class 12 electrostatics Electricity and magnetism made simple notes

7/30/2019 class 12 electrostatics Electricity and magnetism made simple notes

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UNIT I

ELECTROSTATICS (8)

CHAPTER 1 COULOBS LAW

Frictional electricity: The electricity developed on objects, when they are rubbed with each other is called

frictional electricity.

This happens due to actual transfer of electrons from one objects to the other. The object which gains

electrons becomes negatively charged while which loses electrons becomes positively charged.

Electric charge: The additional property of electrons and protons which gives rise to electric forces

between them is called electric charge.

The electric charge is a scalar quantity. Its S.I. unit is coulomb(C). A proton posseses a positive

charge +e and an electron a negative charge e, where e = 1.6 x 10 -19Coulomb.

Properties of charge:

1) Like charges repel each other and unlike charge attract each other.

2) The total charge on an object is the algebraic sum of all the charges located at different points on theobject.

3) Quantization of charge :- All charges are always an integral multiple of the charge on electron (e = 1.6 x

10 19 C)

If q is the total charge on the object q =

+

ne where n = 0, 1, 2, 3,

Coulombs law: The first precise measurement of the force between two electric charges was performed by

the French scientist Charles-Augustin de Coulomb in 1788. Coulomb concluded that:

It states that the electrostatic force between any two charges is directly proportional to the product of

the two charges and inversely proportional to the square of the distance between the charges.

i.e.1 2

2

0

1F

4

q q

r=

where F is the magnitude of the force, q 1 and q2 are the magnitudes of the two charges (with the appropriate

signs), and r is the distance between the two charges. Where 0 is permittivity of free space.The universal constant 0 = 8.854x10-12 N-1m-2 C2 is called the permittivity of free space or the

permittivity of the vacuum. We can also write Coulomb's law in the form

1 2

2F

q qk

r=

where the constant of proportionality0

14

k= takes the value k= 8.988x 10

9 Nm2 C-2

Relative permittivity (Dielectric constant): It is the ratio of the force between two charges placed at a

certain distance in air to the force between the same charges placed at the same distance apart in that

medium.

0

FK

F

vacr

med

or

= =

Coulombs law in vector form

1 212 212

0

1F

4

q qr

r=

r


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2

1 221 122

0

1F

4

q qr

r=

r

Two charges 1 2andq q are placed at a distance r apart.

The force on 1 2due toq q is given by,

1 2212

0

1F

4

q qr

r

=

r--------- (i)

The force on 2 1due toq q is given by,

1 221 212

0

1F

4

q qr

r=

r--------- (ii)

Equation (i) and (ii) are called coulombs law in vector form.

Principle of Superposition: It states that when a number of charges are interacting the total force on a given

charge is the vector sum of all the forces exerted on the given charge by all other charges.

If 12 13 1F ,F ,.... , F x+ are the forces on q1 due to q2, q3 .., qx then the total force on q1 is given by

1 12 13 1F F F F x= + + +L L

1 2 1 2 1 21 21 31 12 2 2

0

1 F ...

4x

q q q q q qr r r

r r r

= + + +

r

Linear charge density: It is defined as

L

q

=Where q is the total charge over the length L. Its S.I. unit is cm 1 or c/m.

Surface charge density: It is defined as A

q= where q is the total charge over the surface area A. Its S.I.

unit is cm 2 or c/m 2.

Volume charge density: It is defined as V

q= where q is the total charge over the volume. Its S.I. unit is

cm 3


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CHAPTER 2ELECTRIC FIELD

Electric field Intensity:

Electric field intensity at a point is defined as the force experience by a unit positive charge without

disturbing the source. It is denoted by E

0o

FE lim

qoq =

ur

Its S.I. unit is Newton per coulomb / 1Ne

Physical significance of Electric field:1) It helps to give the methods for finding force on a charge due to another charge in terms of electric

field.

2) It helps to determine the nature of interaction between charges kept at a distance apart.

Electric field due to point charge:

FE

oq

=

20

1

F 4

oq q

r=

2

0

2

0

1

4E

4

o

o

q q

r q

q r

= =

A point charge + q is kept at p at a distance r from point p where the electric field is to be measured.

According to coulombs law the force between the two charges is given by

2

0

1F

4

oq q

r=

--------- (i)

The electric field at point P is defined as

FEo

q=


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4

2

0

2 2

0 0

1

4F

4 4

o

o

q q

r q q

q r r

= = =

In vector rotation, where

2

0

E4

qr

r=

ur

Where r is unit vector along AP direction.

Electric lines of force: It is the path along which unit positive charge would move it is free to do so.

Properties of electric lines of force:

1) Electric lines of force starts from positive and ends with negative.

2) Two lines of force never intersect each other.

Q. Explain why two electric lines of force never intersect each other?

Ans. If two lines of force intersect each other, it would mean that the electric field has two directions at

that point. Due to this, electric lines of force never intersect each other.

Electric Dipole: Two equal and opposite charges placed at a small distance between them is called an

electric dipole.(or) A system of two equal and opposite charges separate by a certain distance is called an

electric dipole.

Electric dipole moment: It is defined as the product of either charge or a distance between them. It is

denoted by p, i.e. P 2a q= Its S.I. unit is Coulomb meter (Cm).

Electric field on axial line of electric dipole:

Consider an electric dipole consisting of charge + q and q, separated by a distance 2a. Let O be the

centre of the dipole and 2a be the length of the dipole. The electric field due to q at point + p is given by,

2

0

E4

A

q

AP=

OR( )

2

0

E4

A

q

r a=

+ --------- (i)

The electric field due to +q at point P is given by

( )22

0 0

E4 BP 4

B

q q

r a= =

The resultant electric field can be written as

( ) ( )

( ) ( ) ( )

B A 2 2

0 0

2 2 22 2

0 0

E E E

4 4

1 1 4.

4 4

q q

r a r a

q q q

r a r a r a

= =

+

= = +


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( )

( )2

2 2

0

2 2E

4

aq r

r a=

But 2aq = p, electric dipole moment

( )2

2 2

0

2E

4

pa

r a=

If dipole is of short length, a can be neglected

3

0

2E

4

p

r=

Electric field on equatorial line of electric dipole:

Consider an electric dipole consisting of charges q and + q whose length is 2a. Let P be the point

on equatorial line of the dipole at a distance r from O.

The electric field at point p due to charge q is given by

( )( )2 2 20 0

E or E4 4

A A

q q

AP r a= =

+

The electric field at point p due to charge + q is given by

( )2 2 20 0E (or) E

4 BP 4B B

q q

r a= =

+

i.e. EA = EBBy triangle law of vector addition we have,

E EA EB= =

AB AP BP

OrABE = EAAP


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( ) ( )

2 2 22 2

0

21/

4

q ac

r ar a=

+ +

( )2 20

2

34

2

aq

r a

= +

But 2aq = p, electric dipole moment

( )2 2 320E

4

P

r a=

+

If dipole is of short length, a can be neglected.

3

0

E4

P

r=

Torque on electric dipole placed in electric field:

Consider an electric dipole

consisting of charge q and + q whose

length is 2a. Let O be the angle between

the dipole and direction of electric field.Force on charge q at point A

= - qE

Force of charge + q at point

B = + q E

These two force are equal and different

line of action. So, they, form a couple

giving rise to a torque.

Torque, = force x distance.= E = BNq

= E 2 sinq a

= ( )2 Esinaq

Esinp =Where 2aq = P is electric dipole moment.

= P E

Electric field lines due to an isolated charge:

Electric field lines due to a system of two charge:

1) for electric dipole (q1 q2 < O) 2) same charges (q1 q2 > O)


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Q Draw the lines of force to represent the uniform electric field.

Ans: Here E represents the electric field.

Q What is the net force on electric dipole placed in uniform electric field?

Ans: It does not experience any net force.

Q When is the torque acting on a dipole maximum?

Ans: The torque is maximum when the electric dipole is placed perpendicular to the direction of electric

field.

Q When is the torque acting on a dipole minimum?

Ans: The torque is minimum when the electric dipole is placed parallel to the direction of electric field.

CHAPTER 3

GAUSS THEOREM

ELECTRIC FLUX: The electric flux through a small surface is define as the electric lines of force passing

through that area, when held perpendicular to the lines of force. It is denoted by .

E.ds =Where E the electric is field and ds is the area vector.

Its S.I. unit is Nm2 / C.

GAUSS THEOREM: Any close surface around a charge distribution so that the gauss theorem can

be conveniently applied to find electric field due to it is called Gaussian surface.

Q Write the importance of Gaussian surface.

Ans: To find electric field due to a charge distribution using gauss theorem, we have to evaluate the

surface integral. This surface integral can be evaluated easily by choosing a suitable surface.

Electric field due to linear charge :

Consider an infinitely long linear charge having linear charge density. To find electric field at point p

at a distance r from O, we draw a Gaussian surface of radius r.

