnetic dipole moment

(b) Charge of electron

(c) Angular momentum

(d) Magnetic flux density

6. Magnitude of angular magnetic moment associated with a revolving electron in a hydrogen

atom will be

(a) μl = 𝑒

2𝑚𝑒

(b) μl = −𝑒𝑙2

2𝑚𝑒

(c) μl = −𝑒

4𝑚𝑒

(d) μl = −𝑒𝑙

2𝑚𝑒

7. The gyro magnetic ratio of an revolving electron is,

(a) 𝜇

𝑙

(b) 𝑒

2𝑚𝑒

(c) 8.8 × 1010 C kg-1

(d) all of the above

8. Magnetic field lines show the direction (at every point) along which a small magnetised needle

aligns. Do the magnetic field lines also represent the lines of force on a moving charge at every

point?

(a) No

(b) yes

(c) Neither (a) nor (b)

(d) given information is not sufficient

9. The resemblance of magnetic field lines for a bar magnet and a solenoid suggest that

(a) a bar magnet may be thought of as a large number of circulating currents in analogy with a

solenoid

(b) cutting a bar magnet in half is like cutting a solenoid

(c) both (a) and (b)

(d) neither (a) nor (b)

10. Cutting a bar magnet in half is like cutting a solenoid, such that we get two smaller solenoids

with

(a) weaker magnetic properties

(b) strong magnetic properties

(c) constant magnetic properties

(d) both (a) and (b)

11. The strength of the earth's magnetic field varies from place to place on the earth’s surface, its

value being of the order of

(a) 105 T

(b) 10-6 T

(c) 10-5 T

(d) 108 T

12. The magnetic field is now thought to arise due to electrical currents produced by convective

motion of metallic fluids (consisting mostly of molten iron and nickel) in the outer core of the

earth. This is known as the

(a) dynamo effect

(b) Tindal effect

(c) both (a) and (b)

(d) Neither (a) nor (b)

13. The pole near the geographic North-pole of the Earth is called the……….magnetic pole and the

pole near the geographic south pole is called the………...magnetic pole.

(a) South-North

(b) South- East

(c) North-East

(d) North- South

14. The vertical plane containing the longitude circle and the axis of rotation of earth is called

the

(a) geographic meridian

(b) magnetic meridian

(c) magnetic declination

(d) magnetic inclination

15. One can define…… of a place as the vertical plane which passes through the imaginary line

joining the magnetic North and South poles.

(a) geographic meridian

(b) magnetic meridian

(c) magnetic declination

(d) magnetic inclination

16. Which of the following elements need to describe Earth’s magnetic field at a point on the

surface of earth?

(a) Angle of declination D

(b) Angle of Dip

(c) Horizontal component of Earth’s magnetic field

(d) all of the above

17. If a magnetic needle is perfectly balanced about a horizontal axis, so that it can swing in a

plane of the magnetic meridian, the needle would make an angle with horizontal. This is known

as the

(a) angle of dip

(b) angle of inclination

(c) angle of declination

(d) Both (a) and (b)

18. The angle of dip at a certain place where the horizontal and vertical components of earth’s

magnetic field are equal is

(a) 30o

(b) 70o

(c) 50o

(d) 45o

19. The primary origin of magnetism lies in

(a) atomic currents

(b) intrinsic spin of electrons

(c) Pauli exclusion principle

(d) both (a) and (b)

20. At a certain place, the horizontal component of the Earth’s magnetic field is Bo and the angle

of dip is 60o. The total intensity of the field at that place will be

(a) Bo

(b) 2Bo

(c) √2Bo

(d) 1.5 Bo

21. The magnetism of magnet is due to

(a) Earth

(b) Cosmic rays

(c) Due to pressure of big magnet inside the earth

(d) Spin motion of electrons

22. The vertical component of earth magnetic field is zero at

(a) Magnetic poles

(b) Magnetic equator

(c) Geographic poles

(d) Geographic equator

23. Angle on the horizontal plane between magnetic north and true north is

(a) Magnetic declination

(b) Electric declination

(c) Magnetic inclination

(d) Electric inclination

24. A small bar magnet of moment M is placed in a uniform field B. If magnet makes an angle

of 30° with field. The torque acting on the magnet is:-

(a) MB (b) MB/2 (c) MB/3 (d) MB/4

25. A magnetic needle suspended horizontally by a silk fibre, oscillates in the horizontal plane

because of the restoring torque originating mainly from:-

(a) The torsion of the silk fibre

(b) The force of gravity

(c) The horizontal component of earth's magnetic field

(d) All the above factors

26. The period of oscillation of a magnet in vibration magnetometer is 2 sec. The period of

oscillation of a magnet whose magnetic moment is four times that of the first magnet is

(a) 1 sec (b) 4 sec (c) 8 sec (d) 0.5 sec

27. At certain place the angle of dip is 30° and the horizontal component of earth magnetic field

is 0.50 Gauss. The earth's total magnetic field (in Gauss) is:-

(a) 1/4 (b) 1 (c) 1/3 (d) ½

28. The angle of dip at the magnetic equator is:-

(a) 0° (b) 45° (c) 30° (d) 90°

29. At Geo-magnetic poles, the angle of dip is:-

(a) 45° (b) 30° (c) zero (d) 90°

30. The time period of a freely suspended magnet is 4 sec. If it is broken perpendicular to its

length into two equal parts and one part is suspended in the same way, then its time period

in seconds will be :-

(a) 4 (b) 2 (c) 0.5 (d) 0.25

31. At magnetic North Pole of the earth the value of horizontal component BH and angle of dip

q is:-

(a) BH = 0, q = 45° (b) BH = 90, q = 0°

(c) BH = 0, q = 90° (d) BH = 45, q = 45°

32. A magnet of magnetic moment M is rotating through 360° with respect to the magnetic

field B, the work done will be :-

(a) MB (b) 2MB (c) 2MB (d) Zero

33. The work done in rotating a magnet of magnetic moment M by an angle of 90° from the

external magnetic field direction is 'n' times the corresponding work done to turn it

through an angle of 60° where 'n' is :-

(a) 1/2 (b) 2 (c) 1/4 (d) 1

ANSWERS(MULTIPLE CHOICE QUESTIONS):

Q. ANS Q ANS Q ANS Q ANS Q ANS Q ANS Q ANS

1 a 6 d 11 c 16 d 21 d 26 a 31 c

2 d 7 d 12 a 17 d 22 b 27 c 32 d

3 b 8 a 13 d 18 d 23 a 28 a 33 b

4 c 9 c 14 a 19 d 24 d 29 d

5 a 10 b 15 b 20 b 25 c 30 b

ASSERTION & REASON QUESTIONS:

Select the correct answer to these questions from the codes (a), (b), (c) and (d) as given

below:

(a) Both A and R are true and R is the correct explanation of A

(b) Both A and R are true but R is not the correct explanation of A

(c) A is true but R is false

(d) A is false and R is also false

1. Assertion (A): Magnetic moment of an atom is due to both the orbital motion and spin motion

of every electron.

Reason (R): A charged particle produces a magnetic field.

2. Assertion (A): The ends of a magnet suspended freely point out always along north south.

Reason (R): Earth behaves as a huge magnet.

3. Assertion (A): Magnetic moment is measured in joule/tesla or amp m2.

Reason (R): Joule/tesla is equivalent to amp m2.

4. Assertion (A): The true geographic north direction is found by using a compass needle.

Reason (R): The magnetic meridian of the earth is along the axis of rotation of the earth.

5. Assertion (A) : The magnetic poles of earth do not coincide with the geographic poles.

Reason (R) : The discrepancy between the orientation of compass and true north-south

direction is known as magnetic declination.

6.Assertion (A): the magnetic moment of an electron revolving around the nucleus decreases

with increasing principal quantum number (n).

Reason (R): magnetic moment of revolving electron is inversely proportional to the principal

quantum number (n).

7. Assertion (A) : When the radius of a circular loop carrying current is doubled, its magnetic

moment becomes four times.

Reason (R): magnetic moment depends on area of the loop.

8. Assertion (A): The earth’s magnetic field is due to iron present in its core.

Reason (R): At a low temperature magnet losses its magnetic property or magnetism.

9.Assertion (A): A compass needle when placed on magnetic north pole of the earth rotates in

vertical direction.

Reason (R): The earth’s has only horizontal component of its magnetic field at north poles.

10. Assertion (A): Earth’s magnetic field doesn’t affect the working of moving coil

galvanometer.

Reason (R): The earth’s magnetic field is quite weak as compared to magnetic field produced

in the moving coil galvanometer.

ANSWERS(Assertion & Reason):

CASE STUDY BASED QUESTIONS:

1. Components of Earth’s Magnetic Field:

The earth’s magnetic field at a point on its surface is usually characterised by three quantities:

(a) declination (b) inclination or dip and (c) horizontal component of the field. These are known

as the elements of the earth’s magnetic field. At a place, angle between geographic meridian and

magnetic meridian is defined as magnetic declination, whereas angle made by the earth’s

magnetic field with the horizontal in magnetic meridian is known as magnetic dip.

Q. 1 2 3 4 5 6 7 8 9 10

ANS c a a d a d a c d a

(i) In a certain place, the horizontal component of the magnetic field is 1/3 times the vertical

component. The angle of dip at this place is

(a) zero

(b) π/3

(c) π/2

(d) π/6

(ii) The angle between the true geographic north and the north shown by a compass needle is

called as

(a) inclination

(b) magnetic declination

(c) angle of meridian

(d) magnetic pole.

(iii) The angles of dip at the poles and the equator respectively are

(a) 30°, 60°

(b) 0°, 90°

(c) 45°, 90°

(d) 90°, 0°

(iv) A compass needle which is allowed to move in a horizontal plane is taken to a

geomagnetic pole. It

(a) will become rigid showing no movement

(b) will stay in any position

(c) will stay in the north-south direction only

(d) will stay in the east-west direction only.

(v) Select the correct statement from the following.

(a) The magnetic dip is zero at the centre of the earth.

(b) Magnetic dip decreases as we move away from the equator towards the magnetic pole.

(c) Magnetic dip increases as we move away from the equator towards the magnetic pole.

(d) Magnetic dip does not vary from place to place.

2. Earth’s Magnetic Field:

The magnetic field lines of the earth resemble that of a hypothetical magnetic dipole located at

the centre of the earth. The axis of the dipole is presently tilted by approximately 11.3° with

respect to the axis of rotation of the earth.

The pole near the geographic North Pole of the earth is called the North magnetic pole and the

pole near the geographic South Pole is called South magnetic pole.

(i) The strength of the earth’s magnetic field varies from place to place on the earth’s surface,

its value being of the order of

(a) 105 T

(b) 10–6 T

(c) 10–5 T

(d) 108 T

(ii) A bar magnet is placed North-South with its North-pole due North. The points of zero

magnetic field will be in which direction from the centre of the magnet?

(a) North-South

(b) East-West

(c) North-East and South-West

(d) None of these.

(iii) The value of angle of dip is zero at the magnetic equator because

(a) V and H are equal

(b) the values of V and H zero

(c) the value of V is zero

(d) the value of H is zero.

(iv) The angle of dip at a certain place, where the horizontal and vertical components of the

earth’s magnetic field are equal, is

(a) 30°

(b) 90°

(c) 60°

(d) 45°

(v) At a place, the angle of dip is 30°. If horizontal component of earth’s magnetic field is H,

then the total intensity of magnetic field will be

(a) H /2

(b) 2/√3 H

(c) H/3

(d) 2H

3. Gauss’s law of Magnetism:

By analogy to coulomb’s law of electrostatics, we can write Gauss’s law of magnetism as

∮ 𝐁. 𝐝𝐬 = µ0 minside where ʃ B.ds is the magnetic flux and minside is the net pole strength inside

the closed surface.

We do not have an isolated magnetic pole in nature. At least none has been found to exist till

date. The smallest unit of source of magnetic field is a magnetic dipole where the net magnetic

pole is zero. Hence, the net magnetic pole enclosed by any closed surface is always zero.

Correspondingly, the flux of the magnetic field through any closed surface is zero.

(i).. Consider the two idealized systems.

(I) A parallel plate capacitor with large plates and small separation and

(II) A long solenoid of length L>>R , radius of cross section.

