Solution of Book Answers 2nd Physics

SOLUTION OF BOOK QUESTIONS 2 nd YEAR PHYSICS

CHAPTER # 12Q#1

The potential is constant throughout a given region of space .is the electric field is zero or non zero in this region?Ans#1:

Electric field E and potential difference are related by equation, E =-∆ v/∆ rWhen the potential is taken as constant which means v= constant then the potential difference becomes zero ∆ v=0 so, E =0/r ∆ → E = 0This relation tells that electric field is zero in a region of where the potential is constantQ#2: Suppose that you follow an electric field line due to a positive point charge .do electric field and the potential increases or decreases?Ans#2:

Relation for the electric field and and electric potential areE= 1/4π ε0 x q/r2 and v=1/4π ε0 x q/ r this gives for same charge

E α 1/ r2 Vα 1/ r When we follow electric field line due to positive charge, distance from positive charge increases which cases to decrease electric field and potential.Q#3:

How can you identify that which plate of a capacitor is positively charged? Ans#3:

When a freely suspende`d positively charged gold leaf of electroscope is 8brought near to one of the plates of capacitor the divergence to gold leaf shows that the plate of the capacitor is positively charge d otherwise negatively charged..Q#4:

Describe the force or forces on a positive point charge when placed between parallel plates,

(a) With similar and equal charges.(b) With opposite and equal charges.

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Ans#4:The electric force F on a charge q due to eclectic field E is given by

F= qE(A) When a positive point charge is placed between parallel plates with similar and equal charges. it will experience no electric force because electric field between the plate is zero . F=qE

F=q(0) =0 → F=0(B) When a positive pint charge is placed between the plates of the Scapacitor it will experience electric force , F=qE S

Due to electric field set up from positive to negative .Q#5:

Electric lines of forces never cross each other. Why? Ans#5:

No, two electric lines of forces never cross each other because “E” only one direction at a given point .if two lines cross each other then there will be more directions at a point which is impossible. Q#6:

If appoint charge q of mass m is released in a non uniform electric field with field lines pointing in the same direction, will it make a rectilinear motion?Ans#6:

Motion along a straight line is called rectilinear motion when a charge “q” of mass “m” is released in a non uniform electric field. It will not move in a straight line and adopts irregular path due to variable electric field.Q#7:

Is E necessarily zero inside a charged rubber balloon if balloon is spherical? Assume that charge is distributed uniformly over the surface ?


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Ans#7: Yes E is necessarily zero inside a charged rubber balloon if balloon is spherical and enclosed no charge but chage is distributed uniformly over its surface. According to Gauss’s law.

Фe =q/ε0=0/ ε0=0 → Фe = 0 --------1From definition, Фe = E.A--------2 Putting equation 1 in eq 2 0= E.A A # 0 so, E=0 Q#8:

Is it true that Gauss’s law states that the total number of lines of forces crossing any closed surface in the outward direction is proportional to the net positive charge enclosed within surface? Ans#8:

Yes, it is true that total no of lines of forces crossing any closed surface in outward direction proportional to the net positive charge according with surface in the outward direction proportional to the net positive charge enclosed with in the surface .According to Gauss’s law, Фe =q/ε0 = constant x q

→ Фe α q here q = net positive charge . Фe =no of electric lines of force Hence no of electric lines of forces are proportional to net positive charge Q#9:

Do electrons tend to go to region of high potential or of low potential?Ans#9:

Electrons tend to go to a region of high potential because high potential is more positive than the lower potential as the charge on electron is negative.

CHAPTER # 13QUESTIONS AND ANSWERS:-Q#1:


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A potential difference is applied across the ends of a copper wire .what is the effect of drift velocity(1) Increasing the potential difference.(2) decreasing the length and the temperature of the wire.Ans#1 The uniform velocity that the free electrons acqure opposite to the electric field of battery is called the drift velocity.

(1) the increase of potential difference makes the electric field strong which cases to increase the drift velocity

(2) the decrease in length an temperature of the wire which causes to to increase the drift velocity of the free electron in the wire.