According to gauss Theorem.

0

E.q

ds =

ur uur

0

. cosq

ds O =

We have , LL

qq= =

And 2 Lds r=


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8

0

LE 2 Lr =

0

E

2 r=

Electric field due to an infinite plane sheet of charge :Consider an infinite in plane sheet of charges having surface charge density on both sides. To find

electric field, we draw Gaussian surface (cylinder) having cross sectional area A.

According to gauss theorem

0

0

0

E. S

E S cos

E

qd

qd

qds

=

=

=

rr

But 2ds A=And

q

A=

Aq =

0

0

AE2A

AE

2

=

=

If the plane sheet has finite thickness

Then,0

0

q

2A

2AE2A

E

=

=

=

Electric field due to uniformly charge spherical shell:

Consider a thin spherical shell of radius R centre O. Let +q be the charge of the spherical shell(1) When point P lies outside the spherical shell

Consider small area ds at point p at a distance r from centre of the spherical shell.

According to Gausss Theorem

0

0

E . S

E cos0

qd

qds

=

=

ur r

0

Eq

ds =

But,24ds r=


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And, surface charge density,

2

4R

q=

24Rq =

22 4RE 4

o

r =

2

2

E

Ro

r

=

(2) When P lies inside the spherical shell

Let P be point inside the spherical shell at a distance r from O.

According to gauss theorem

0

E.q

ds =

ur uur

But, the Gaussian surface does not enclose any charge.i.e. q = 0

E. 0ds =E = O

i.e. electric field inside a spherical shell is 0.

CHAPTER 4

ELECTRIC POTENTIAL

Q Show that the work done in moving a unit charge along a close path is). (3m)

OR

Show that the integral of electric field.

A point charge +q is located at origin and let AP1, BP2, A be a closed path.

Then represents the work done in moving a unit positive charge from point A to B.

Then,

B

A BA

1 1 1E.

4dl q

o r r

=

uur r

(along P1)

and

B

B AA

1 1 1E.

4dl q

o r r

=

uur r

(along P2

1 1 1 1 1 1E. E.

4 4

B A

A B B AA B

dl d s q qo r r o r r

+ = +

ur r ur ur


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11

( )W F. E E cos180

E ( 1) E

W E

o o

o o

o

d dl q dl q dl

q dl q dl

d q dr

dl dr

= = =

= = = =

We have, 2E 4 o

q

r=

0

0 2

0

W E4

q qd q dr dr

r

= =

Hence, work done in moving test charge from infinity to point A is given by

A A

A

20 0A 2

0 0

0 0

0 A 0 A

W W4 4

1 1 1

4 4

r rA

r

q q q qd dr r dr

r

q q q q

r r

= = =

= =

or0

A

0 A

W4

q qr

=

0AA

0 A0

0

WV

4

q q

rq

q

= = = A

0 A

V4

q

r=

ELECTRIC POTENTIAL AT ANY POINT DUE TO AN ELECTRIC DIPOLE:

Consider an electric dipole consisting of charge q and + q whose length is 2a. Let P be the point at

a distance R from O such that POB = The potential at point P due to q is given by,

A

0

V --------- (i)4 AP

q=

The potential at point P due to + q is given by,

B

0

V --------- (ii)4 BP

q=

To find AP & BP draw BN || OB and AN || OP such that

AP = OP + ON= r + a COS

And BP = OP ON

= r a COS

Now, the net potential at point P is given by


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V = VA + VB

0

0

2 2 2

2 2 2

0

1 1

4 BP AP

1 1

4 cos cos

cos coscos

2 cos

4 cos

q

q

r a r a

r a r ar a

q a

r a

=

= +

+ +=

=

But, 2 aq = P, electric dipole moment.

( )2 2 20

PcosV

4 cosr a

=

Cases

1. When p lies on axial line of dipole ( = O)

( )2 2P

V4

or a

=

2. When P lies on equitorial line of electric dipole ( )90 =

V = O

Q Show that electric field at a point is equal to negative gradient of electrostatic potential at that point.

VCE

d

dr

= Ans: Consider a point charge +q at O. Let V and V + dV be the potential at point P and Q at a distance r

and r dr.

If dW is small work done in moving test charge qo through distance dl

( )

( )

WV V V

WV i

o

o

dd

q

dd

q

+ =

=

Now,

W F .

E cos 180

W E

o

o

d dl

q dl

d q dl

=

= =

But,

W Eo

dl rd

d q dr

= =

Putting this value in equation (i)

EV

VE

o

o

q drd

q

d

dr

=

=

Q Define equipotential surface.

Ans: Any surface which has the same electrostatic potential at every point is called equipotential surface.

Q Show that no work is done in moving a test charge over equipotentia surface.


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Ans: Let point A and B be two points lying on equipotential surface. The potential difference is given by

WVB VA AB

oq

=

But VA = VB (In equipotential surface)

AB

WO

oq

=

= WAB = O

i.e. no work is done in moving a test charge over equipotential surface.

EQUIPOTENTIAL SURFACE FOR UNIFORM ELECTRIC FIELD

EQUIPOTENTIAL SURFACE FOR ISOLATED POINT CHARGE


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EQUIPOTENTIAL SURFACE FOR SYSTEM OF TWO CHARGES

POTENTIAL ENERGY OF AN ELECTRIC DIPOLOE PLACED IN UNIFORM ELECTRIC

FIELD

The torque experienced by an electric dipole (having dipole moment p) placed inside uniform

electric field E is given by

E sinp =If the dipole is rotated through small angle d against the torque.

Then, small work done is,

W

W E sin

d d

d p d

= =

Thus, work done in rotating the dipole from its orientations making an angle 1 to 2 is given by

[ ]

[ ]

2 2 2

2

1

1 1 1

2 1

W W E sin E sin E cos

W E . cos

d p d p d p

p cos

= = = =

=

This work done is stored as potential energy (U) of the dipole.


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( )

( )

2 1

1 2

PE cos cos

if 90 and

PE cos cos90

PE cos

In vector notation

P.E

=

= =

=

=

=

o

o

ur ur

CHAPTER 5

POLAR AND NON-POLAR DIELECTRICS

Dielectric in the atoms or molecules of which the centres of gravity of positive and negative charges

coincide is called non-polar dielectric.

A dielectric in the atoms or molecules of which the centres of gravity of +ve and ve charges do not

coincide is called polar dielectrics.

Q. Define polarization.

Ans: The stretching of dielectric atoms due to displacement of charges in the atoms under the action of

applied electric field is called polarization.

Polarization density:

Then, P NP

Let A = Area of capacitor, D = distance between plates of capacitor. Volume of dielectric slab = Ad-q1 and +q2 are developed on two face, the total dipole moment is

They induce dipole moment develop per unit

volume in a dielectric slab on placing it inside the electric

field is called polarization density. It is denoted by P.If P is

induced dipole moment acquired by atoms of dielectric and

N is the number of atoms per unit volume.


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P

P

dP

Ad A

, polarization surface charge density.A

P

i i

i

q q

q

= =

=

=

The reduce value of electric field is given by

P P P0 0

0 0 0 0 0

PE E E= = = =

We know that 0P E= When called electric susceptibility of the dielectric

00 0

0

0

EE E E E

E E E

= =

=

0 E E E+ =

( ) 0E

E 1 E+ =

0E

But k, dielectric constantE

k 1

=

= +i.e the relation between dielectric constant and electric susceptibility of the dielectric.

Dielectric strand: The dielectric strand of a dielectric is defined as the maximum value of electric field that

can be applied to the dielectric without its electric breakdown.

Its SI unit is volt per metre.

CAPACITANCE: Capacitance of a conductor may be defined as the ratio of the electric charge on it to its

electric potential due to that charge. It is denoted by C.

CV

q=

Where V is the electric potential due to the charge. Its SI unit is Farad (F)

PRINCIPLE OF CAPACTOR

A capacitor is a device for storing a large quantity of charge.

An insulated metallic plate A is given +ve charge till its potential becomes maximum. If another

metallic plate B is placed near plate A1 ve charge will be induced on nearer face of plate B and +ve charge

on farther face as shown in Fig (ii). The potential of plate A gets lower due to induced charge and to make

the potential of plate A the same, additional charge has to be given, i.e. its capacitance increases.


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If A plate B is connected to earth as shown in Fig (iii), +ve charge on plate B flows to the earth and

-ve charge remains on the plate due to attraction from plate A.

Thus, its potential is lower and a large amount of charge has to be given to make the potential the

same, i.e. its capacitance is greatly increased. This is the principle of capacitor.

CAPACITANCE OF PARALLEL PLATE CAPACITOR

If +ve charge (+q) is given on plate A, - ve charge ( q) is induced on plate B. The electric field is setup between the two plates which is given by

VE (in magnitude)

d

dr=

Since the electric field is uniform, we can write

VE

d=

Where V is the potential between the two plates.