In (I) E is ideally treated as a constant between plates and zero outside. In(II) magnetic field is

constant inside the solenoid and zero outside. Thse idealised assumptions, however,

contradict fundamental laws as

(a) case (I) contradicts Gauss’s law for electric fields

(b) case (II) contradicts gauss’s law for magnetic fields.

(c) case(I) agrees with ʃ E.dl = 0.

(d) case (II) contradicts ʃ H.dl = Ien.

(ii) The net magnetic flux through any closed surface, kept in a magnetic field is

(a) Zero (b) µ0/4π (c) 4 πµ0 (d) 4 µ0/π

(iii) A closed surface S encloses a magnetic dipole of magnetic moment 2ml. The magnetic flux

emerging from the surface is.

(a) µ0m (b) Zero (c) 2 µ0m (d) 2m/ µ0

(iv) Which of the following is not a consequence of Gauss’s law?

(a) The magnetic poles always exists as unlike pairs of equal strength.

(b) If several magnetic lines of force enter in a closed surface, then an equal number of lines of

force must leave that surface.

(c) There are abundant sources of sinks of the magnetic field inside a closed surface.

(d) Isolated magnetic poles do not exist.

(v) The surface integral of a magnetic field over a surface

(a) is proportional to mass enclosed (b) is proportional to change enclosed

(c) is Zero (d) equal to its magnetic flux through that surface.

4. Dipole:

In electromagnetism, there are two kinds of dipoles:

An electric dipole deals with the separation of the positive and negative charges found in

any electromagnetic system. A simple example of this system is a pair of electric charges of

equal magnitude but opposite sign separated by some typically small distance.

A magnetic dipole is the closed circulation of an electric current system. A simple example is a

single loop of wire with constant current through it. A bar magnet is an example of a magnet

with a permanent magnetic dipole moment.

Dipoles, whether electric or magnetic, can be characterized by their dipole moment, a vector

quantity. For the simple electric dipole, the electric dipole moment points from the negative

charge towards the positive charge, and have a magnitude equal to the strength of each charge

times the separation between the charges

For the magnetic (dipole) current loop, the magnetic dipole moment points through the loop

(according to the right hand grip rule), with a magnitude equal to the current in the loop times the

area of the loop.

(i) The SI unit of magnetic dipole moment is

(a,) N-m (b) J-T (c) A-m2 (d) A/m2

(ii) The direction of Magnetic dipole moment is

(a) From South Pole of the magnet to North pole

(b) From North Pole of the magnet to South pole

(c) From South pole to any arbitrary direction

(d) From North Pole to any arbitrary direction

(iii) The magnetic dipole moment of a current carrying Loop is

(a) M = NIA (b) M = I/A (c) M= NA (d) M = NI/A

(iv) The Direction of magnetic dipole moment is decide by

(a) Left Hand screw rule (b) Right hand screw rule

(c ) Gauss’s law (b) Ampere’s Circuital law

(v) Magnetic field at any axial point of dipole is given by

(a) µ0/4π . 4M/r3 (b) µ0/4π . M/2r3 (c) µ0/4π . M/r3 (d) µ0/4π . 2 M/r3

5. Magnetic Moment and Magnetic Field :

Observe the following figures

(i) (ii) (iii)

(iv) (v)

(i) If the magnet 1 has a magnetic moment 3 times that of magnet 2,are pivoted so that it is free

to rotate in the horizontal plane. In equilibrium what angle will the magnet 1 subtend with the

magnetic meridian?

(a) tan−1(1/2) (b) tan−1(1/3) (c) tan−1(1) (d) 00

(ii) In figure 2 in equilibrium position. The ratio of distance d1 and d2 will be

(a) (2tanθ)1/3 (b) (2tanθ)−1/3 (c) (2cotθ)1/3 (d) (2cotθ)−1/3

(iii) In figure 3 The magnitude of the magnetic field at a distance d from the centre on the

bisector of the right angle will be

(a) µ0/4π . M/d3 (b) µ0/4π . M √2 /d3 (c) µ0/4π . 2√2 M/d3 (d) µ0/4π.2 M/d3

(iv) In figure 4 two identical bar magnets with a length 10 cm and weight 50 gm-weight are

arranged freely with their like poles facing in a inverted vertical glass tube. The upper magnet

hangs in the air above the lower one so that the distance between the nearest poles of the magnet

is 3mm. Pole strength of the poles of each magnet will be

(a) 6.64 amp - m ( b) 2 amp-m (c ) 10.25 amp-m (d) 0 amp-m

(v) In figure 5 A small coil C with N = 200 turns is mounted on one end of a balance beam and

introduced between the poles of an electromagnet as shown in figure. The cross sectional area of

coil is A= 1.0 cm2, length of arm OA of the balance beam is l=30cm. When there is no current in

the coil the balance is in equilibrium. On passing a current I = 22 mA through the coil the

equilibrium is restored by putting the additional counter weight of mass Δm=60mg on the balance

pan. Find the magnetic induction at the spot where coil is located.

(a) 0.4T (b) 0.3T (c) 0.2T (d) 0.1T

Solution: (i) - For equilibrium of the system torques on M1 and M2 due to BH must counter

balance each other i.e. M1×BH=M2×BH. If q is the angle between M1 and BH will be (90−θ); so

M1BHsinθ=M2BHsin(90−θ) ⇒tanθ=M2/M1=M/3M=1/3 ⇒θ=tan−1(1/3)

(ii) In equilibrium B1=B2tanθ

⇒μ0/4π.2Md13=μ0/4π.Md23tanθ ⇒d1d2=(2cotθ)1/3

(iii) Resultant magnetic moment of the two magnets is Mnet=√M2+M2−−−−−−−−= M √2

Imagine a short magnet lying along OP with magnetic moment equal to M√2. Thus point P lies on

the axial line of the magnet. ∴Magnitude of magnetic field at P is given by B=μ0/4π.2M√2/d3

(iv) The weight of upper magnet should be balanced by the repulsion between the two

magnet

∴ μ04π.m2/r2=50gm−wt ⇒10−7×m2/(9×10−6)=50×10−3×9.8

⇒m=6.64amp×m

(v) On passing current through the coil, it acts as a magnetic dipole. Torque acting on magnetic

dipole is counter balanced by the moment of additional weight about position O. Torque acting

on a magnetic dipole τ=M B sinθ=(NIA)Bsin90o=NIAB Again τ=Force x Lever arm =Δmg ×l

⇒NIAB=Δmg l ⇒B=Δmg l/NIA=60×10−3×9.8×30×10−2 /200×22×10−3×1×10−4 = 0.4 T

ANSWERS(Case Study Questions):

ANS.(i) ANS.(ii) ANS.(iii) ANS.(iv) ANS.(v)

Q.1 b b d a c

Q.2 c b c d b

Q.3 b a b c d

Q.4 c a a b d

Q.5 b c c a a

CHAPTER 6

ELECTROMAGNETIC INDUCTION

GIST OF LESSON: 1. Magnetic Flux: The magnetic flux linked with any surface is equal to total number of

magnetic lines of force passing normally through it. It is a scalar quantity.

2. The phenomenon of generation of current or emf by changing the magnetic flux is known as Electromagnetic Induction EMI. 3. Faraday’s Law of Electromagnetic Induction First Law: Whenever magnetic flux linked with the closed loop or circuit changes, an emf induces in the loop or circuit which lasts so long as change in flux continuous. Second Law : The induced emf in a closed loop or circuit is directly proportional to the rate of change of magnetic flux linked with the closed loop or circuit

where, N = number of turns in loop. Negative sign indicates the Lenz’s law. 4. Lenz’s Law The direction of induced emf or induced current is such that it always opposes the cause that produce it. NOTE: Lenz’s law is a consequence of the law of conservation of energy. 5. If N is the number of turns and R is the resistance of a coil. The magnetic flux linked with its each turn changes by dФ in short time interval dt, then induced current flowing through the coil is

6. If induced current is produced in a coil rotated in a uniform magnetic field, then

7. Motional Emf : The potential difference induced in a conductor of length l moving with velocity v, in a direction perpendicular to magnetic field B is given by

8. Fleming’s Right Hand Rule : If the thumb, forefinger and middle finger of right hand are stretched mutually perpendicular to each other such that the forefinger points the direction

of magnetic field, thumb points towards the direction of magnetic force, then middle finger points towards the direction of induced current in the conductor.

9. The induced emf developed between two ends of conductor of length l rotating about one end with angular velocity ω in a direction perpendicular to magnetic field is given by,

10. The induced emf can be produced in a coil by (i) putting the coil/loop/circuit in varying magnetic field. (ii) changing the area A of the coil inside the magnetic field, (iii) changing the angle 0 between B and A. 11. Rotation of rectangular coil in a uniform magnetic Field:

Magnetic flux linked with coil

= BANcos

= BANcost

a) Induced emf in thecoil

E = (𝑑Ø/𝑑𝑡)

= BANwsinwt

= E0sinwt

b) Induced current in the coil. I = E/R

= 𝐸0𝑠𝑖𝑛𝑤𝑡

𝑅

c) Both Emf and current induced in the coil are alternating. 12. Self Induction and Self inductance:

a) The phenomenon in which an induced emf is produced by changing the current in a

coil is called self induction.

∅ 𝛼 I or ∅ = L I

Or L = ∅

𝐼

Now 𝐸 = −𝐿𝑑𝐼

𝑑𝑡

𝐿 = −𝐸

𝑑𝐼𝑑𝑡

where L is a constant, called self inductance or coefficient of self – induction.

b) Self inductance of a solenoid.

𝐿 = 𝜇0𝑁2𝐴

𝑙 , where symbols have their usual meanings.

CHAPTER 6

ELECTROMAGNETIC INDUCTION

MULTIPLE CHOICE QUESTIONS:

1. In the given figure current from A to B in the straight wire is increasing. The direction of the induced current in the loop is

(a) clockwise. (b) anticlockwise. (c) straight line. (d) No current is induced.

2. In a coil of self-induction 5 H, the rate of change of current is 2 As-1. Then emf induced in the coil is (a) 10 V (b) -10 V (c) 5 V (d) -5 V

3. The magnetic flux linked with a coil of N turns of area of cross section A held with its plane parallel to the field B is

4. Two identical coaxial coils P and Q carrying equal amount of current in the same direction are brought nearer. The current in (a) P increases while in Q decreases (b) Q increases while in P decreases (c) both P and Q increases (d) both P and Q decreases

5. Faraday’s laws are consequence of the conservation of (a) charge (b) energy (c) magnetic field (d) both (b) and (c)

6. Direction of current induced in a wire moving in a magnetic field is found using (a) Fleming’s left hand rule (b) Fleming’s right hand rule (c) Ampere’s rule (d) Right hand screw rule

7. Which of the following statements is not correct? (a) Whenever the amount of magnetic flux linked with a circuit changes, an emf is induced in circuit. (b) The induced emf lasts so long as the change in magnetic flux continues. (c) The direction of induced emf is given by Lenz’s law. (d) Lenz’s law is a consequence of the law of conservation of momentum.

8. Lenz’s law is a consequence of the law of conservation of (a) charge (b) energy (c) induced emf (d) induced current

9. A solenoid is connected to a battery so that a steady current flows through it. If an iron core is inserted into the solenoid, the current will (a) increase (b) decrease (c) remain same (d) first increase then decrease

10. There is a uniform magnetic field directed perpendicular and into the plane of the paper. An irregular shaped conducting loop is slowly changing into a circular loop in the plane of the paper. Then (a) current is induced in the loop in the anti-clockwise direction. (b) current is induced in the loop in the clockwise direction. (c) ac is induced in the loop. (d) no current is induced in the loop.

11. In the given figure current from A to B in the straight wire is decreasing. The direction of induced current in the loop is A

(a) clockwise (b) anticlockwise (c) changing (d) nothing can be said

12. Which of the following does not use the application of eddy current? (a) Electric power meters (b) Induction furnace (c) LED lights (d) Magnetic brakes in trains

13. The north pole of a bar magnet is rapidly introduced into a solenoid at one end (say A). Which of the following statements correctly depicts the phenomenon taking place? (a) No induced emf is developed. (b) The end A of the solenoid behaves like a south pole. (c) The end A of the solenoid behaves like north pole. (d) The end A of the solenoid acquires positive potential.