( )V Ed --------- i =If is surface charge density, then,

0 0

A

also, E

A

q

q

=

= =


0

dV

A

q=

We know that capacitance, CV

q=

0

Cd

A

qq=

0A

Cd

=

ENERGY STORED IN A CAPACITOR

Consider a capacitor of capacitance C, which does not contain any charge initially. If it is connected

to a battery its potential rises to V and if q is the amount of charge on the capacitor then,

( )

C V

CV ---------- i

q

q

= =


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If the battery supplies infinite simal charge dq to constant potential V. Then through the small

amount of work done is given by CV

q=

W VC

qd dq dq= =

Then, the total work done in delivering charge q is given by

22

0 0 0 0

2

1 1 1 1W W 0

C C C 2 2 C

1W

2 C

qq q qq qd dq qdq q

q

= = = = =

=

This work done is stored as electric potential energy (u)

[ ]

2

2

1U (ii)

2 C

1V C

2 V

or,

1CV CV

2

q

qq

u q

=

= =

= =

Q

Q

We know that,

0A

Cd

=

Where A is area of the plate and d is distance between the two plates.

An electric field0

E =

Where is surface charge density

Now,

0

A

A

EA

q

q

=

==

Putting this value in eqn. (ii)

2 2 2

0

0

2

0

E A1

A2

1U E A

2

m

dd

=

=

Where Ad = Volume of capacitor

Energy store per unit volume2

0

1E

2=

CAPACITANCE OF PARALLEL PLATE CAPACITOR WHEN DIELECTRIC IS INSERTED

BETWEEN THE TWO PLATES:

A parallel plate capacitor having plates of area A separated by a distance d is placed in vacuum and

its capacitance is given by,


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0V E E (d )t t= +

Since, 0E

Ek=

0

00

0

EE

EV E (d )

V E (d )

k

t tk

tt

k

=

= +

= +

00 0

E

A

q= =

But,0

V (d )A

q tt

k= +

Capacitance,

0

0

A

C V (d ) d (1 )A

q q

q t tt tk k

= = = +

When dielectrics field the whole space between the two plates

i.e t = d

0 0 0

A A AC

d dd d(1 ) d d

k

t

k k

= = =

0C Ck =

CAPACITANCE OF PARALLEL PLATE CAPACITOR WHEN CONDUCTING SLAB ISINSERTED BETWEEN TWO PLATES:

A parallel plate capacitor having plates of area A and separated by a distance d is placed in vacuum

and its capacitance is given by

00

A

dC

=

When a dielectrics of dielectric constant k (kappa) is

insulted between the two plates, the electric field reduces due

to polarization of dielectrics. If V is the potential difference

between the two plates then,


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When conducting slab fills the whole space between the two plates,

t = d

0 0A A

C(d d) 0

C

= =

=

VAN-DE GRAFF GENERATOR

PRINCIPLE:

1) Electric discharge takes place in air or gases readily at pointed conductor.

2) A hollow conductor continues excepting charge through its inner surface irrespective of how large its

potential may grow.

CONSTRUCTION AND WORKING:

It consists of a large hollow metallic sphere placed on two insulating column. The metal comb (C1) is

kept at high potential and the collecting comb (C 2) is kept in contact with the inner sphere which is

connected by two pulleys.

Positive ions are sprayed on the belt due to repulsive action of C 1, the comb C2 collects these ions

and transfer them to the metallic sphere. As the belt goes moving the positive ions gets accumulated.

00

A

dC

=

When conducting slab is insulted between the two

plates, the potential difference between the two plates reduces

and since the electric field inside the conducting slab is 0.

0

0

0

0

V E (d )

But, E

A

V (d )A

t

q

qt

=

=

=

=

Capacitance CV

q=

0

(d )A

q

qt

=

0 A(d )t=


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Electric discharge also takes place from the sphere. When the maximum potential is reach the charge leaksto the surrounding air inside the conductor and rate of laws of charge is equal to the rate of transfer of charge

to the sphere to prevent this leakage, the generator is completely enclosed in each connected steal tank if

projectile such as proton, deuteron etc. are generated in the discharge tube D. With lower and earth upper

and inside the sphere. They accelerated in downward direction to hit a target with large kinetic energy to

bring about nuclear disintegration.

CAPACITOR IN SERIES:

1 2 3

1 1 1 1

C C C C= + +

CAPACITOR IN SERIES:

1 2 3C C C C= + +UNIT II CHAPTER 1

ELECTRIC CURRENT RESISTANCE AND E.M.F.

ELECTRIC CURRENT: It is defined as the rate of flow of charge through a conductor. It is denoted by

I. Iq

t=

Where q is the amount of charge flowing through a conductor in time t. It is a scalar quantity. It SI unit is

ampere.

Q Define one ampere.OR

Define the SI unit of electric current.

Ans: The current through a conductor is called one ampere. If 1 coulomb of charge flows through the

conductor in 1 sec.


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DRIFT VELOCITY

When a potential difference V is applied to a conductor of length l. The electric field set up is given

by

VEr

VE

dd

l

=

=

The force experienced by the electrons is given by

F Ee= The acceleration experienced by the electrons due to this force can be written as,

E

m

ea

=

r

This acceleration of electron takes placed only for a short time due to deflection in their collision

with +ve ions in the conductor and the time for which they accelerate is called relaxation time.

It an electron having thermal velocity u, accelerates for time t, and it will attain velocity.

1 1 1v u a= +

Velocity acquired by other electrons is

2 2 2v u a= +

And 3 3 3v u a= +The drift velocity can be written as

1 1........

nd

v v v

v n

+ + +=

r

1 1 2 2( ) ( ) ........ ( )

n nu a u a u a

n

+ + + + + +=

1 2 1 2........ ........

n nd

u u uv a

n n

+ + + + + += +

r r

Where 1 2 ........ nu u u+ + + is called average thermal velocity which is equal to zero. And 1 2........

n

n

+ + +

is called average relaxation time ().

Therefore, 0dv a= +

E

d

ev

m

=

uur

Q Define drift velocity.

Ans: Drift velocity is defined as the average velocity with which the free electrons in the conductor gets

drifted under the influence of the applied electric field.

RELATION BETWEEN DRIFT VELOCITY AND ELECTRIC CURRENT:

Consider a conductor length l and cross sectional area a. If n is the number of electrons per unit

volume, then the total charge on the conductor is,Aq n le=If dv is the drift velocity of the free electrons. Then,


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23

d

d

lv

t

l

t v

=

=

The current flowing through the conductor can be written as,

AI

I A

d

d

q n le

t l v

n ev

= =

=

2

We know that,

E

AI E

d

ev

m

n e

m

=

=

ELECTROMOBILITY:

The mobility of electrons in a conductor is defined as the drift velocity acquired per unit applied

electric field. It is denoted by

E

dv

=

Its SI unit is m2 V 1 S 1 .

OHMS LAW:

It states that physical conditions remaining unchanged, the current flowing through a conductor is

directly proportional to the potential difference across its two ends.

Mathematically

V I

V RI ,

=

Where, R is the proportionality constant known as resistance of the conductor.

RESISTANCE:

Resistance of a conductor is defining as the ratio of potential difference across the conductor to the

current flowing through it. It is the opposition offered by the conductor to the flow of current.

V

RI

=

It SI unit is ohm ( )

RESISTIVITY (SPECIFIC RESISTANCE):

The resistively of the material of a conductor is define as the resistance offered by this material of

unit length and unit area. It is denoted by .

A R

l=

It SI unit is ohm metre.( m)

FACTORS AFFECTING ELECTRICLA RESISTIVITY:

Consider a conductor having length (l) and cross sectional area A. If electric field E is applied across

the conductor. Then the drift velocity is


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24

E(i)

d

ev

m=

And the current flowing through the conductor is

I A dn v e=

2A E

In e

m

=

We know that,

V

El

=

2

2

A VI

V(ii)

I A

n e

ml

ml

ne

=

=

According to ohms law,

V

R

I

=

2

R . (iii)A

m l

ne =

If is the resistively of the conductor,

R (iv)A

l=

Comparing equation (iii) and (iv)

2

m

ne

=

CONDUCTANCE:

The reciprocal of the resistance of a conductor is called each conductance. It is denoted by G.

1G

R=

Its SI unit is ohm 1 1( )

CONDUCTIVITY:

The reciprocal of the resistively of a conductor is called each conductivity. It is denoted by

Its SI unit is ohm 1 m 11 1( m )

1

=

ELECTRICAL ENERGY:

The total work done by the source of e.m.f. in maintaining the electric current in the circuit for a

given time is called electrical energy consume in the circuit.

W = V I t

Its SI unit is Joule (J).

Electric Power: The rate at which work is done by the source of e.m.f. in maintaining the electric current in

the circuit is called electric power.2

2 VP VI I R R

= = =

Its SI unit is Watt (W).


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25

Q Define 1 kilowatt hour on board of trade unit ( 0B )Ans: The energy consume is called 1 kilowatt hour. It a device of electric power of 1 kilowatt is used for 1

hour. (1 kwh = 3.6 x 106J)

COLOOUR CODE FOR CARBON RESISTOR :

B B R O YGreat Britain Very Good Wife

Latter as an aid

to memoryColour Figure Multiplier Colour Tolerance

B Black 0 010 Gold 5%

B Brown 1 110 Silver 10%

R Red 2 210 No colour 20%

O Orange 3 310

Y Yellow 4 410

G Green 5 510

B Blue 6 610

V Violet 7 710

G Grey 8 810

W White 9 910

Q A carbon resistor has three colors Blue, Yellow and Red respectively. What will be the resistance?

Ans: 264 10 20% Q A carbon resistor is marked in colour bands of red, black, orange and silver. What is the reistance

and tolerance value of the resistor ?

Ans: 320 10 10% Q Draw the colour code scheme of 342 10 10% + carbon resistance.

Ans:

34.5 10 10% +

Q Draw 34.5 10 k 5%

Ans:

3

2

4.5 10 k 5%

45 10 k 5%

=

=

RESISTORS IN SERIES:

Three resistor R1, R2 and R3 are connected in series to a battery of e.m.f. E. Let V 1, V2 and V3 be the

potential drop across the three resistors.

Such that E = V1 + V2 + V3 ---------- (i)

According to ohms law

V1 = IR1, V2 = IR2, V3 = IR3Therefore, E = IR1 + IR2 + IR3

E = I (R1 + R2 + R3)

1 2 3

E= R + R + R ---------- (ii)

I

If R is the equivalent resistance of the combination. Then,


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26

E = IR

E

= RI

Putting this value in equation (ii)

R = R1 + R2 + R3.

RESISTORS IN PARALLEL:Three resistors R1, R2 and R3 are connected in parallel to a battery of e.m.f.. Let I1, I2 and I3 be the

current flowing through its resistors.

Such that I = I1 + I2 + I3 ----------- (i)

According to ohms law,

1 2 3

1 2 3

E E EI , I , I

R R R= = =


1 2 3

E E EI

R R R= + +

1 2 3

I 1 1 1 ---------- (ii)E R R R

= + +

If R is the equivalent resistance of the combination,

E I R

I 1

E R

=

=

Putting this value in equation (ii)

1 2 3

I 1 1 1

R R R R = + +

Q Define internal resistance of a cell.

Ans: The resistance offered by the electrolytes of the cell when electric current flows through it is known

as its internal resistance. Its SI unit is ohm. It is denoted by r.

Q Define electromotive force (e.m.f.).

Ans: The potential difference between the two poles of the cell when no current is drawn from the cell

(open circuit) is called electromotive force of the cell.

Q Define terminal potential difference.

Ans: The potential difference between the two poles of the cell when current is drawn from the cell(closed circuit) is called terminal potential difference.

EXPRESSION FOR INTERNAL RESISTANCE OF A CELL:

A source of E having internal resistance r is connected to an external resistance R. If I is the current

flowing through the circuit, then

EI

R r=

+---------- (i)

The terminal potential difference is equal to the potential drop across resistor R.

V = IR ---------- (ii)

From equation (i) and (ii)


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27

EV R

R r=

+

E

R RV

r+ =

E

1 RV

r =

CELLS IN SERIES:

1) When the cells are of the same e.m.f. and internal resistance. A combination of n cells having

e.m.f. E and internal resistance r. Each is connected in series to an external resistance R.Total e.m.f. of the cell = n E

Total internal resistance of the cell = nr

If I is the current flowing through the circuit, then

total e.m.fI

total resistance=

NI

R

e

nr=

+a) If R > nr, i.e R + nr R

E

I times the currenr due to a single cell.R

n

n = =2) When the cells are of different e.m.f. and external resistance.

Two cells having internal resistance r1 and r2 and e.m.f. E1 and E2 are connected in series to an

external resistance R.

The terminal potential difference due to the first cell is,

V1 = E1, - I r1


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28

For the second cell,

V2 = E2 I r2The potential difference between two points A and B is given by

V = V1 + V2V1 = E1 I r1 + E2 I r2V = (E1 + E2) I (r1 + r2) ---------- (i)

If E and r are the equivalent e.m.f and internal resistance. Then,

V = E I r ---------- (ii)

Comparing equation (i) and (ii)

E = E1 + E2 and

R = r1 + r2

total e.m.f

Itotal resistance

=

E

R r=

+1 2

1 2

E E

R r r

+=

+ +

CELLS IN PARALLEL:

1) When the cells are of same e.m.f and total internal resistance.

A combination of n cells having e.m.f E and internal resistance r, each, is connected to an external

resistance R. If r1 is the combined internal resistance of the cell

then,

1 1 1 1......... times

nr r r r

n

r

rr

n

= + + +

=

=

Total e.m.f of the cell = E

Total resistance = R + r

If I is the current flowing through the circuit.

total e.m.fI

total resistance

E

R E

R

r

r n

=

=

+=

+a) If R , i.e R R r n r n+ ?

E

I currenr due to a single cell.R

= =

b) If R , i.e R r n r n r n+ =E E

I times the currenr due to a single cell.n

nr n r

= =

2) When the cell are of different e.m.f and total external resistance.


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29

Two cells having e.m.f E1 and E2 and internal resistance R1 and R2 are connected in parallel to an

external resistance R. Let I1 and I2 be the current from the two cells such that,

I = I1 + I2 ----------- (i)

The potential difference between the two points A and B is given by

V = E1 I1 r1I1 r1 = E1 V

1

1

1

E V

I r

=

Similarly,

22

2

E VI

r

=


1 2

1 2

1

1

E V E VI

E VI

r r

r

= +

=

1 2 2 2 1 1

1 2

1 2

E V E VI

r r r r r r

r r

+ =

( )( )

( )1 2 2 11 2 1 21 2

E EV

r rr r I r r

r r

++ =

+

( ) ( )1 2 2 1 1 2

1 2 1 2

E EV

r r I r r

r r r r

+=

+ +----------(ii)

If E is the total e.m.f and r is the total internal resistance. Then

V = E Ir ---------- (iii)

From equation (ii) and (iii)

( )1 2 2 1

1 2

2 1

1 2

E EE

r r

r r

r rr

r r

+ = +

= +

---------- (iv)

total e.m.f

Itotal resistance

=

E

R r=

+Where the values of E and r are given by equation (iv)

CHAPTER 2


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30

ELECTRICAL MEASUREMENT

KIRCHOFFS LAW:

First Law: It states that the algebraic sum of a current meeting at a point in an electrical circuit is always

zero. It is also known as function rule.

Second Law: It states that in any close part of an electrical circuit, the algebraic sum of e.m.f is equal to the

algebraic sum of the product of resistance and current flowing through them.

PRINCIPLE OF WHEATSTONE BRIDGE:

Four rsistance PQR and X are connected across the arms of a Wheatstone Bridge. Let I be the

current flowing through the circuit.

Applying Kirchhoffs second law on the close loop ABDA,

- I1 P + I2 R I3 G = 0 ---------- (i)

In the close loop BCDB,

- (I1 I3) Q + (I2 + I3) X + I3 G = 0 ---------- (ii)

If points B and D are at the same potential.

I3 = 0

And the Wheatstone bridge is said to be balanced.-I1 P + I2 R = 0

I1 P = I2 R ---------- (iii)

and I1 Q + I2 X = 0

I1 Q = I2 X ---------- (iv)

Dividing equation (iii) and (iv)

1 2

1 2

I P I R

I Q I X=

P R

Q X=

This is the condition for balanced Wheatstone bridge.


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31

SLIDE WIRE BRIDGE (METRE BRIDGE)

Metre Bridge is a device for measuring unknown

resistance.

Principle: It is based on the principle of Wheatstone

bridge i.e. when the bridge is balanced.

P R

Q X=

Where P and Q are called ratio arms. R is variable resistance and X is unknown resistance.

Construction: It consists of a wooden board over which constantan wire AC (100cm length) and a

metre scale are fitted. A copper strip is fitted on the board to provide two gaps for connecting resistance box

R and unknown, resistance X. A galvanometer is connected to terminal D and a jockey.

Applications:

1) To measure an unknown resistance:When the bridge is balanced . Let AB = l and BC = 100 l. The resistance of the wire between AB is

taken as P and between B and C is taken as Q. Then,

P

Q 100

P(i)

Q 100

l

l

l

l

=

From the principle of Wheatstone bridge.

P R

Q X=

From equation (i) and (ii)

R

X 100

l

l=

Knowing the values of R and l, the unknown resistance X can be determine.