14. A metal plate can be heated by (a) passing either a direct or alternating current through the plate. (b) placing in a time varying magnetic field. (c) placing in a space varying magnetic field, but does not vary with time. (d) both (a) and (b) are correct.

15. Identify the wrong statement. (a) Eddy currents are produced in a steady magnetic field. (b) Eddy currents can be minimized by using laminated core. (c) Induction furnace uses eddy current to produce heat. (d) Eddy current can be used to produce braking force in moving trains.

16. If number of turns in primary and secondary coils is increased to two times each, the mutual inductance (a) becomes 4 times (b) becomes 2 times (c) becomes A times (d) remains unchanged 4

17. When the rate of change of current is unity, the induced emf is equal to (a) thickness of coil (b) number of turns in coil (c) coefficient of self-inductance (d) total flux linked with coil

18. Two inductors of inductance. L each are connected in series with opposite? magnetic fluxes. The resultant inductance is (Ignore mutual inductance) (a) zero (b) L (c) 2L (d) 3L

19. A square of side L meters lies in the x-y plane in a region, where the magnetic field is given by B = B0 i + 3j + 4k) T, where Bo is constant. The magnitude of flux passing through the square is (a) 2BoL² Wb. (b) 3BoL² Wb. (c) 4BoL² Wb. (d) √29 BoL² Wb.

20. A loop, made of straight edges has six comers at A(0, 0, 0), B(L, 0, 0) C(L, L, 0), D(0, L, 0), E(0, L, L) and F(0,0, L). A magnetic field B = Bo (i+k)T is present in the region. The flux passing through the loop ABCDEFA (in that order) is (a) BoL² Wb. (b) 2BoL² Wb. (c) √2BoL² Wb. (d) 4BoL² Wb.

21. An e.m.f is produced in a coil, which is not connected to an external voltage source. This is not due to (a) the coil being in a time varying magnetic field. (b) the coil moving in a time varying magnetic field. (c) the coil moving in a constant magnetic field. (d) the coil is stationary in external spatially varying magnetic field, which does not change with time.

22. There are two coils A and B as shown in Figure. A current starts flowing in B as shown, when A is moved towards B and stops when A stops moving. The current in A is counterclockwise. B is kept stationary when A moves. We can infer that

(a) there is a constant current in the clockwise direction in A. (b) there is a varying current in A. (c) there is no current in A. (d) there is a constant current in the counterclockwise direction in A.

23. Same as question 4 except the coil A is made to rotate about a vertical axis (Figure). No current flows in B if A is at rest. The current in coil A, when the current in B (at t = 0) is counterclockwise and the coil A is as shown at this instant, t = 0, is

(a) constant current clockwise. (b) varying current clockwise. (c) varying current counterclockwise. (d) constant current counterclockwise.

24. A cylindrical bar magnet is rotated about its axis (Figure). A wire is connected from the axis

and is made to touch the cylindrical surface through a contact. Then (a) a direct current flows in the ammeter A. (b) no current flows through the ammeter A. (c) an alternating sinusoidal current flows through the ammeter A with a time period T = 2πω (d) a time varying non-sinusoidal current flows through the ammeter.

25. Eddy currents do not cause (a) damping (b) heating (c) sparking (d) loss of energy

26. Magnetic flux of 5 mwb is linked with a coil, when current of 1 mA flows through it, Self-induced will be a) 1 H b) 5 H c) 20 H d) 50 H

27. According to Faraday’s laws of electromagnetic induction, an emf is induced in a conductor whenever? a) The conductor is perpendicular to the magnetic field b) Lies in the magnetic field c) Cuts magnetic lines of flux d) Moves parallel to the magnetic field

28. Direction of induced current is determined by __________ a) Fleming’s left-hand rule b) Fleming’s right-hand rule c) Faraday’s law d) Right hand thumb rule

29. “The direction of an induced e.m.f. is always such that it tends to set up a current opposing the motion or the change of flux responsible for inducing that e.m.f.”, this is the statement for? a) Fleming’s left-hand rule b) Fleming’s right hand rule c) Faraday’s law d) Lenz’s law

30. According to Fleming’s right-hand rule, the thumb points towards? a) Current b) E.M.F. c) Motion of the conductor d) Magnetic flux

31. According to Fleming’s right-hand rule, the middle finger points towards? a) Current b) E.M.F. c) Motion of the conductor d) Magnetic flux

32. The relation between the direction of induced emf and the direction of motion of the conductor is? a) Parallel b) Equal c) Not related d) Perpendicular

33. According to _________________ induced emf is equal to rate of change of magnetic flux. a) Newton’s law b) Lenz law c) Faraday’s law d) Coulomb’s law

34. What is the principle of the transformer? a) Gauss law b) Coulomb’s law c) Mutual induction d) Ampere’s law

35. As the number of turns in the coil increases, what happens to the inductance of the coil? a) Increases b) Decreases c) Remains the same d) Becomes zero

36. What happens to the self inductance of a coil when its length increase? a) Increases b) Decreases c) Remains the same d) Becomes zero

37. What happens to the inductance when the current in the coil becomes double its original value? a) Becomes half b) Becomes four times c) Becomes infinity d) remains same

38. If the current changes from 20A to 10A in 5 seconds and the value of inductance is 1H, calculate the emf induced. a) 8V b) 6V c) 4V d) 2V

39. The phenomenon due to which there is an induced current in one coil due to the current in a neighboring coil is? a) Eddy currents b) Self-induction c) Mutual induction d) Steady current

40. If the current in one coil becomes steady, back emf in the coil is? a) Zero b) Infinity c) Doubles d) Halves

ANSWERS(Multiple Choice Questions):

Q. 1 2 3 4 5 6 7 8 9 10 11 12 13

ANS a b d d b b d b b a b c c

Q. 14 15 16 17 18 19 20 21 22 23 24 25 26

ANS d a a c c c b d d b b c b

Q. 27 28 29 30 31 32 33 34 35 36 37 38 39

ANS c b d c a d c c a b d d c

Q. 40

ANS a

ASSERSION AND REASON QUESTIONS:

Directions: These questions consist of two statements, each printed as Assertion and Reason. While answering these questions, you are required to choose any one of the following four responses. (a) If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion. (b) If both Assertion and Reason are correct but Reason is not a correct explanation of the Assertion. (c) If the Assertion is correct but Reason is incorrect. (d) If both the Assertion and Reason are incorrect.

1. Assertion: Induced emf will always occur whenever there is change in magnetic flux. Reason : Current always induces whenever there is change in magnetic flux.

2. Assertion: Faraday’s laws are consequence of conservation of energy. Reason : Self inductance of a coil depends on current flowing through the coil.

3. Assertion : Only a change in magnetic flux will maintain an induced current in the coil. Reason : The presence of large magnetic flux through a coil maintain a current in the coil of the circuit is continuous. 4. Assertion: Lenz’s law violates the principle of conservation of energy. Reason: Induced emf always opposes the change in magnetic flux responsible for its production. 5. Assertion : In the phenomenon of mutual induction, self-induction of each of the coils

persists.

Reason : Self induction arises when strength of current in same coil changes. In mutual

induction, current is changing in both the individual coils.

6. Assertion : Acceleration of a magnet falling through a long solenoid decreases. Reason : The induced current produced in a circuit always flow in such direction that it opposes the change to the cause that produced it.

7. Assertion : Figure shows a horizontal solenoid connected to a battery and a switch. A copper ring is placed on a smooth surface, the axis of the ring being horizontal. As the switch is closed, the ring will move away from the solenoid.

Reason : Induced emf in the ring, e=-dΦ/dt

8. Assertion : An emf can be induced by moving a conductor in a magnetic field. Reason : An emf can be induced by changing the magnetic field.

9. Assertion : Figure shows a metallic conductor moving in magnetic field. The induced emf across its ends is zero.

Reason : The induced emf across the ends of a conductor is given by e = Bvℓ sinθ.

10. Assertion : Eddy currents are produced in any metallic conductor when magnetic flux is changed around it. Reason : Electric potential determines the flow of charge.

ANSWERS (Assertion & Reason Questions):

Q. 1 2 3 4 5 6 7 8 9 10

ANS c c c a b a a b a b

CASED STUDY QUESTIONS:

1. Motional EMF : Conductor PQ of length L moving towards right with velocity V perpendicular

to magnetic field B directed into the plane of paper as the conductor pk moves it its free electrons

also move in the same direction and experience magnetic Lorentz force F=evB. By Fleming's left

hand rule electrons move from P to Q within the conductor the end P becomes positive and end

Q becomes negative. An electric field is set up in the conductor from P to Q this exerts a force

F=eE on the free electrons. The accumulation of charges at the two ends continues till these two

forces balance each other ie Fm=Fe or evB=eE or vB=E. Potential difference between the ends P

and Q .

is V=El=vBl. It is the magnetic force on the free electrons that maintains potential difference and

produces the EMF E=Blv. This MF is produced due to the motion of conductor it is called a

motional EMF.

(i) A conducting wire sits on smooth metal rails as shown in the figure a variable magnetic field points out the page the strength of this field is increased linearly from zero immediately after the field start to increase what will be the direction of the current in the wire and the direction of the wires motion

(ii) A 50 cm long bar ab is moved with a speed of 4 metre per second in a magnetic field

B=0.01 T as shown in the figure the EMF generated is a) 01V b) 0.02V c) 0.03 V

d) 0.04 V

(iii) An aeroplane having a wing span of 35 M flies due north with a speed of 90 metre per

second given B= 4 x10-⁵ T. The potential difference between the tips of the wings will be

a). 0.126V

b). 1.26V

c). 12.6V

d). 0.013V

(iv) If the vertical component of Earth's magnetic field be. 6x10-5Wbm-2, then what will be

induced potential difference between the rails of a meter-gauge when a train is running on

them with a speed of 36 km/h

(v) A wheel with 10 metallic spokes each 4.5 m long is rotated with a speed of 120 rev/min

in a plane normal to the horizontal component of Earth's magnetic field BH at a place. if BH =

0.4 G at the place. the magnitude of induced EMF between the axle and the rim of the wheel

is

2. Eddy Currents:

The motion of a copper plate is damped when it is allowed to oscillate between the magnetic

pole pieces. Magnetic flux associated with the plate keeps on changing as the plate moves in

and out of the region between the magnetic poles. The changing flux sets up induced currents

in the copper plates which flow along closed path in planes perpendicular to the magnetic

field the current looks like adress aur whirlpools in water. So they are known as eddy currents.

They also oppose the change in magnetic flux. Eddy currents develop heat at the cost of

kinetic energy of the plates as the plate slows down. This is called electromagnetic damping.

To reduce electromagnetic damping one can cut slots in the plate which reduces the possible

parts of Eddy currents considerably.

(i) A metallic pendulum oscillating in a uniform magnetic field directed perpendicular to the

plane of oscillation

a. slows down

b. Becomes faster

c. Unaffected

d. Oscillates with the changing frequency

(ii). A metallic cylinder is held vertically. A small bar magnet is dropped along its Axis will fall

with acceleration such that

a. a>g

b. a<g

c. a=g

d. a=0, velocity

(iii). A coil is suspended in a uniform magnetic field with the plane of the coil parallel to the

magnetic lines of force when a current is passed through the coil it starts oscillating, it is very

difficult to stop. But if an aluminium plate is placed near the coil is. This is due to

a. Development of air current when the plate is placed

b. Induced of electric charge on the plate

c. Shielding of magnetic lines of force as element aluminium is a paramagnetic material

d. Electromagnetic induction in the aluminium plate giving rise to electromagnetic damping

(iv). Eddy currents are produced when

a. A metal is kept in a steady magnetic field

b. A metal is kept in a varying magnetic field

c. A circular coil is placed in a magnetic field

d. A current is passed through a circular coil

(v). The core of a transformer is laminated so that

a. Rusting of iron core may be stopped

b. Ratio of voltage in primary and secondary coil may be increased

c. Rate of change of flux is increased

d. Energy losses due to Eddy currents may be reduced.

3. Self Induction:

When a current passing through a coil changes the magnetic flux linked with the coil changes.

As a result and EMF is induced in the coil. This phenomenon is called self induction.

At any instant the flux linked with that n number of turns of the coil is proportional to the

current I through the coil that is N directly proportional to I or and Nɸ =LI.

The proportionality constant L is called self inductance or simply inductance. Inductance is

the ratio of flux linkage to the current. It is equal to Nɸ/I lenz law the induced EMF opposes

the change of current so inductance L is measure of the inertia of the coil against the change

of current through it. In contrast to it a resistor r opposes the flow of current I through the

circuit. A solenoid made from a thick wire has a negligible resistance but a sufficiently large

inductance. Such an element is called an ideal inductor. Inductance of a coil depends upon

the geometry of the coil, number of turns in the coil and the permeability of the medium

inside the coil.

(i). What is the unit of self inductance of a coil?

(ii). If the number of turns per unit length of a coil of a solenoid is doubled, its self

inductance will be

a. remains constant

b. be doubled

c. be halved

d. be four times

(iii). If both the number of turns and core length of an inductor are doubled keeping other

factors constant self inductance will be

a. unaffected

b. doubled

c. halved

d. quadrupled

(iv). In a coil of self inductance 5 Henry, the rate of change of current is 2 amp per second.

The EMF induced in the coil is

a. -5V

b. 5V

c. -10V

d. 10V

(v). A Coil of 100 turns carries a current of 5 mA and creates a magnetic flux of 10-5 wb. The

inductance is

a. 0.2mH

b. 2.0 mH

c. 0.02mH

d. None of these

4. Eddy Currents:

Eddy Currents and their Effects Currents can be induced not only in conducting coils, but also in conducting sheets or blocks. Current is induced in solid metallic masses when the magnetic flux threading through them changes. Such currents flow in the form of irregularly shaped loops throughout the body of the metal. These currents look like eddies or whirlpools in water so they are known as eddy currents. Eddy currents have both undesirable effects and practically useful applications. For example, it causes unnecessary heating and wastage of power in electric motors, dynamos and in the cores of transformers.

(i) The working of speedometers of trains is based on

(a) wattles current

(b) eddy currents

(c) alternating current (d) pulsating currents

(ii) Identify the wrong statement.

(a) Eddy currents are produced in a steady magnetic field.

(b) Induction furnace uses eddy currents to produce heat.

(c) Eddy currents can be used to produce braking force in moving trains.

(d) Power meters work on the principle of eddy currents.

(iii) Which of the following is the best method to reduce eddy currents?

(a) Laminating core (b) Using thick wires (c) By reducing hysteresis loss

(d) None of these

(iv) The direction of eddy currents is given by

(a) Fleming's left hand rule (b) Biot-Savart law (c) Lenz's law (d) Ampere-circuital law

(v) Eddy currents can be used to heat localised tissues of the human body. This branch of medical

therapy is called

(a) Hyperthermia (b) Diathermy (c) Inductothermy (d) none of these.

ANSWERS(Case Study Questions):

ANS.(i) ANS.(ii) ANS.(iii) ANS.(iv) ANS.(v)

Q.1 c b a b d

Q.2 a b d b d

Q.3 d d b c c

Q.4 b a a c C

Solutions of Case Study Questions:

1(i)

1(ii)

1(iii)

1(iv)

1(v)

2(i)

2(ii)

2(iii)

2(iv)

2(v)

3(i)

3(ii)

3(iii)

3(iv)

3(v)

4(i) B

4(ii) A

4(iii) A

4(iv) C

4(v) C

CHAPTER 7

ALTERNATING CURRENT

GIST OF LESSON: An electric current whose magnitude changes continuously with time and changes its direction periodically, is called an alternating current.

The instantaneous value of alternating current at any instant of time t is given by

I = Io sin ωt where, I0 = peak value of alternating current.

The variation of alternating current with time is shown in graph given below

Mean or average value of alternating current for first half cycle

Im = 2Io / π = 0.637 Io Mean or average value of alternating current for next half cycle

I’m = – 2Io / π = – 0.637 Io Mean or average value of alternating current for one complete cycle = O.

Root mean square value of alternating current

Iv = Irms = Io / √2 = 0.707 Io Where, Io = peak value of alternating current. Root mean square value of alternating voltage

Vrms = Vo / √2 = 0.707 Io = 0.707 Vo

Reactance The opposition offered by an inductor or by a capacitor in the path of flow of alternating current is called reactance.

Reactance is of two types

(i) Inductive Reactance (XL) Inductive reactance is the resistance offered by an inductor. Inductive reactance (XL) = Lω = L2πf = L2π / T Its unit is ohm. XL ∝ f For direct current, XL = 0 (f = 0)

(ii) Capacitive Reactance (Xc) Capacitive reactance is the resistance offered by an inductor Capacitive reactance,

Xc = 1 / Cω = 1 / C2πf = T / C 2π Its unit is ohm Xc ∝ 1 / f For direct current, Xc = ∞ (f = 0)

Impedance The opposition offered by an AC circuit containing more than one out of three components L, C and R, is called impedance (Z) of the circuit.

Impedance of an AC circuit, Z = √R2 + (XL – XC)2 Its SI unit is ohm.

Power in an AC Circuit The power is defined as the rate at which work is being in the circuit.

The average power in an AC circuit,

Pav = Vrms irms cos θ = V / √2 i / √2 cos θ = Vi / √2 cos θ

where, cos θ = Resistance(R) / Impedance (Z) is called the power factor 0f AC circuit.

Current and Potential Relations Here, we will discuss current and potential relations for different AC circuits.

(i) Pure Resistive Circuit (R circuit)

(a) Alternating emf, E = Eo sin ωt (b) Alternating current, I = Io sin ωt (c) Alternating emf and alternating current both are in the same phase.

(d) Average power decay, (P) = Ev . Iv (e) Power factor, cos θ = 1

(ii) Pure Inductive Circuit (L Circuit)

(a) Alternating emf, E = Eo sin ωt (b) Alternating current, I = Io sin (ωt – π / 2) (c) Alternating current lags behind alternating emf by π / 2.

(d) Inductive reactance, XL = Lω = L2πf (e) Average power decay, (P) = 0

(f) Power factor, cos θ = cos 90° = 0

(iii) Pure Capacitive Circuit

(a) Alternating emf, E = Eo sin ωt (b) Alternating current, I = Io sin (ωt + π / 2) (c) Alternating current lags behind alternating emf by π / 2.

(d) Inductive reactance, XL = Cω = C2πf (e) Average power decay, (P) = 0

(f) Power factor, cos θ = cos 90° = 0

(iv) R – C Circuit

E = Eo sin ωt I = Eo / 2 sin (ωt – φ) Z = √R2 + (1 / ωC)2 tan φ = – 1 / ωC / R

Current leading the voltage by φ

V2 = V2R = V2

C

(v) L – C Circuit

(vi) L – C – R Circuit

(a) Alternating emf, E = Eo sin Ωt (b) Alternating current, I = Io sin (Ωt ± θ) (c) Alternating current lags leads behind alternating emf by ω.

(d) Resultant voltage, V = √V2R + (VL – VC)2

(e) Impedance, Z = √R2 + (XL – XC)2 (f) Power factor, cos θ = R / Z = R / √√R2 + (XL – XC)2 (g) Average power decay, (P)= EVIV cos θ Resonance in AC Circuit The condition in which current is maximum or impedance is minimum in an AC circuit, is called resonance.

(i) Series Resonance Circuit

In this circuit components L, C and R are connected in series.

At resonance = XL = XC Resonance frequency f = 1 / 2π√LC

A series resonance circuit is also known as acception circuit.

(ii) Parallel Resonance Circuit

In this circuit L and C are connected in parallel with each other.

At resonance, XL = XC Impedance (Z) of the circuit is maximum.

Current in the circuit is minimum.

Wattless Current Average power is given by

Pav = Erms = Irms cos θ Here the Irms cos φ contributes for power dissipation. Therefore, it is called wattless current. Transformer It is a device which can change a low voltage of high current into a high voltage of low current and vice-versa.

Its working is based on mutual induction.

There are two types of transformers.

(i) Step-up Transformers It converts a low voltage of high current into a high voltage of low current.

In this transformer,

Ns > NP, Es > EP and IP > IS (ii) Step-down Transformer It converts a high voltage of low current into a low voltage of high current. In this transformer,

NP > NS, EP > ES and IP < IS Transformation Ratio Transformation ratio,

K = NS / NP = ES / EP = IP / IS For step-up transformer, K > 1

For step-down transformer, K < 1

Energy Losses in a Transformer The main energy losses in a transformer are given below

1. Iron loss

2. Copper loss

3. Flux loss

4. Hysteresis loss

5. Humming loss

Important Points Transformer does not operate on direct current. It operates only on alternating voltages at input as well as at

output.

Transformer does not amplify power as vacuum tube.

Transformer, a device based on mutual induction converts magnetic energy into electrical energy.

Efficiency, η = Output power / Input power

Generally efficiency ranges from 70% to 90%.

CHAPTER 7

ALTERNATING CURRENT

MULTIPLE CHOICE QUESTIONS:

1. In A.C. circuit having only capacitor, the current

(a) Lags behind the voltage by 900 in phase

(b) Leads the voltage by 900 in phase

(c) Leads the voltage by 1800 in phase

(d) Lags behind the voltage by 1800 in phase

2. The ratio of peak value and r.m.s. value of an alternating current is

(a) 1 (b) 1

2

(c) √2 (d) 1

√2

3. The maximum value of a.c. voltage in a circuit is 770V. Its r.m.s. Value is

(a) 70.7 V b) 100 V

(c) 500 V d) 704 V

4. In a pure inductive circuit, the current

(a) Leads the e.m.f. by 900

(b) Lags behind the e.m.f. by 900

(c) Sometimes leads and sometime legs behind the e.m.f.

(d) Is in phase with the e.m.f.

5. Voltage and current in an ac circuit are given by

𝑉 = 5 sin(100𝜋𝑡 −𝜋

6), 𝐼 = 4 sin (100𝜋𝑡 +

𝜋

6)

(a) Voltage leads the current by 300

(b) Current leads the voltage by 300

(c) Current leads the voltage by 600

(d) Voltage leads the current by 600

6. An inductor L of reactance XL is connected in series with a bulb B to an ac source as shown in the figure.

How does the brightness of the bulb change when Number of turns of the inductor is increased?

(a) Increases (b) Decreases (c) Remains same (d) none of these

7. The instantaneous current in an ac circuit is I=2.0 sin314t, what is its frequency?

(a) 100 Hz (b) 60 Hz (c) 50 Hz (d) 2π

8. An A.C source is connected to a resistive circuit. Which of the following is true?

(a) Current leads the voltage and both are in same phase

(b) Current lags behind the voltage and both are in same phase

(c) Current and voltage are in same phase

(d) Any of the above may be true depending upon the value of resistance

9. What is the effective value of current?

(a) RMS current (b) Average current (c) Instantaneous current (d) Total current

10. In a sinusoidal wave, average current is always _______ rms current.

(a) Greater than (b) Less than (c) Equal to (d) Not related

11. Using an A.C. voltmeter, the potential difference in the electrical line in a house is read to be 234 volts.

If the line frequency is known to be 50 cycles per second, the equation for the line voltage is

(a) V = 165 sin(100πt) (b) V = 331 sin(100πt)

(c) V = 234 sin(100πt) (d) V = 440 sin(100πt)

12. What are the dimensions of impedance?

(a) ML2T-3I-2 (b) M-1L-2T-3I2

(c) ML3T-3I-2 (d) M-1L-3T3I2

13. A voltage across a pure inductor is represented in figure.

Which one of the following curves in the figure will represent the current?

14. The reactance of a capacitor C is X. If both the frequency and capacitance be doubled, then new

reactance will be

(a) 2X (b)X

(c) 4X (d)𝑋

4

15. An Inductor may store energy in

(a) Its electric field (b)Its coil

(c)Its magnetic field (d) Both in electric and magnetic fields.

16. For high frequency, capacitor offers

(a)More reactance (b) Zero reactance

(c) Less reactance (d) None of these.

17. Choose the correct statement.

(a) The capacitor can conduct in a dc circuit but not an inductor

(b) In dc circuit the inductor can conduct but not a capacitor

(c) In dc circuit both the inductor and capacitor cannot conduct

(d) The inductor has infinite resistance in a dc circuit

18. If a capacitor is connected across a source of AC voltage then

(a) voltage and current are in same phase.

(b) voltage and current are out of phase.

(c) voltage leads current by 90 degree phase angle.

(d) Current leads voltage by 90 degree phase angle.