2) To compare two unknown resistance:

First the resistance R1 is connected in place of X and the balance point is obtained by moving the

jockey over the wire.

Let AB = l1 and BC = 100 l1. The resistance of the wire between AB is taken as P and between BC

is taken as Q. Such that

And

1

1

1

1

P

Q 100

P

Q 100

l

l

l

l

=

From the principle of Wheatstone bridge

1

1

1 1

11

1

P R

Q R

R

R 100

R R (i)100

l

l

l

l

=

=

=


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32

The experiment is repeated with R2 replacing R1 and let AB = l2 and BC = 100 l2

1

2

2

R R (ii)100

l

l

=

Dividing equation (i) by equation (ii) we get

1 1 2

2 2 1

R 100

R 100

l l

l l

=

Knowing the values of l1 and l2 the ratio 1 2R R can be determine.

POTENTIOMETRE:

Potentiometer is a device for measuring internal resistance of the cell or comparing emfs of two

cells.

Principle: When a constant current is passed through a wire of uniform area of cross section, the

potential drop across any portion of the wire is directly proportional to the length of that portion.Construction: It consists of a number of segments of wire (manganin or constantan) of uniform

cross section (each 100cm long) stretched on a wooden board. A metre rod is fixed parallel to its length as

shown in figure..

A constant current is maintained through the potentiometer wire with the help of the rheostat

Theory: Let V be the potential difference across certain portion of wire of resistance R Then,

V = IR

Since R

A

l=

V IA

l =

Where is the resistivity of the wire.

If a constant current is passed through the wire. Then,V (constant )

i.e., V

l

l

=

i.e. potential drop along the wire is directly

proportional to the length of that portion.

Applications:

1) To measure e.m.f of the cell:

The circuit connection is shown in the figure.

The potentiometer wire is first calibrated by using

a standard cadmium cell of e.m.f 1.018 V. The jockey


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33

is put at 203.6cm mark and current is adjusted by using rheostat so that galvanometer shows no deflection.

The potential gradient along the potentiometer wire is given by

3 11.018V 5 10 V203.6

dvcm

dl cm

= =

The cadmium cell is now replaced by the unknown cell of e.m.f ,say E. Keeping the current through

the potentiometer wire constant, the balancing length l is found. Then the e.m.f of the unknown cell is then

given by

3

V

5 10 V

dE l

dl

l

=

=

2) To measure internal resistance of a cell:

The circuit connection is shown in the figure. A constant current I is maintain through the

potentiometer wire by using Rheostat. The key K2 is kept out and the balancing length l1 is obtained. Then,

Let x be the resistance per unit length of the wire.

1E ( )I ---------- (i)xl=

After introducing the resistance , say S, the key K2 is

inserted and the balancing length l2 is obtained to find

terminal potential difference,

2V ( )I ---------- (ii)xl=

Dividing equation (i) by (ii)

1

2

E

V

l

l=

The internal resistance of a cell is given by

1

2

1 2

2

E1 S

V

, 1 S

S

r

lor r

l

l lr

l

=

=

=

Knowing the values of l1 and l2, the internal resistance of a cell can be measure.

3) To compare e.m.f of two cells:

The circuit connection is shown in the figure. A constant current I is maintained through the

potentiometer wire between points A and B by using Rheostat.

When terminals 1 and 2 of two way key are connected, let l be the balancing length between point A

and jockey J. If x is the resistance per unit length of the wire, then,

1 1E ( )I ---------- (i)xl=

If terminals 2 and 3 are connected (after removing

it from 1 and 3), let the balancing length be l2. Then

2 2E ( )I ---------- (ii)xl=

Dividing equation (i) by (ii), we get1 1

2 2

E

E

l

l=


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34

Knowing the values of l1 and l2, the ratio of the emfs of two cells can be found.

UNIT V

CHAPTER I

ELECTROMAGNETIC WAVES

ELECTROMAGNETIC WAVE:

It consists of sinusoidally time varying electric and magnetic field acting at right angle to each other

and at right angle to the direction of the propagation of the waves.

ELECTROMAGNETIC SPECTRUM:

The orderly distribution of electromagnetic waves (according to wavelength or frequency) in theform of distinct groups having widely differing properties is known as electromagnetic spectrum.

It is divided into the following main parts:

1) Radio waves 2) Microwaves 3) Infra-red

4) Visible light 5) Ultra-violet rays 6) X-rays 7) - rays

The speed of electromagnetic waves in free space is given by

0 0

1c

=

Where 6 -10 ( 1.257 10 T A )m= and 12 2 1 20 ( 8.854 10 C N )m = are respectively the absolutepermeability and absolute permittivity of the free space.

The speed of electromagnetic waves in a medium is given by


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35

1

v =

Where is the absolute permeability and is the absolute permittivity of that medium.

RADIO WAVES 3 11(3 10 - 3 10 Hz)Properties: 1) Radio waves obey laws of reflection and refraction.

2) Radio waves are electromagnetic waves and travels with a speed of 8 -13 10 ms .Uses: 1) Radio waves are used in radio astronomy.

2) They are used to transmit radio and T.V. signals.

Microwaves: 8 11(3 10 - 3 10 Hz)Properties: 1) Microwaves are electromagnetic waves and travels with a speed of 8 -13 10 ms

2) Microwaves produced heat, when absorbed by matter.

Uses: 1) Microwaves are used in radar system.

2) Microwaves are used in long distance telephone communication systems.

Infra-red ray: 11 14(3 10 - 4 10 Hz)Properties: 1) Infra-red rays obey laws of reflection and refraction.

2) Infra-red rays effect a photographic plate.

Uses: 1) Infra-red rays are used in solar water heaters and cookers.

2) Infra-red rays are used for producing dehydrated fruits.

Visible Light: The frequency of visible light ranges from 14 144 10 - 8 10 Hz

Ultra-violet rays: 14 16(8 10 - 8 10 Hz)Properties: 1) They obey the laws of reflection and refraction.

2) They can effect a photographic plate.Uses: 1) Ultra violet rays are used for checking the mineral samples by making use of its property

of causing fluorescence.

2) As ultra violet rays can cause photoelectric effect, they are used in burglars alarm.

X-rays: 16 19(10 - 3 10 Hz)Properties: 1) X-rays are electromagnetic waves of very short wavelength ranging from 0.01A 10Ao o

2) They affect the photographic plate very intensely.

Uses: 1) X-rays are used in detecting charge in old oil paintings.

2) X-rays are used in surgery for the detection of fractures, diseased organs and foreign

bullets.

Gamma rays: 19 20(3 10 - 5 10 Hz)Properties: 1) -rays are electromagnetic waves and have velocity equal to that of light.

2) -rays are not deflected by electric and magnetic fields.

Uses: 1) -rays are used in radiotherapy.

2) -rays are used to produce nuclear reactions.


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UNIT III

CHAPTER 1

MAGNETIC EFFECTS OF CURRENT

BIOT-SAVARTS LAW:

According to Biot Savarts law the magnetic field (in magnitude) at a point at a distance r from the

current element of length el carrying a current I is given by

2

I sinB

4

dld

r

=

Where is the angle between direction of the current and the line joining the current element to the

observation points and is the absolute permeability of free space.

RIGHT HAND THUMB RULE:If we grasp the conductor in the palm of right hand so that the thumb points in the direction of flow

of current, then the direction in which the fingers and represents the direction of magnetic field lines.

MAGNETIC FIELD AT THE CENTRE OF THE CIRCULAR LOOP:

Consider a circular loop of radius r carrying a current I whose centre is O. To find magnetic field at

the centre of the loop, consider a small element AB = dl.

According to Biot-Savarts Law, the magnetic field due to the whole circular loop is given by

0

2

I sinB

4

dld

r

=

Since = 90, sin 90 = 1

02

I sinB

4

dld

a

=

The magnetic field due to the whole circular loop is given by

0 02 2

I IB B

4 4

dld dl

a a= = =

But 2 , the circumference of the loopdl a= 0

2IB

4 a = If the loop has n number of turns, then

0 2 I

B4

n

a=


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37

MAGNETIC FIELD AT A POINT ON THE AXIS OF CIRCULAR LOOP:

A circular loop of radius a, centre o carries a current I. Let P be the point on the axis of the loop at a

distance X from O. To find magnetic field at point P, consider a small element AB = dl and CP = r.