19. What is power factor of an ac circuit:

(a) cosφ

(b) tanφ

(c) cotφ

(d) sinφ

20. If the value of C in a series RLC circuit is decreased, the resonant frequency

(a)Is not affected

(b) Increases

(c)Is reduced to zero

(d)Decreases

21. In a series RLC circuit that is operating above the resonant frequency, the current

(a) Lags the applied voltage

(b) Leads the applied voltage

(c) Is in phase with the applied voltage

(d) zero

22. The impedance at the resonant frequency of a series RLC circuit with L = 20 mH C = 0.02 μ F and R = 90

Ω is

(a)0 Ω (b) 90 Ω

(c) 20 Ω (d) 40 Ω

23. A 24 Ω resistor, an inductor with a reactance of 120 Ω, and a capacitor with a reactance of 120 Ω are in

series across a 60 V source. The circuit is at resonance. The voltage across the inductor is

(a) 60 V (b) 660 V

(c) 30 V (d) 300 V

24. How many types of power can be defined in an AC circuit?

(a) 3 (b) 2

(c) 1 (d) 5

25. Which among the following varies in both magnitude and sign over a cycle?

(a) Apparent power (b) Effective power

(c) Instantaneous power (d) Average power

26. What is the power factor for a series LCR circuit at resonance?

(a) Infinity (b) -1

(c) 0 (d) 1

27. In A.C generator increasing the no. of turns in the coil

(a) Decreases the Electromotive force (EMF)

(b) Electromotive force (EMF) remains same

(c) Increases the Electromotive force (EMF)

(d) Electromotive force (EMF) becomes zero

28. The metal detectors installed at airports and other places for security purpose are based on the principle

of

(a) Potential difference (b) Electromagnetic difference

(c) Electromagnetic induction (d) Potential energy

29. In alternative current generator in India, AC current reverses its direction

(a) 10 times per second (b) 20 times per second

(c) 60 times per second (d) 50 times per second

30. The direction of induced Emf or current is indicated by

(a) Fleming’s left-hand rule (b) Fleming’s right-hand rule

(c) Faraday’s law of electromagnetic induction (d) None of these

31. An iron core transformer with a turns ratio of 8 : 1 has 120 V applied across the primary. The voltage

across the secondary is

(a) 15 V (b) 120 V (c) 180 V (d) 960 V

32. Transformer works on the principle of:

(a) Convertor (b) Inverter

(c) Mutual induction (d) Self-induction

33. Which of the following effects is not shown in alternating current ?

(a) Chemical effect (b) Magnetic effect

(c) Heating current (d) All of these

34. The figure shows variation of R, XL and XC with frequency f in a series LCR circuit. Then for what

frequency point, the circuit is inductive.

(a) A (b) B

(c) C (d) A and B

35. The core of a transformer is laminated, so as to

(a) Make it light weight (b) Make it robust and strong

(c) Increase the secondary voltage (d) Reduce energy loss due to eddy current

36. In a transformer, the no. of turns of primary and secondary coil are 500 and 400 respectively. If 220

V is supplied to the primary coil, then ratio of currents in primary and secondary coils is

(a) 4:5 (b) 5:4 (c) 5:9 (d) 9:5

37. In an LC oscillator, the frequency of oscillator is ……………. L or C.

(a)Proportional to square of

(b)Directly proportional to

(c)Independent of the values of

(d)Inversely proportional to square root of

38. EMF generated in an alternator depends on which of the following factors?

(a) Number of turns in the coil

(b)Speed of the rotating field

(c)Magnetic field strength

(d)All of these

ANSWERS(Multiple Choice Questions):

Q. NO. ANSWER Q. NO. ANSWER Q. NO. ANSWER Q. NO. ANSWER

1 B 11 A 21 A 31 A

2 C 12 A 22 B 32 C

3 C 13 D 23 C 33 A

4 B 14 D 24 A 34 C

5 C 15 C 25 C 35 D

6 B 16 C 26 D 36 A

7 C 17 B 27 C 37 D

8 C 18 D 28 C 38 D

9 A 19 A 29 D

10 B 20 C 30 B

ASSERTION & REASON QUESTIONS:

Directions.

In each of the following questions 2 statements are given. One is assertion and the other is reason.

Examine the statements carefully mark the correct answer as

(a) If both assertion and Reason are true and reason is the correct explanation of assertion.

(b) If both assertion and Reason are true and but reason is not the correct explanation of assertion.

(c) If assertion is true but reason is false.

(d) If both assertion and reason is false.

1. Assertion(A): - A capacitor of suitable capacitance can be used in a circuit in place of the choke coil.

Reason(R): - A capacitor blocks DC and allow AC only.

2. Assertion(A): -The sum of the instantaneous current values over one complete cycle is zero, and the

average current is zero.

Reason(R): - The applied voltage and the current varies sinusoidally and has corresponding positive and

negative values during each cycle.

3. Assertion(A): -The inductive reactance limits the current in a purely inductive circuit in the same way as

the resistance limits the current in a purely resistive circuit.

Reason(R): - The inductive reactance is directly proportional to the inductance and to the frequency of

the current.

4. Assertion(A): Transformers are used only in alternating current source not in direct current.

Reason(R) : Only a.c. can be stepped up or down by means of transformers.

5. Assertion(A): An important application of electromagnetic induction is ac generator.

Reason(R): The direction of current changes periodically and therefore the current is called alternating

current.

6. Assertion(A): In a series LCR circuit at resonance condition, power consumed by the circuit is maximum.

Reason(R): At resonance, the effective resistance of the circuit is maximum.

7. Assertion (A): We use a thick wire in the secondary coil of a step down transformer to reduce the

production of heat.

Reason (R): When the plane of the armature is parallel to the line of force of magnetic field, the

magnitude of induced e.m.f. is maximum

8. Assertion (A): Soft iron is used as a core of transformer.

Reason (R): Area of hysteresis loop for soft iron is small.

9. Assertion (A): Capacitor serves as a block for D.C. and offers an easy path to A.C.

Reason (R): Capacitive reactance is inversely proportional to frequency.

10. Assertion (A): In the phenomenon of mutual induction self-induction of each of coils persists

Reason(R)- Self-induction arises when strength of current in same coil change in the mutual induction,

current is changing in both the individual

ANSWERS(Assertion & Reason Questions):

Q.No. 1 2 3 4 5 6 7 8 9 10

ANS b a a a a c b a a b

CASE STUDY QUESTIONS:

1. Transformer:

A transformer is an electrical device which is used for changing the a.c. voltages. It is based on the

phenomenon of mutual induction i.e. whenever the amount of magnetic flux linked with a coil changes, an

e.m.f. is induced in the neighboring coil. For an ideal transformer, the resistances of the primary and

secondary windings are negligible

It can be shown that Es/Ep=Is/Ip=Ns/Np=K

For a step-up transformer Ns>Np: Es>Ep: K>1 ((Is<Ip)

For a step-down transformer Ns<Np: Es<Ep: K<1

The above relations are on the assumptions that efficacy of transformer is 100%

Infact efficiency= out power/input power=Es Is/Ep Ip

(i) Which of the following quantity remains constant in an ideal transformer?

(a) Current (b) Voltage

(c) Power (d) All of these

(ii) Transformer is used to

(a) Convert ac to dc voltage (b) Convert dc voltage into ac

(c) Obtain desired dc power (d) Obtain desired ac voltage and current

(iii) The number of turns in primary coil of a transformer is 20 and the number of turns in a secondary

is 10. If the voltage across the primary is 220V ac what is the voltage across the secondary?

(a) 100 V (b) 120 V

(c) 110 V (d) 220 V

(iv) In a transformer the number of primary turns is four times that of the secondary turns. Its primary

is connected to an a.c. source of voltage V. Then

(a) Current through its secondary is about four times that of the current through its primary.

(b) Voltage across its secondary is about four times that of the voltage across its primary.

(c) Voltage across its secondary is about two times that of the voltage across its primary.

(d) Voltage across its secondary is about ?& times that of the voltage across its primary.

(v) A transformer is used to light 100 W-110 V lamp from 220 V mains. If the mains current is 0.5 A, the

efficiency of the transformer is

(a) 95% (b) 99% (c) 90% (d) 96%

2. Power Associated with LCR Circuit:

In an a.c. circuit, values of voltage and current change every instant. Therefore, power of an a.c. circuit

at any instant is the product of instantaneous voltage (E) and instantaneous current (I). The average

power supplied to a pure resistance R over a complete cycle of a.c. is P=Ev Iv. When circuit is inductive,

average power per cycle is EvIv Cosφ

In an a.c. circuit, 600 mH inductor and a 50 pf capacitor are connected in series with 10 ohm resistance.

The a.c. supply to the circuit is 230 V, 60 Hz.

(i) The average power transferred per cycle to resistance is

(a) 10.42 W (b) 15.25 W (c) 17.42 W (d) 13.45 W

(ii) The average power transferred per cycle to capacitor is

(a) zero (b) 10.42 W (c) 17.42 W (d) 15W

(iii) The average power transferred per cycle to inductor is

(a) 25 W (b) 17.42 W (c) 16.52 W (d) zero

(iv) The total power transferred per cycle by all the three circuit elements is

(a) 17.42 W (b) 10.45 W (c) 12.45 W (d) zero

(v) The electrical energy spend in running the circuit for one hour is

(a) 7.5 x 105 Joule (b) 10 x 103 Joule (c) 9.4 x 103 Joule (d) 6.2 x 104 Joule

3. AC Voltage applied to a capacitor:

Let a source of alternating e.m.f 𝐸 = 𝐸0sin (𝑤𝑡) be connected to a capacitor of capacitance C. If I is

the instantaneous value of current in the circuit at instant t, then 𝐼 = 𝑤𝑐 𝐸0 sin (𝑤𝑡 +𝜋

2). The capacitive

reactance limits the amplitude of current in a purely capacitive circuit and It is given by 𝑋𝐶 = 1

𝑊𝐶.

(i) What is the unit of capacitive reactance?

(a) Farad (b) Ampere (c) ohm (d) ohm-1

(ii) The capacitive reactance of a 5 µF capacitor for a frequency of 106 Hz is

(a) 0.032 Ω (b)2.52 Ω

(c) 1.25 Ω (d) 4.51 V

(iii) In a capacitive circuit, resistance to the flow of current is offered by

(a) Resistor (b) capacitor

(c)Inductor (d) frequency

(iv) In a capacitive circuit, by what value of phase angle does alternating current leads the e.m.f.?

(a) 450 (b) 900 (c) 750 (d) 600

(v) One microfarad capacitor is joined to a 200V, 50 Hz alternator. The r.m.s. current through capacitor

is

(a) 6.28 × 10-2 A (b) 7.5 × 10-4 A

(c) 10.52 × 10-2 A (d) 15.25 × 10-2 A

4. Step Down Transformer in the Transmission of Electric Power:

Step down transformers is used to decrease or step-down voltages. These are used when voltage need to

be lowered for use in homes and factories.

A small town with a demand of 800 kW of electric power at 220 V is situated 15 km away from an electric

plant generating power at 440 V. The resistance of the two-wire line carrying power is 0.5Ω per km. The

town gets power from the line through a 4000-220 V step-down transformer at a sub-station in the town.

(i) The value of total resistance of the wires is

(a) 25 Ω (b) 30 Ω

(c) 35 Ω (d) 15 Ω

(ii) The line power loss in the form of heat is

(a) 550 KW (b) 650 kW

(c) 600 kW (d) 700 kW

(iii) How much power must the plant supply, assuming there is negligible power loss due to leakage?

(a) 600 KW (b) 1600 KW

(c) 500 kW (d) 1400 KW

(iv) The voltage drop in the power line is

(a) 1700 V (b) 3000 V

(c) 2000 V (d) 2800 V

(v) The total Value of voltage transmitted from the plant is

(a) 500 V (b) 4000 V

(c) 3000 V (d) 7000 V

5. AC Voltage applied to an Inductor:

Let a source of alternating e.m.f 𝐸 = 𝐸0sin (𝑤𝑡) be connected to a capacitor of capacitance C. If I is he

instantaneous value of current in the circuit at instant t, then 𝐼 = 𝑤𝑐 𝐸0 sin (𝑤𝑡 −𝜋

2). The inductive

reactance limits the amplitude of current in a purely capacitive circuit and it is given by 𝑋𝐿 = 𝑤𝐿.

(i) A 100 Hz a.c. is flowing in a 14 mH coil. The reactance of the coil is

(a) 15 Ω (b)7.5 Ω

(c) 8.8 Ω (d) 10 Ω

(ii) In a pure inductive circuit, resistance to the flow of current is offered by

(a) Resistor (b) inductor

(c) Capacitor (d) resistor and inductor

(iii) In a inductive circuit, by what value of phase angle does alternating current legs behind e.m.f.?

(a) 450 (b) 900

(c) 1200 (d) 750

(iv) How much inductance should be connected to 200 V, 50 Hz a.c. supply so that a maximum current of

0.9 A flows through it?

(a) 5 H (b) 1 H

(c) 10 H (d) 4.5 H

(v) The maximum value of current when inductance of 2 H is connected to 150 volt, 50 Hz supply is

(a) 0.337 A (b) 0.721 A

(c) 1.521 A (d) 2.522 A

ANSWERS(Case Study Questions):

ANS.(i) ANS.(ii) ANS.(iii) ANS.(iv) ANS.(v)

Q.1 c d c a c

Q.2 c a d a d

Q.3 c a b b a

Q.4 d c d b d

Q.5 c b b b a

SAMPLE PAPER

&

ANSWER KEY

CLASS XII

PHYSICS(042)

2021-22

Sample Question Paper 2021-22 Term 1

Subject: Physics (042)

Time: 90 Minutes Max. Marks 35 General Instructions: 1. The Question Paper contains three sections. 2. Section A has 25 questions. Attempt any 20 questions. 3. Section B has 24 questions. Attempt any20 questions. 4. Section C has 6 questions. Attempt any 5 questions. 5. All questions carry equal marks. 6. There is no negative marking.

SECTION A

This section consists of 25 multiple choice questions with overall choice to

attempt any 20 questions. In case more than desirable numberof questions are

attempted, ONLY first 20 will be considered for evaluation.

Q1.Which of the following is NOT the property of equipotential surface?

(i) They do not cross each other.

(ii) The rate of change of potential with distance on them is zero.

(iii) For a uniform electric field they are concentric spheres.

(iv) They can be imaginary spheres.

Q2. Two point charges +8q and -2q are located at x=0 and x=L respectively. The point on x axis at which net electric field is zero due to these charges is-

(i) 8L

(ii) 4L

(iii) 2 L

(iv) L

Q3. An electric dipole of moment p is placed parallel to the uniform electric field. The amount of work done in rotating the dipole by 900 is-

(i) 2pE

(ii) pE

(iii) pE/2

(iv) Zero

Q4. Three capacitors 2µF, 3µF and 6µF are joined in series with each other. The equivalent capacitance is-

(i) 1/2µF

(ii) 1µF

(iii) 2µF

(iv) 11µF

Q5. Two point charges placed in a medium of dielectric constant 5 are at a distance r between them, experience an electrostatic force ‘F’. The electrostatic force between them in vacuum at the same distance r will be-

(i) 5F

(ii) F

(iii) F/2

(iv) F/5

Q6. Which statement is true for Gauss law-

(i) All the charges whether inside or outside the gaussian surface contribute to the electric flux.

(ii) Electric flux depends upon the geometry of the gaussian surface.

(iii) Gauss theorem can be applied to non-uniform electric field.

(iv) The electric field over the gaussian surface remains continuous and uniform at every point.

Q7.A capacitor plates are charged by a battery with ‘V’ volts. After charging battery is disconnected and a dielectric slab with dielectric constant ‘K’ is inserted between its plates, the potential across the plates of a capacitor will become

(i) Zero

(ii) V/2

(iii) V/K

(iv) KV

Q8.The best instrument for accurate measurement of EMF of a cell is-

(i) Potentiometer

(ii) metre bridge

(iii) Voltmeter

(iv) ammeter and voltmeter

Q9.An electric current is passed through a circuit containing two wires of same material, connected in parallel. If the lengths and radii of the wires are in the ratio of 3:2 and 2:3, then the ratio of the current passing through the wire will be

(i) 2:3

(ii) 3:2

(iii) 8:27

(iv) 27:8

Q10.By increasing the temperature, the specific resistance of a conductor and a semiconductor-

(i) increases for both.

(ii) decreases for both.

(iii) increases for a conductor and decreases for a semiconductor.

(iv) decreases for a conductor and increases for a semiconductor.

Q11.We use alloys for making standard resistors because they have

(i) low temperature coefficient of resistivity and high specific resistance

(ii) high temperature coefficient of resistivity and low specific resistance

(iii) low temperature coefficient of resistivity and low specific resistance

(iv) high temperature coefficient of resistivity and high specific resistance

Q12. A constant voltage is applied between the two ends of a uniform metallic wire, heat ‘H’ is developed in it. If another wire of the same material, double the radius and twice the length as compared to original wire is used then the heat developed in it will be-

(i) H/2

(ii) H

(iii) 2H

(iv) 4H

Q13.If the potential difference V applied across a conductor is increased to 2V with its temperature kept constant, free electrons in a conductor

(i) remain the same(ii) become half of its previous value(iii) be double of its initial value(iv) become zero.

Q14.The equivalent resistance

(i) 3 ohms

(ii) 5.5 ohms

(iii) 7.5 ohms

(iv) 9.5 ohms

Q15. The SI unit of magnetic field intensity is

(i) AmN-1 (ii) NA-1m-1

(iii) NA-2m-2

(iv) NA-1m-2

Q16.The coil of a moving coil galvanometer is wound over a metal frame in order to

(i) reduce hysteresis(ii) increase sensitivity

(iii) increase moment of inertia (iv) provide electromagnetic damping

Q17.Two wires of the same length are shaped into a square of side 'a' and a circle with radius 'r'. If they carry same current, the ratio of their magnetic moment is

(i) 2 : π (ii) π : 2 (iii) π : 4 (iv) 4 : π

potential difference V applied across a conductor is increased to 2V with its temperature kept constant, the drift velocity of the

in a conductor will -

emain the same. become half of its previous value. e double of its initial value.

quivalent resistance between A and B is-

. The SI unit of magnetic field intensity is

The coil of a moving coil galvanometer is wound over a metal

reduce hysteresis increase sensitivity increase moment of inertia provide electromagnetic damping

.Two wires of the same length are shaped into a square of side 'a' and a circle with radius 'r'. If they carry same current, the ratio of their

potential difference V applied across a conductor is increased the drift velocity of the

.Two wires of the same length are shaped into a square of side 'a' and a circle with radius 'r'. If they carry same current, the ratio of their

Q18. The horizontal component of earth’s magnetic field at a place is √3 times the vertical component. The angle of dip at that place is (i) π/6 (ii) π/3

(iii) π/4 (iv) 0

Q19. The small angle between magnetic axis and geographic axis at a place is-

(i) Magnetic meridian (ii) Geographic meridian (iii) Magnetic inclination (iv) Magnetic Declination

Q20.Two coils are placed close to each other. The mutual inductance of the pair of coils depends upon the

(i) rate at which current change in the two coils (ii) relative position and orientation of the coils (iii) rate at which voltage induced across two coils (iv) currents in the two coils

Q21. A conducting square loop of side 'L' and resistance 'R' moves in its plane with the uniform velocity 'v' perpendicular to one of its sides. A magnetic induction 'B' constant in time and space pointing perpendicular and into the plane of the loop exists everywhere as shown in the figure. The current induced in the loop is

(i) BLv/R Clockwise

(ii) BLv/R Anticlockwise

(iii) 2BLv/R Anticlockwise

(iv)Zero

Q22. The magnetic flux linked with the coil (in Weber) is given by

theequation –

Փ = 5t2 + 3t + 16

The induced EMF in the coil at time, t=4 will be-

(i) -27 V

(ii) -43 V

(iii) -108 V

(iv) 210 V

Q23. Which of the following graphs represent the variation of current(I) with frequency (f) in an AC circuit containing a pure capacitor?

(i) (ii) (iii) (iv)

Q24. A 20 volt AC is applied to a circuit consisting of a resistance and a coil with negligible resistance. If the voltage across the resistance is 12 volt, the voltage across the coil is-

(i) 16 V

(ii) 10 V

(iii) 8 V

(iv) 6 V

Q25. The instantaneous values of emf and the current in a series ac circuit are-

E = Eo Sin ωt and I= Io sin( ωt+π/3) respectively, then it is

(i) Necessarily a RL circuit

(ii) Necessarily a RC circuit

(iii)Necessarily a LCR circuit

(iv) Can be RC or LCR circuit

SECTION B

This section consists of 24 multiple choice questions with overall choice to

attempt any 20 questions. In case more than desirable number of questions

are attempted, ONLY first 20 will be considered for evaluation.

Q26. A cylinder of radius r and length l is placed in an uniform electric field parallel to the axis of the cylinder. The total flux for the surface of the cylinder is given by-

(i) zero

(ii) π r2

(iii) E π r2

(iv) )2 Eπ r2

Q27. Two parallel large thin metal sheets have equal surface densities

26.4x10-12 C/m2of opposite signs. The electric field between these sheets is-

(i) 1.5N/C

(ii) 1.5 x 10-16 N/C

(iii) 3 x 10-10N/C

(iv) 3N/C

Q28. Consider an uncharged conducting sphere. A positive point charge is placed outside the sphere. The net charge on the sphere is then,

(i) negative and uniformly distributed over the surface of sphere

(ii) positive and uniformly distributed over the surface of sphere

(iii) negative and appears at a point the surface of sphere closest to point charge.

(iv) Zero

Q29. Three Charges 2q, -q and -q lie at vertices of a triangle. The value of E and V at centroid of triangle will be-

(i) E#0 and V#0

(ii) E=0 and V=0

(iii) E#0 and V=0

(iv) E=0 and V#0

Q30. Two parallel plate capacitors X and Y, have the same area of plates and same separation between plates. X has air and Y with dielectric of constant 2 , between its plates. They are connected in series to a battery of 12 V. The ratio of electrostatic energy stored in X and Y is-

(i) 4:1

(ii) 1:4

(iii) 2:1

(iv) 1:2

Q31.Which among the following, is not a cause for power loss in a transformer-

(i) Eddy currents are produced in the soft iron core of a transformer. (ii) Electric Flux sharing is not properly done in primary and secondary

coils.

(iii) Humming sound produed in the tranformers due to magnetostriction. (iv) Primary coil is made up of a very thick copper wire.

Q32.An alternating voltage source of variable angular frequency ‘w’ and fixed amplitude ‘V’ is connected in series with a capacitance C and electric bulb of resistance R(inductance zero). When ‘w’ is increased-

(i) The bulb glows dimmer. (ii) The bulb glows brighter. (iii) Net impedance of the circuit remains unchanged. (iv) Total impedance of the circuit increases.

Q33. A solid spherical conductor has charge +Q and radius R. It is surrounded by a solid spherical shell with charge -Q, innerradius 2R, and outer radius 3R. Which of the following statements is true?

(i)The electric potential has a maximum magnitude at C and the electric field has a maximum magnitude at A

(ii) The electric potential has a maximum magnitude at D and the electric field has a maximum magnitude at B.

(iii) The electric potential at A is zero and the electric field has a maximum magnitude at D.

(iv). Both the electric potential and electric field achieve a maximum magnitude at B.