According to Biot Savarts Law.0

2

0

2

I sinB

4

IB ( 90 )

4

dld

r

dld

r

=

= = oQ

Consider a small element A B = dl opposite to AB and B = B .d d Both dB and dB1 can be resolvedinto two components: B cosd and B cosd cancel each other and the effective component is only

B sin .d

Thus the magnetic field at point P due to the whole circular loop is

0

2

0

2

B B sin

Isin

4

Isin

4

d

dl

r

dlr

=

=

=

But 2 , the circumference of the loop of radiusdl a a=

0

2

IsinB 2 (i)

4a

r

=

From OCP,V

( )

12 2 2 2 2

2r a x r a x= + = +

And

( )1

2 2 2

sin

a

r

a

a x

=

=+

Putting this value in eqn. (i)

( ) ( )

( )

0

12 22 2 2

20

32 2 2

IB 2

4

2I

4

aa

a xa x

a

a x

= + +

= +


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38

AMPERES CIRCUITAL LAW:

Amperes circuital law states that the line integral of magnetic field around any close path or circuit

is equal to 0m times the total current (I) threading the close circuit.

0B Idl =

Where 0 is the absolute permeability of free space.

Application:

Magnetic field due to infinitely long straight wire:

Consider an infinitely long straight wire carrying current I. Consider a circular path of radius r, such

that the wire pass through its centre.

Let PQ=dl be a small element of circular path and B be msgnetic field at point P.

According to amperes circuital law,

0

0

0

0

B I

B cos 0 I

or B I ( cos 0 1)

or B I

dl

or dl

dl

dl

= =

= =

=

o

oQ

But 2 ,dl a= circumference of circular path.

0

0

B 2 I

2IB

4

r

r

=

=

This equation gives strength of magnetic field due to a straight conductor.

Magnetic field due to a Toroidal Solenoid:

Consider a toroidal solenoid consisting of an anchor of radius r. Let n be the number of turns per unit

length of the solenoid.


0

0

0

0

0

B total current threading the ring

B ( 2 I)

B ( 2 I)

2 ( 2 I) ( cos 0 1)

or B I (since 2 )

dl

or dl n r

or dl n r

or B r n r

n dl r

=

=

= = =

= =

o

rr

Q


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39

Magnetic field due to a long straight solenoid:

Consider a long solenoid having n turns per

unit length. Let I be the current flowing in the

solenoid. Consider a close path ABCD and let AB =

L. The line integral of B along path ABCD is given

byB C D A

ABCD A B C D

B B B B Bdl dl dl dl dl = + + + r r r r rr r r r r

. i

Let AB=L. For path AB, and dlB are along the same direction.

B B

A A

B B BLdl dl = = rr

. ii

For path BC and DA, and dlB are perpendicular to each other.

C A

B D

B B 0dl dl = = r rr r

.iii

For path CD, 0=B since field outside the solenoid is zero.

D

C

B 0dl =rr

. iv

Combining i, ii, iii, and iv, we getB

ABCD A

B B BLdl dl = = r rr r


0

ABCD

0

ABCD

0

0

B Total current enclosed by the path ABC D

B ( L I)

or BL ( L I)

B I

dl

or dl n

n

or n

=

=

= =

rr

This equation gives the magnetic field due to a solenoid.


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40

CHAPTER 2

FORCE ON A CHARGE

FORCE ON A CHARGE IN ELECTRIC FIELD:

A charge q inside an electric field E experienced a force F given by

F Eq=FORCE ON A CHARGE MOVING INSIDE AN ELECTRIN FIELD:

A charge q moving with velocity v inside a magnetic field B experience a force F given by

F ( B)q v= +or, F B sinqv =

LORENTZ FORCE:

The total force experience by a charge moving inside the electric and magnetic field is called Lorentzforce.

F E ( B)q q v= + +r

Q Define the SI unit of magnetic field. (Tesla)

Ans: The strength of magnetic field is called Tasla if a charge of 1 C moving with a velocity of 1 ms 1

(along the direction perpendicular to the magnetic field), experience a force of 1 Newton.

MOTION OF A CHARGE PARTICLE INSIDE A UNIFORM MAGNETIC FIELD:

When a charge particle having charge q moves inside a magnetic field B with velocity v is experiencea force given by

F ( B)q v= r

1) When velocity of the charge particle is perpendicular to the magnetic field:

When a charge particle moves perpendicular to the magnetic field, it moves in a circular path as it

experience towards its centre. The necessary centripetal force is provided by the lorentz magnetic force.2

B

B

mvqv

r

mvr

q

=

=

The period of the circular motion is given by


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41

2T

2

B

2

B

r

v

r mv

v q

m

q

=

=

=

The angular frequency,2

T=

2

2 B

B(Cyclotron frequency)

r

mv q

q

m

=

=

2) When v andB are inclined to each other:A charge particle moves inside a

magnetic field making an angle with the

direction of the magnetic field. The velocity v of the

charge particle can be resolved into two

components: V cos and V sin . As the charge

particle experienced an inward force.

The period of the circular path is2

B

sinB

mvqv

r

mvrq

=

=

The period of the circular path is

2T

2 sin

sin B

2mT

B

n

r

v

mv

v q

q

=

=

=

cos 2Pith of helican path

B

2 cos

B

v n

q

nv

q

=

=


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CYCLOTRON:

Principle: A positive ions can acquire sufficiently large

energy with cooperatively smaller alternating potential

difference by making it to cross the same electric field time

and again by making used of a strong magnetic field.

Construction: It consists of two D-shaped hollow semi

circular metal chambers D1 and D2 (called dees) separated by a

small gap and connected to high frequency electric field. The

whole arrangement in enclosed in a metallic box at low

pressure between two poles of strong electromagnet. Positive

ions are produced in the gap of ionization of gas.

Theory: Positive ion is produced at the centre gap when D 1 is

at positive potential and D2 at negative potential. Hence the ion

will move from D1 to D2. The force on the positive ion

provides the necessary centripetal force and it is deflectedalong a circular path

.2

B

(i)B

mvqv

r

mvr

q

=

=

When the ions reach the gap after passing through D 2, the polarity of the dees changes and it is

attracted by D1 and this process repeats itself time and again.

The time spent inside the dee,

(ii)

rtv

m

Bq

=

=

This time t is equal to half of the time period of the electric fieldT

2

T

2 B

2T

B

m

q

m

q

=

=

Angular frequency,2

T

=

2

2 B

B

m q

q

m

=

=

Cyclotron frequency,1

Tv =

12 B

B

2

vm q

qv

m

=

=


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If maxv is the maximum velocity acquired by the ions corresponding to radius R

max

max

RB

B R

2

mv

q

qv

m

=

=

Hence maximum kinetic energy

2

max

2

2

max

2 2 2

1E

2

1 1 B R

2 2

1 B R

2

mv

qmv m

m

q

m

=

= =

=


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44

CHAPTER 3

FORCE ON A CURRENT

FORCE ON A CONDUCTOR PLACED IN A MAGNETIC FIELD :

A conductor of length l carries a current I and is placed inside a magnetic field. If the conductor has

cross sectional area A, number of electrons per unit volume n and drift velocity v d then,

I Ad

n ev=Multiplying both sides by l, we get

I Ad

l n lev=

I and dl vr

are opposite in direcrion

I A (i)d

l n lev= r

The magnetic lorentz force can be written as

( B) (ii)d

f e v= + r

Each three electrons experience a force and let N be the total number of electrons then,F (iii)nf=

N A we haven l =

F A ( B)

A ( B)

F I B

d

d

n l e v

n le v

l

= +

= +

=

rr

rr r

or, F BI sinl= a) If 0 or 180 then, sin 0= =o o

F 0 (minimum) =

b) If 90 or 180 then, sin 1= =o o

F BI (maximum)l =

FLEMINGS LEFT HAND RULE:

It states that if the four finger, central finger and thumb of left hand are stre mutually perpendicular

to each other, such that the four fingers point in the direction of magnetic field and the central finger in the

direction of current and the thumb points in the direction of the force experienced by the conductor.


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FORCE BETWEEN INFINITELY LONG PARALLEL CONDUCTOR:

Consider two infinitely long parallel conductors X1Y1 and X2Y2 separated by a distance r and carrying

currents I1 and I2 .

The magnetic field at point P due to current I2 flowing through the conductor X2 Y2 is

0 22

2IB

4 r=

The direction of 2B at point P is perpendicular to the plane of paper and in inward direction.

Hence the conductor X1 Y1 lies inside the magnetic field B2 and experienced a force given by

2 1

0 1 2

F B I

2I IF (towards left)

4 r

=

=

Similarly the conductor X2 Y2 experience an equal force but towards right. Torque on a current loopplaced in a magnetic field.

TORQUE ON A CURRENT LOOP PLACED IN A MAGNETIC FIELD:

A rectangular coil ABCD carrying current I is suspended in a magnetic field B. Let l be the length

and b be the breadth of the coil. Let 1 2 3 4F ,F ,F and F and be the force acting on the arm DA, BC, AB and CD

of the coil. Now,


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46

( )1

F I(DA) B

I(DA)Bsin 90

I B cosb

=

= +

=

o

1F BI cos (i)b =

and, 2F I(BC) B=

( )

2

I(BC)Bsin 90

I B cos

F BI cos (ii)

b

b

=

=

=

o

r

The forces and are equal in magnitude and opposite in direction. Hence they cancel each other.