Q34. A battery is connected to the conductor of non-uniform cross section area. The quantities or quantity which remains constant is-

(i) electric field only

(ii) drift speed and electric field

(iii)electric field and current

(iv) current only

Q35. Three resistors having values R battery. Suppose R1 carries a current of 2.0 A, R ohms, and R3 dissipates 6.0 watts of power. Then the voltage across R is-

(i) 1V

(ii) 2V

(iii) 3V

(iv) 4V

Q36.A straight line plot showing the terminal potential difference (V) of a cell as a function of current (I) drawn from it, is internal resistance of the cell would be then

(i) 2.8 ohms

(ii) 1.4 ohms

(iii) 1.2 ohms

(iv) zero

Q37. A 10 m long wire of uniform cross a potentiometer. The wire is connected in series with a battery of 5 V along with an external resistance of 480 balanced at 6.0 m length of the wire

(i) 1.2 V

(ii) 1.02 V

(iii) 0.2 V

. Three resistors having values R1, R2, and R3 are connected in series to a carries a current of 2.0 A, R2 has a resistance of 3.0

dissipates 6.0 watts of power. Then the voltage across R

Q36.A straight line plot showing the terminal potential difference (V) of a cell as a function of current (I) drawn from it, is shown in the figure. The internal resistance of the cell would be then-

A 10 m long wire of uniform cross-section and 20 Ω resistance is used in a potentiometer. The wire is connected in series with a battery of 5 V along with an external resistance of 480 Ω. If an unknown emf E is balanced at 6.0 m length of the wire, then the value of unknown emf is

are connected in series to a has a resistance of 3.0

dissipates 6.0 watts of power. Then the voltage across R3

Q36.A straight line plot showing the terminal potential difference (V) of a cell shown in the figure. The

resistance is used in a potentiometer. The wire is connected in series with a battery of 5 V

. If an unknown emf E is , then the value of unknown emf is-

(iv) 0.12 V

Q38.The current sensitivity of a galvanometer increases by 20%. If its resistance also increases by 25%, the voltage sensitivity will

(i ) decrease by 1%

(ii) increased by 5%

(iii) increased by 10%

(iv ) decrease by 4%

Q39. Three infinitely long parallel straight current carrying wires A, B and C are kept at equal distance from each other as shown in the figure . The wire C experiences net force F .The net force on wire C, when the current in wire A is reversed will be

(i) Zero

(ii) F/2

(iii) F

(iv) 2F

Q40. In a hydrogen atom the electron moves in an orbit of radius 0.5 Ao

making 10 revolutions per second, the magnetic moment associated with the orbital motion of the electron will be

(i) 2.512 x 10-38 Am2

(ii) 1.256 x 10-38 Am2

(iii) 0.628 X10-38 Am2

(iv) zero

Q41. An air-cored solenoid with length 30 cm, area of cross-section 25 cm2 and number of turns 800, carries a current of 2.5 A. The current is suddenly switched off in a brief time of 10-3s. Ignoring the variation in magnetic field near the ends of the solenoid, the average back emf induced across the ends of the open switch in the circuit would be

(i) zero

(ii)3.125 volts

(iii) 6.54 volts

(iv) 16.74 volts

Q42. A sinusoidal voltage of peak value 283 V and frequency 50 Hz is applied to a series LCR circuit in which R = 3 Ω, L = 25.48 mH, and C = 796 µF, then the power dissipated at the resonant condition will be-

(i)39.70 kW

(ii) 26.70 kW

(iii)13.35 kW

(iv)Zero

Q43. A circular loop of radius 0.3cm lies parallel to much bigger circular of radius 20 cm. The centre of the small loop is on the axis of the bigger loop. The distance between their centres is 15 cm. If a current of 2.0 A flows through the smaller loop, then the flux linked with the bigger loop is

(i) 3.3 X 10-11weber

(ii) 6 X 10-11weber

(iii) 6.6 X 10-9weber

(iv) 9.1 X 10-11weber

Q44.If both the number of turns and core length of an inductor is doubled keeping other factorsconstant, then its self-inductance will be-

(i) Unaffected (ii) doubled

(iii) halved

(iv) quadrupled

45. Given below are two statements labelled as Assertion (A) and Reason (R)

Assertion (A): To increase the range of an ammeter, we must connect

a suitable high resistance in series to it.

Reason (R): The ammeter with increased range should have high resistance.

Select the most appropriate answer from the options given below:

(i) Both A and R are true and R is the correct explanation of A

(ii) Both A and R are true but R is not the correct explanation of A.

(iii)A is true but R is false.

(iv) A is false and R is also false.

46. Given below are two statements labelled as Assertion (A) and Reason (R)

Assertion (A): An electron has a high potential energy when it is at a location associated with a more negative value of potential, and a low potential energy when at a location associated with a more positive potential.

Reason (R):Electrons move from a region of higher potential to region of lower potential.

Select the most appropriate answer from the options given below:

(i) Both A and R are true and R is the correct explanation of A

(ii) Both A and R are true but R is not the correct explanation of A.

(iii)A is true but R is false.

(iv) A is false and R is also false.

47. Given below are two statements labelled as Assertion (A) and Reason (R)

Assertion(A): A magnetic needle free to rotate in a vertical plane, orients itself (with its axis) vertical at the poles of the earth.

Reason (R): At the poles of the earth the horizontal component of earth’s magnetic field will be zero.

Select the most appropriate answer from the options given below:

(i) Both A and R are true and R is the correct explanation of A

(ii) Both A and R are true but R is not the correct explanation of A.

(iii)A is true but R is false.

(iv) A is false and R is also false.

48. Given below are two statements labelled as Assertion (A) and Reason (R)

Assertion(A): A proton and an electron, with same momenta, enter in a magnetic field in a direction at right angles to the lines of the force. The radius of the paths followed by them will be same.

Reason(R): Electron has less mass than the proton.

Select the most appropriate answer from the options given below:

(i) Both A and R are true and R is the correct explanation of A

(ii) Both A and R are true but R is not the correct explanation of A.

(iii)A is true but R is false.

(iv) A is false and R is also false.

49. Given below are two statements labelled as Assertion (A) and Reason (R)

Assertion (A):On Increasing the current sensitivity of a galvanometer by increasing the number of turns, may not necessarily increase its voltage sensitivity.

Reason(R ): The resistance of the coil of the galvanometer increases on increasing the number of turns.

Select the most appropriate answer from the options given below:

(i) Both A and R are true and R is the correct explanation of A

(ii) Both A and R are true but R is not the correct explanation of A.

(iii)A is true but R is false.

(iv) A is false and R is also false.

SECTION C

This section consists of 6 multiple choice questions with an overall choice to attempt any 5. In case more than desirable number of questions are attempted, ONLY first 5 will be considered for evaluation.

Q50. A small object with charge q and weight mg is attached to one end of a string of length ‘L’ attached to a stationary support. The system is placed in a uniform horizontal electric field ‘E’, as shown in the accompanying figure. In the presence of the field, the string makes a constant angle θ with the vertical. The sign and magnitude of q-

(i) positive with magnitude mg/E

(ii) positive with magnitude (mg/E)tanθ

(iii) negative with magnitude mg/E tanθ

(iv) positive with magnitude E tanθ/mg

Q51.A free electron and a free proton are placed between two oppositely charged parallel plates. Both are closer to the positive plate than the negative plate.

Which of the following statements is true?

I. The force on the proton is greater than the force on the electron.

II. The potential energy of the proton is greater than that of the electron.

III. The potential energy of the proton and the electron is the same.

(i) I only

(ii) II only

(iii) III and I only

(iv)II and I only

Case study : Read the following paragraph and answers the questions:

The large-scale transmission and distribution of electrical energy over long distances is done with the use of transformers. The voltage output of the generator is stepped-up. It is then transmitted over long distances to an area sub-station near the consumers. There the voltage is stepped down. It is further stepped down at distributing sub-stations and utility poles before a power supply of 240 V reaches our homes.

Q52. Which of the following statement is true?

(i) Energy is created when a transformer steps up the voltage

(ii) A transformer is designed to convert an AC voltage to DC voltage

(iii) Step–up transformer increases the power for transmission

(iv) Step–down transformer decreases the AC voltage

Q53. If the secondary coil has a greater number of turns than the primary,

(i) the voltage is stepped-up (Vs >Vp ) and arrangement is called a step-up transformer

(ii) the voltage is stepped-down (Vs <Vp ) and arrangement is called a step- down transformer

(iii) the current is stepped-up (Is >Ip ) and arrangement is called a step-up transformer

(iv) the current is stepped-down (Is <Ip ) and arrangement is called a step- down transformer

Q54. We need to step-up the voltage for power transmission, so that

(i) the current is reduced and consequently, the I2R loss is cut down

(ii) the voltage is increased , the power losses are also increased

(iii) the power is increased before transmission is done

(iv) the voltage is decreased so V2/R losses are reduced

Q55. A power transmission line feeds input power at 2300 V to a step down transformer with its primary windings having 4000 turns. The number of turns in the secondary in order to get output power at 230 V are

(i) 4

(ii) 40

(iii) 400

(iv) 4000

Physics

Marking Scheme

For SQP – 45

XII – I Term

Q.1 Which of the_____________________

Ans. 1 (iii)

As all other statements are correct. In uniform electric field equipotential surfaces are never

concentric spheres but are planes to Electric field lines.

Q.2 Two Point charges_____________________

Ans. 2 (iii)

Let P is the observation point at a distance r from -2q

and at (L+r) from +8q.

Given Now, Net EFI at P = 0

1E = EFI (Electric Field Intensity) at P due to +8q

2E = EFI (Electric Field Intensity) at P due to -2q

1 2E = E

2 2

k(8q) k(2q)

(L r) r

2 2

4 1

(L r) (r)

4r2 = (L+r)2

2r = L+r

r = L

P is at x = L + L = 2L from origin

Correct Option is (iii) 2L

Q3. An electric______________________

Ans. 2 (ii)

W = pE (cos1 – cos2)

1 = 0

2 = 90

W = pE (cos 0 - cos 90)

= pE (1 – 0) = pE

Q4. Three Capacitors______________

Ans.4. (ii)

series 1 2 3

series

1

F

1 1 1

C C C C

1 1 1 1

C 2 3 6

3+2+1 6

6 6

Cseries 1

Q5. Two Point Charges___________________

Ans.5. (i)

Q2

r K = 5 1 2

2

1 Q QF

4 ok r

Q2 Force in the charges in the air is

1 2

2

1 Q QF

4 o r

= K F

= 5 F

Q6. Which statement is true________________

Ans.6. (iv)

All other statements except (iv) are in correct

The electric field over the Gaussian surface remains continuous and uniform at every point.

Q7. A capacitor plates

Ans.7. (iii)

C

+ –

+ –

+ –

Battery is disconnected`ʹ

Ck

+ –

+ –

+–

Q = Charge remains context

Cʹ = K C

Qʹ = Cʹ Vʹ

Q = Cʹ Vʹ

Q = K C Vʹ

Vʹ= 𝑄

𝐾 𝐶=

𝑉

𝐾

Q.8. The best instrument for_________________

Ans.8. (i)

Potentiometer

Q9. An electric current___________________

Ans.9. (iii) 8:27

R1

+ -

R2

l : = 3 : 2

r r 2 3

I I ?

1 2 l

: = :

: =

1 2

1 2

1

2

11

2

22

2

lR

r

lR

r

2 21 1 2 1 2

2 22 2 1 2 1

2 3

3

1 1 2

2 2 1

R l r l r =

R l r l r

3 3 (3) 27 = =

2 2 (2) 8

I V / R R = = 8 / 27

I V / R R

Q.10. By increasing the temperature___________________

Ans. 10. (iii) Specific resistance of a conductor increases and for a semiconductor decreases with

increase in temperature because for a conductor, a temperature.

coefficient of resistivity = + ve

and for a semiconductor, = -ve

Q.11. We use alloys___________________

Ans. 11 (i) Alloys have low temperature coefficient of resistivity and high specific resistance. If

= low , the value of ‘R’ with temperature will not change much and specific resistance is high

then required length of the wire will be less.

Q.12. A constant Voltage___________________

Ans. 12. (iii)

2

2 2

2 2

1

2

2

2

2

l 2lR R

A (2r)

l 2lR = R

r 4r

V VH t & H t

R R

V = constant

H V R t

H R V t

R l 2 r =

R r l

H 2 =

H 1

H 2H

Correct

option is (iii)

Q.13. If the potential diff____________________

Ans.13. We know

deE

Vml

V =e

ml

If temperature is kept constant, relaxation time - will remain constant, and e, m are also constants.

d

d

V V

V 2V

Correct option is (ii)

Q.14. The equivalent resistance _________________

Ans. 14. (iii)

Redrawing the circuit, we get

B

3 8

30

6

1

1

2 1

2

2

2

3 & 6 are in parallel.

3 6 18 R 2

3 6 9

Now R and 8 in series

R = R + 8 = 2 + 8 = 10

Now R and 30 in parallel

R 30 10 30 Rep =

R 30 10 30

300 30 15 =

40 4 2

= 7.5 (iii) correct option

Q.15. The SI unit of magnetic field intensity is ______________

Ans.15. We know

1 1

FB

Il sin

NSI Unit of B = NA m

Am

Correct option is (ii)

Q16. The coil of _____________________

Ans. (iv) Correct Option

The coil of a moving coil galvanometer is wound over metallic frame to provide electromagnetic

damping so it becomes dead beat galvanometer.