3F I(AB) B=

3

I(AB)Bsin 90

I B

F BI (iii)

b

b

==

=

o

r

Then, 4F I(CD) B=

4

I(CD)Bsin 90

I B

F BI (iv)

b

b

==

=

o

r

The forces 3F and 4F are equal in magnitude, opposite in direction and have different line of action.

Hence the form a torque given by Force er distance

BI sin

BIAsin

l b

= =

=Where A is area of the coil

In vector notation

M B = r

Since IA M= , magnetic dipole moment of the loop

1) If the coil has n turns BIAsinn =

2) If is the angle between the coil and the magnetic field, we can write. BIA cosn=


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MOVING COIL GALVANOMETER:

Principle: When a current carrying coil is placed in a

magnetic field it experienced a torque.

Construction: It consists of a coil wounded on a

non-metallic frame and has a central soft iron core. The coil is

suspended between poles of permanent magnet (whose poles are

made concave so as to produce radial magnetic field)from a

torsion head. The other end of the coil is attached to a hair

spring and a small concave mirror is attached to a small

suspension wire. The whole arrangement is enclosed in a

non-magnetic case which can be leveled by leveling screws.

Theory: The coil ABCD of length l and breath b

having n turns is suspended inside a magnetic field. If current is

passed through the coil in the direction of ABCD.

Force on arm AB, F = nBIl (normally

outwards) Force on arm CD, F = nBIl (normally inwards)

The two forces are equal, parallel and opposite and acting at

different points. So they given rise to a torque.

In case of radial magnetic field,Deflecting torque nBIl b nBIA= =

If k is restoring torque per unit twist.Restoring torque ka=

In equilibrium, Defecting torque = restoring torque.

,

,

,

or nBIA k

kor I

nBIA

or I G

a

a

a

=

=

=

Wherek

GnBIA

= is called Galvanometer constant.

. .,i e I a

Deflection produced is directly proportional to the current passed through the Galvanometer.

a) Current sensitivity: It is defined as the deflection produced in the galvanometer on

passing unit current through its coil current sensitivity, BA

I

n

k=

b) Voltage sensitivity: It is defined as the deflection produced in the galvanometer

when a unit voltage is applied across its coil.

BA

V IR R

n

k= =


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TO CONVERT A GALVANOMETRE INTO AMMETR:

To convert an galvanometer into an ammeter of range 0 to I, a small resistance S is connected in parallel to

the coil of the galvanometer. Let G be the resistance of the galvanometer.

Potential difference across shunt = potential difference across galvanometer.

( )

( )

g g

g

g

I I S I G

IS G

I I

=

=

Knowing the value of gI , I and G, the value of S can be find out.

TO CONVERT GALVANOMETRE INTO VOLT METRE:

To convert a galvanometer into volt meter of range 0 to V volt, a resistance R is connected in series

with the galvanometer. When a current gI flows through the coil, the galvanometer gives full scale

deflection. If G is the resistance of the galvanometer, then,

g

g

V I (G R)

VR G

I

= +

=

Knowing the value of gI , I and G the value of R van be found out.


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49

Magnetism is a force generated in matter by the motionof electrons within its atoms. Magnetism

and electricity represent different aspects of the force of electromagnetism , which is one part of Nature's

fundamental electroweak force. The region in space that is penetrated by the imaginary lines of magnetic

force describes a magnetic field. The strength of the magnetic field is determined by the number of lines of

force per unit area of space. Magnetic fields are created on a large scale either by the passage of an electric

current through magnetic metals or by magnetized materials called magnets. The elemental metalsiron,

cobalt, nickel, and their solid solutions or alloys with related metallic elementsare typical materials that

respond strongly to magnetic fields. Unlike the all-pervasive fundamental force field of gravity, the

magnetic force field within a magnetized body, such as a bar magnet, is polarizedthat is the field is

strongest and of opposite signs at the two extremities or poles of the magnet.

Magnetism arises from two types of motions of electrons in atomsone is the motion of the

electrons in an orbitaround the nucleus, similar to the motion of the planets in our solar system around the

sun, and the other is the spin of the electrons around its axis, analogous to the rotationof Earth about its

own axis. The orbital and the spin motion independently impart a magnetic moment on each electron

causing each of them to behave as a tiny magnet. The magnetic moment of a magnet is defined by the

rotational force experienced by it in a magnetic field of unit strength acting perpendicular to its magnetic

axis. In a large fraction of the elements, the magnetic moment of the electrons cancel out because of thePauli exclusion principle, which states that each electronic orbit can be occupied by only two electrons of

opposite spin. However, a number of so-called transition metalatoms, such as iron, cobalt, and nickel, have

magnetic moments that are not cancelled; these elements are, therefore, common examples of magnetic

materials. In these transition metal elements the magnetic moment arises only from the spin of the electrons.

In the rare earth elements (that begin with lanthanum in the sixth row of the periodic tableof elements),

however, the effect of the orbital motion of the electrons is not cancelled, and hence both spin and orbital

motion contribute to the magnetic moment. Examples of some magnetic rare earth elements are: cerium,

neodymium, samarium, and europium. In addition to metals and alloys of transition and rare earth elements,

magnetic moments are also observed in a wide variety of chemical compounds involving these elements.

Among the common magnetic compounds are the metal oxides, which are chemically bonded compositions

of metals with oxygen.

Earth's geomagnetic field is the result of electric currents produced by the slow convective motion of its

liquid core in accordance with a basic law of electromagnetism which states that a magnetic field is

generated by the passage of an electric current. According to this model, Earth's core should be electrically

conductive enough to allow generation and transport of an electric current. The geomagnetic field generated

will be dipolar in character, similar to the magnetic field in a conventional magnet, with lines of magnetic

force lying in approximate planes passing through the geomagnetic axis. The principle of the compass

needle used by the ancient mariners involves the alignment of a magnetized needle along Earth's magnetic

axis with the imaginary south pole of the needle pointing towards the magnetic north pole of the earth. The

magnetic north pole of Earth is inclined at anangle of 11 away from its geographical north pole.

You were introduced to the concept of an electric charge and studied some properties of charges at

rest. You learnt that a static distribution of charge produces a static electric field. Similarly, steady flow of

charge (i.e., a steady current) produces a static magnetic field, which is, infact, the topic of this unit.

However, there are some major differences between the two fields which you will discover in this unit. In

the science laboratory, during your school days, you must have been fascinated with magnets. Recall, when

you tried to push two magnets together in a way they didn't want to go, you felt a mysterious force! In fact,

magnetic fields or the effect of such fields have been known since ancient times when the effect of the

naturally occurring permanent magnet (Fe304) was first observed. The north and south seeking properties of

such materials played a large role in early navigation and exploration. Except for this application,

magnetism was a little known phenomenon until the 19th century, when Oersted discovered that an electric

current in a wire deflects a compass needle. This discovery showed that electric current has something to do

with the magnetic field because a compass needle gets deflected and finally points inathe north-south


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direction only when placed in a magnetic field. In this Unit, we shall consider in detail the production of the

magnetic fields due to steady currents, and the forces they exert on circuits carrying steady currents and on

isolated moving charge. A good way of gaining a better understanding of the nature of fields is to know how

they affect the charged particles on which they act. Hence, in the next unit, you will study the behaviour of

charged particles in both electric and magnetic fields

CHAPTER 4

MAGNETIC DIPOLE AND EARTH MAGNETISM

MAGNETIC DIPOLE: An arrangement of two magnetic poles of equal and opposite strength separated by

a finite distance is called magnetic dipole.

MAGNETIC DIPOLE MOMENT: The product of the strength of either magnetic pole and the magnetic

length of the magnetic dipole. It is denoted by M

M (2 )m l=Its SI unit is Ampere metre square.

CURRENT LOOP AS A MAGNETIC DIPOLE: A circular coil carries current I as shown in fig. Themagnetic field lines due to and portion will be the shape of circular loop. Using right hand thumb rule it

follow that the lower face behaves as south pole while the upper face behaves as north pole.

The magnetic dipole moment of the current loop of area A carrying current I is given by

M IA=

If the coil has n turns

M IAn=

ATOM AS MAGNETIC DIPOLE (Magnetic dipole moment of a revolving electron):

In an atom electron revolves around the nucleus and may be considered as small current loop. Due to

this an atom posses magnetic dipole moment and behaves as a magnetic dipole.

The angular momentum of electron on due to its orbital motion is given by

L M (i)evr=

Where Me is the mass of electron and r is the radius of the orbit.

The orbital motion of electric is equivalent to current.


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1I

Te

=

Where I is the time period of the electron

2T

I (ii)2

r

v

ev

r

=

= Area of the loop

2A (iii)r= Magnetic dipole moment of the atom

2M IA2

2

evr

r

evr

= =

=

2

M L2

e

e

e

em vr

m

e

m

=

=

In vector rotation

M L (iv)2

e

e

m

=

r r

According to Bohrs theory, the angular momentum of an electron is an integral multiple of the2

n

L (v)2

nh=

Where n =1, 2, 3, . and so on.

From equation (iv) and (v), we have

M2 2

M4

e

e

ev nh

m

ehn

m

=

=

Q. Define Bohr magnetron.

Ans: It is define as the magnetic dipole moment associated with an atom due to orbital motion of an

electron in the first orbit of hydrogen atom. It is denoted by B .

24 2

B 9.27 10 A

4e

ehvm

m

= =

MAGNETIC FIELD ON AXIAL LINE OF A BAR MAGNET:

Consider a bar magnet NS having pole strength m1 magnetic length 2l and centre O. Let P be the

point on the axial line at a distance r from O.

The magnetic field axial r at point P is given by


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1 2B B B

axail= +

then( )

0 01 22

B (outwards)

4 NP 4 -

m m

r l= =

r

and( )

0 02 22

B (inwards)

4 SP 4

m m

r l= =

+

r

Since 1 2B B>

( ) ( )

1 2

0 0

2 2

B B B

4 4-

axail

m m

r l r l

=

= +

( ) ( )

( ) ( )

( )

2 2

0 0

2 2 22 2

- 1 1

4 4- -

r l r l m m

r l r l r l

+ = = +

( )

0

22 2

4

4 -

rlm

r l

=

Since, m (2l) = M, magnetic dipole moment.

( )0

22 2

2MB

4 -

r

r l =

If the bar magnet is of very short length l can be neglected.

0

3

2MB

4axail

r=

r

MAGNETIC FIELD ON EQUITORIAL LINE OF A BAR MAGNET:

Consider a bar magnet NS having pole strength m magnetic length 2 l and centre O. Let P be the

point on the equitorial line at a distance r from O.

88

Let 1 2B and B be magnetic field at point P due to N-pole and S-pole of the bar magnet.


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( )0 0

1 2 2 2

B

4 NP 4

m m

r l= =

+

and ( )0 0

1 2 2 2

B

4 SP 4

m m

r l= =

+

Using triangle law of vector, we have,

( ) ( )

1

0

1 2 2 22 2

SNB BNP

2

4

equi

l m

r lr l

=

= ++

( )0

3 22 2

M( m (2l) = M, magnetic dipole moment)

4 r l=

+Q

If the bar magnet is of short length, l can be neglected.

( )

0 0

3 2 32

M MB

4 4equi

r

r

= =

TORQUE ON A BAR MAGNET IN A MAGNETIC FIELD:

A bar magnet NS having pole strength m and of length 2l is placed in uniform magnetic field B

making an angle .

Force on N-pole of magnet = mB (along)

Force on S-pole of magnet = mB (opposite to)

The two forces are equal, parallel and opposite and constitute a torque. ()

Force distance

B KNm

= =

In NKSV

KN KNsin

NS 2KN 2 sin

ll

= =

=Since, m (2l) = M, magnetic dipole moment.

MBsin =In vector rotation

M B= r


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BAR MAGNET AS EQUIVALENT SOLENOID:

For a solenoid having cross sectional area A and carrying current I. Its turns behaves as small

magnetic dipole having magnetic dipole moment I A. Therefore, solenoid can be treated as an arrangement

of small magnetic dipole placed in line which can be treated as a bar magnet having magnetic dipole

moment n I A.

CHAPTER 5

CLASSIFICATION OF MAGNETIC MATERIAL

MAGNETIC INTENSITY (Also called H-field or Magnetic field strength)


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Consider that inside vacuum, a magnetic field Bo exist. The magnetic intensity is given by the

relation

0

0

BH=

Where7 1

04 10 TmAm p - -=

Its SI unit is Ampere per metre (Am-1)

INTENSITY OF MANETISATION: It is defined as the magnetic moment developed per unit volume

when a magnetic specimen is subjected to magnetizing field. It is denoted by I. Its SI unit is 1Am .

MI=

V

MAGNETIC INDUCTION: The magnetic induction is defined as the number of magnetic lines of

induction (magnetic force lines inside the material) crossing per unit area normally through the magnetic

substance. It is denoted by B. Its SI unit is Tesla (T)

Magnetic induction B is the sum of B0 and the magnetic field 0I produced due to magnetization.

0 0

0 0

0

B = B + I

, B = H+ I

, B = (H+I)

or

or

MAGNETIC SUSCEPTIBILITY: It is defined as the ratio of the intensity of magnetization to the

magnetic intensity.

I

Hm

= (It has no unit)

MAGNETIC PERMEABILITY: It is defined as the ratio of the magnetic induction (B) to the magnetic

intensity (H)

B

H=

Its SI unit is Tesla metre per Ampere (TmA -1)

Q. Derive the relation 1r mm c= + where symbols have their usual meaning.

Ans: We have

0, B = (H+I)or

Dividing both sides by H, we get

0

B= I+

H

I

H

1

Also, 0I B

andH

mH

= =

( )

( )

( )

0

0

1

, 1

, 1 ......................................2

m

m

r m

or

or

= +

= +

= +

Where 0r

=

is called relative permeability of the magnetic substance.

CLASSIFICATION OF MAGNETIC SUBSTANCE


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DIAMAGNETIC SUBSTANCE: A substance, which when placed in a magnetic field is feebly magnetized

in a direction opposite to that of the magnetizing field is called diamagnetic substance.

e.g.: Copper, Zinc, water, Bismuth, etc

Properties:

1) It does not obey Curies law.

2) The magnetic susceptibility of a diamagnetic substance has a small negative value.

3) For a diamagnetic substance, the intensity of magnetization (I) has small negative value.

PARAMAGNETIC SUBSTANCE: A substance, which when placed in a magnetic field is feebly

magnetized in the direction of the magnetizing field is called paramagnetic substance.

e.g.: Aluminum, Manganese, Sodium, antimony, etc.

Properties:

1) A magnetic substance is feebly attracted by a magnet.

2) It obey Curies law i.e. the susceptibility of paramagnetic substance is inversely proportional to its

absolute temperature.

3) The magnetic susceptibility of a paramagnetic substance has small positive value.

FERROMAGNETIC SUBSTANCE:Those substances, which when placed in a magnetic field are strongly magnetized in the direction of

magnetizing field are called ferromagnetic substances.

E.g.: Iron, Nickel, Cobalt, etc.

Properties:

1) A ferromagnetic substance is strongly attracted by a magnet.

2) The magnetic susceptibility ( mc ) of a ferromagnetic substances has a large positive.

3) The ferromagnetic substance does not obey Curies law.

CURIE LAW IN MAGNETISM:

For paramagnetic material, intensity of magnetization of the material is

1) directly proportional to magnetic intensityi.e., I Ha

2) inversely proportional to the absolute temperature of the material,

i.e.,1

IT

a

Combining, we get,

HI

T

H, I C

Tor

a

=

Where C is called curies constant.

I C,

H Tm

or c = =

This is called Curies law in magnetism.

i.e., The magnetic susceptibility of paramagnetic material is inversely proportional to its

absolute temperature.

CURIES POINT (CURIES TEMPERATURE):

The Curie point (or Curie temperature) of a ferromagnetic material is the temperature above which it

loses its ability to posses magnetism in the absence of an enternal magnetic field.

EFFICIENCY OF TRANSFORMER:

It is the ratio of output power to input power. It is denoted by y,


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E I

E I

s s

p p

output powery

input power= =

ENERGY LOSS IN A TRANSFORMER:

1) Copper losses: Due to resistance of the copper coil, some electrical energy is converted into heat

energy.

2) Humming losses: When a.c. is supplied to the primary coil, the iron core starts vibrating andproducing humming sound thereby losing some electrical energy in the form of sound.

3) Hysteresis losses: When the ion core is magnetized, the iron core gets heated due to hysteresis. The

energy loses in the form of heat is equal to the area of hysteresis loop.

USES OF TRANSFORMER:

1) They are used as voltage regulator and stabilized water supply.

2) Small transformer are used in radio sets, TV and telephones etc.

Q. Which is the material used in making the core of transformer or a moving coil galvanometer?

Ans: Soft iron is used in making the core of galvanometer. It makes the magnetic field strong as magnetic

field lines tend to cross through the paramagnetic.

Q. Why soft iron is used in making the core of a transformer?Ans: The area of the hysteretic loop for soft iron is very small. Since energy dissipated during a complete

cycle of magnetization and demagnetization is proportional to the area of the hysteresis loop, a small

amount of energy will be wasted, when the core of the transformer is made of soft ir

Documents

class 12 electrostatics Electricity and magnetism made simple notes