Q.17. Two wires of__________________

Ans.17. Correct option (iii)

lA

lB

aa

Square r Circle

Area of a Square

1

2

2

2

a

Also here l = 4a

l a =

4

l Area =

16

l A =

16

Area of a Circle

2

2

2

2

r

Also here, 2 r l

l r =

2

l Now Area =

2

l A =

4

Now Magnetic moment = I A

M1 = IA, & M2 = I A2

Since I (current) is same in both

21 1

22 2

1 2

M A l 4 =

M A 16 l 4

M M = : 4

Correct option is (iii)

Q.18. The horizontal comp________________

Ans.18. Correct option (i)

Target law Bv = BH tan

V

H

H v

0

Btan =

B

Given B = 3 B

Bv 1tan =

3 Bv 3

= 30 or radians.6

Q.19. The small___________________

Ans. 19. Correct option is Magnetic declination or Angle of declination. It is the small angle

between geographic axis & magnetic axis.

Q.20. Two coils___________________

Ans.20. Correct option is (ii)

Mutual inductance of a pair of two coils depends on the relative position and orientation of two

coils, other statements are incorrect.

Q.21. A conducting__________________

Ans. 21. Correct option is (iv)

Current induced is I = lel

R

d

Now lel = dt

But there is no change of flux with time, as B, A & all remain constant with time.

No current is induced

Q22. The magnetic flux___________

Ans.22. 2

2

t=4

5t 3t 16

de

dt

d 5t 3t 16

dt

10t + 3

e = 10(4)+3 = 43V

e = -43Volts

Correct option is (ii)

Q23 Which of the following_______________

Ans.23. Correct option is (iii)

C

VI in Pure Capacitor

X

V = V 2 fc

1

2 fc

I f

other parameters kept cosntant

Q24. A 20 Volt AC________________

Ans.24. Correct option is (i)

VR VL

R

R

L

L

eff

eff

2 2R L

2 2 2 2eff eff

2 2L

2 2 2L

V = Effective Voltage across R

V I R

V = Effective Voltage across L

V I L

Net V = V V

= I R I L

20 = (12) V

(20) (12) V

2L

L

400 = 144 + V

V = 400 144 = 256 16 Volts

Q25. The instantaneous_________________

Ans. 25.

0

0

E = E sin t

I = I sin t + 3

Correct option is (iv)

as I can lead the Voltage in RC and LCR circuit, so it can be RC or LCR circuit.

(iv) is correct option.

fStraight line paragraph

Section - B

Q26.

r

l

Q27. Two Parallel______________________

Ans. 27. (iv) Correct option.

++

E

12

2

0 0

0 0

12

12

CSurface Charge density, = 26.4 10

m

E = + 2 2

2 = =

2

26.4 10 N =

8.85 10 C

N = 3

C

Correct option is (iv)

Q28. Consider__________________

Ans.28.

-Q

+Q

+

Equal and Opposite charges appear on the nearby conductor due to

induction, but still net charge on the conductor is zero. Correct option (iv)

Correct option is (i)

Since –ve electric flux

= + ve flux electric flux enclosed with a cylinder

here

Total Electric

Flux = 0.

Q29. Three Charges_________________

Ans.29.

r

-q E1-q

E2rE2 Gr

Q30. Two parallel______________

Ans.30.

k=2

12V

A = Samed = Same

Q = Same in Series

0 0x y

2 2

x y

x y

x y x

y x x

A 2 AC = C =

d d

Q QU = U =

2C 2C

U C 2C 2 = =

U C C 1

Correct Option is (iii)

Q31. Which among_________________

Ans.31. Correct statement is option (iv) as Primary coil made of Thick Coper wire has very less

R. Therefore negligible power loss. Rest all options are reasons for power losses in a transformer.

Q32. An alternating Voltage_______________

Ans.32.

B C

V

Net E F I at G O

Net Potential at G,

K2Q KQV =

r r

KQ -

r

= 0

Correct option is (iii)

c c1 1

X i.e. X2 fc c

I Brightness of the bulb will .

Correct option is (ii)

Q.33. A solid Sphere______________

Ans.33. Correct option is (4)

As all other statements seem incorrect in context with the given figure.

Q. 34. A battery is connected …..

Ans. Correct option is (iv).

Rest all quantities change with area of cross-section of a conductor.

Q. 35. Three resistors……

Ans. + –

R1 R2 R3

Given

I = 2 A, R2 = 3 Ω, P3 = 6 W

Power across R3 = V3I

6 W = I2 R3

6

4 = R3 =

3

2 = 1.5 Ω

V3 = I R3 = 2 (1.5) = 3 V

Correct option is (iii).

Q. 36. A straight line……

Ans. I = O, V = E, E = 5.6 V

r = 5.6

2.82.0

E

I

Correct option is (i).

Q. 37. A 10 m long potentiometer …..

Ans. Let PQ is a potentiometer wore of length 10m,

I = 5 5

480 20 500

E

R R

= 1

0.01 A100

VPQ = I RPQ = 0.01 × 20

= 0.2 V

If 10 m potentiometer wire balances 0.2 V

Then 1 m potentiometer wire balances 0.2

10 V

Then 6 m potentiometer wire balances 0.2

610

V

= 1.2

0.12 V10

Correct option is (iv).

Q. 38. The current sensitivity……

Ans. Given, gI = 20

100g gI I

= 120

1.2100

g gI I

R = 25 125

100 100 R R R

= 1.25 R

gV = ?

gV = 1.2

1.25

g gI I

R R

= 120 25

125 25g gV V

% change = 100

g g

g

V V

V

=

24

25100

g g

g

V V

V

= (24 25)

10025

= 1

10025

= 4%

Decrease by 4%. Correct option is (iv).

Q. 39. Three infinitely long parallel …..

Ans.

F1

r r

F2

2I I

F2

I

A B C

Let F1 is force per unit, length between A & C

1. . i e F = 0 2

4 2

I I

r

And F2 is force per unit, length between B & C

2F = 0

4

I I

r

Now net force on ‘C’ is per unit length

1 2F F = 2

(1 1)4

I

r

= 2

02

4

IF

r (given)

Now Fig.

B C

I

2I

F1

1

r r

F2

1F2

1

I

F1

1

b

1F = Repulsive force between A & C

= 2

0 2

4 2

I

r

2 2 = F F = A reactive force between B & C

Net force on ‘C’ 1 2 F F = 0

1F = 2

F = 22

4 2

I

r

Net Force on ‘C’ is zero.

Correct option is (i).

Q. 40. In a H-atom …….

Ans. R = 0.5 A°

= 10 rps = 10 × 2 rad/s

= 10 Hz

M = I A = e r2

= 1.6 × 10–19 × 10 × 3.14 × 0.5 × 0.5 × 10–10 × 10–10

= 1.256 × 10–38 Am2

Ans. (ii).

Q. 41. An air-cored solenoid …..

Ans. Magnetic field inside a solenoid

B = 0N

Il

'

Flux linked with ‘N’ turns

Initial flux 1 = N B A = 0N

N I Al

= 2

0N

I Al

= 7 44 10 800 800 2.5 2.5 10

0.30

= 16.74 × 10– 3 Wb

Final flux 2 = 0

Average back emf | e | = d

dt =

3

3

16.74 10 0

10

= 16.74 V

Correct option is (ii).

Q. 42. Vo = 283 V, f = 50 Hz

R = 3 Ω, L = 25.48 mH

C = 796 F

P |at resonance = ?

Power dissipated P = I2 R

I = 0 1 283

32 2

I

= 66.7 A

P = I2 R

= (66.7)2 3

= 13.35 kW

Correct option is (iii).

Q. 43. A circular loop …..

Ans. Let flux linked with smaller loop is 1 and with bigger loop is 2.

Fig.

1 R1

I1

2

Given R2 = 0.2 m

R1 = 0.003 m

x = 15 cm = 0.15 m

Now 1 = B2 A1

= 2

20 2 212 2 3/ 2

2

2

4 ( )

R IR

R x

M = 2 2

01 2 12 2 3/ 2

2 2

2

4 ( )

R R

I R x

Now 2 = M I1

= 2 2

0 2 112 2 3/ 2

2

2.

4 ( )

R RI

R x

= 9.1 × 10–11 Weber

Correct answer is (iv).

Q. 44. If both the no. of turns…..

Ans. L = 2

0N

Al

L = 2

0

(2 )

2

NA

l

= 2

02 2 N

A Ll

Correct answer is (ii). Doubted.

Q.45. Given below___________________

To increase the range

Ans. 45. Correct option is (iv) as both statements are false. To increase the range of an ammeter, suitable

low R (or shunt) should be connected in parallel to it. The ammeter with increased range has low

resistance.

Q.46. An electron____________________

Ans.46. Correct option is (iii)

Statements correct but reason is wrong because electrons move from a region of low potential to high

potential.

Q. 47. A magnetic needle ……..

Ans. The given statement is correct and reason is the correct explanation of the above statement. At

poles, magnetic needle orients itself vertically because horizontal components of earth’s field is zero

there. (correct option is (i))

Q. 48. A proton and an electron, …….

Ans. we know 𝑚𝑣2

𝑟= 𝐵𝑞𝑣𝑠𝑖𝑛 𝜃 = 𝐵𝑞𝑣 𝑆𝑖𝑛 𝜃

Centripetal force = magnetic Lorentz force

sin𝜃 = sin90𝜃 = 1 (∠ between & = 90°)

𝑚𝑣2

𝑟= 𝐵𝑞𝑣

𝑚𝑣

𝑟= 𝐵𝑞

𝑟 = 𝑚𝑣

𝐵𝑞=

𝑝

𝐵𝑞=

𝑙𝑖𝑛𝑒𝑎𝑟 𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚

𝐵𝑞

Since 𝑟 = 𝑝

𝐵𝑞

Given p, B are same

Also q for proton & electron is same except its sign

Radius is same. So statement is correct but

reason is not the correct explanation of the given assertion.

correct option is (ii)

Q. 49. On increasing…………..

Ans. 49. When we increase current sensitivity by increasing no. of turns, then resistance of coil also

increases. So increasing current sensitivity does not necessarily imply that voltage sensitivity will

increase because 𝑉𝑔 = 𝐼𝑔

𝑅

∴ if 𝐼𝑔 ↑ & 𝑅 ↑ by different amounts, then 𝑉𝑔 may increase or decrease.

Correct option is (i).

Q.50. A small object………

Ans. 50. Ans is (ii)

F𝑒 = 𝑚𝑔 𝑡𝑎𝑛𝜃

𝑞𝐸 = 𝑚𝑔 𝑡𝑎𝑛𝜃

𝑞 = (𝑚𝑔

𝐸) tan 𝜃

tan 𝜃 = 𝐹𝑒

𝑚𝑔

Tq

E

qE

Correct ans. is (ii)

Q. 51. A free electron…………

Ans. 51. Correct ans. (ii) i.e. II only

∵ 𝐹𝑝 = 𝐹𝑒 ∵ 𝐹 = 𝑞𝐸

𝐸 = 𝑠𝑎𝑚𝑒

‘𝑞’ = 𝑠𝑎𝑚𝑒

Now, 𝑃𝜀 = 𝑞 𝑉(𝑟)

(𝑃. 𝜀)𝑝 > (𝑃. 𝜀)𝑒

Q. 52. Correct ans is (iv) i.e. step down transformer decreases the ac voltage.

Q.53. correct ans is (i)

i.e. 𝑁𝑠

𝑁𝑝=

𝐸𝑠

𝐸𝑝

i.e. if no. of turns in secondary coil are more than no. of turns in primary, then voltage is increased or

stepped up in secondary, so called step up transformer.

Q.54 Correct ans. is (i).

i.e. current is reduced if voltage is stepped – up so corresponding 𝐼2𝑅 losses are cut down.

Q. 55. Correct ans is (iii)

Given 𝐸𝑖 = 2300𝑉

𝐸𝑜 = 230𝑉

𝑁𝑝 = 4000

𝑁𝑠 = ?

𝐸𝑖

𝐸𝑜=

𝑁𝑝

𝑁𝑠

2300

230=

4000

𝑥

𝑥 = 400 = 𝑁𝑠 = No of turns in secondary coil