Q#2:Do bends in wire effects it’s electrical resistance?

Ans#3 We know that R=ρ l/A with bends in a wire , its length (L) and area of cross section (A) remains the same .hence bends in a wire do not affect the electrical resistance (R) of the wire . Q#4:

Why does the resistance of a conductor rise with temperature? Ans#4

As temperature of the conductor rises the irrational amplitude of the atoms of the conductor increases and hence the probability of collisions of free electrons with them increases .at high temperature , the atoms offer a bigger target area to free electrons to collide with them and resistance of conducer

increases. Q5#:

What are the difficulties in testing whether the filament of lighting bulb obeys ohm law?Ans#5

At the beginning, when bulb is turned ON its filament is at low low temrature and it obeys Ohm’s law. I α V Later on when current rises to its maximum value .the maximum power P=I2R dissipated across the filament which increases resistance but current increases at lower rate and does not obey ohm’s law


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Q#6:Is the filament resistance lower or higher in a 500W, 220V light bulb

than in 100W, 220V bulb? Ans#6 Resistance of 1st bulb= R1=?Resistance of 2nd bulb =R2=? Power of first bulb =P1=500 wattsPower of 2nd bulb=P2=100 wattsVoltage =V=220 voltsBy applying the formula, P=V2/R → R=V2/P For1stbulb = R1=V2/P1 (220)2/500 = 48400/500 = 96.8ΩFor 2nd bulb =R2=V2/P2 (220)2/500 = 48400/100 = 484Ω Q#7:

Describe a circuit which will giva a continuously varying potential?

Ans#7 Potentiometer is an instrument which gives continuously varying

potential. Its circuit diagram is given as. When the sliding contact C is moved from A to B ,r varies from O to

R and potential varies from O to E the current following through is given by ; I=E/R …………1) The potential drops across ‘r” is given by V=Ir as I=E/R so V=E/R xr V= r/R E ………………2) This gives us continuously varying potential Q#8:

Explain why the terminal potential difference of a battery decreases when a current drawn from it is increased?


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Ans#8 The terminal potential difference is given by /Vt= E –Ir ………1) The emf E of cell is constant .when current I through is increases ,the product Ir increases which cases to decrease the the terminal potential difference Vt. Q#9:What is the wheat stone bridge? How can it be used to determine an unknown resistance?Ans#9 An electrical circuit devised by Professor Charles Wheatstone used to determine unknown resistance is called wheat stone bridge. It consists of four resistances R1, R2, R3, and R4. Connected in the form of mesh ABCDA . a battery of emf E is connected between points A and C through a switch S .a sensitive galvanometer is of resistance Rg is connected between points B and D. Its equation is , R1/R2 = R3/R4 ……….1)

Determination of unknown resistance:To determine unknown resistance in the arm containing R4 (R4 =Rx

Then known value of R1, R2. And R3are so adjusted that galvanometer no deflection .then from equation no i

R1/R2 = R3/Rx x = R2/R2 x R3

Chapter #14Q#1

A plan conducting loop is located in a uniform magnetic field that is directed along the x-axis .for what orientation of the loop is the flux a maximum?. For what orientation is the flux a minimum?Ans#1:

When plane of loop is placed perpendicular to the magnetic field the flux will be maximum Фe = EAcos0 = EA (max)When plane of loop is placed parallel to the magnetic field, the flux will be minimum as, Фe = EAcos90 = 0 (min)Q#2

A current in a conductor produces a magnetic field, which can be calculated using Ampere’s law. Since current is defined as the rat of flow


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of charges, what can you conclude about the magnetic field due to stationary charges? What about moving charges?Ans#2:

Magnetic field due to stationary charges is zero. Moving charges produce magnetic field around them.Q#3 Describe the magnetic field inside a solenoid carrying a steady current I” if (a) the length is solenoid is doubled but the number of turns remains the same and (b)the number of turns doubled but the length remains same Ans#3:. The magnetic field due to solenoid is given by ,B = μ0 n I As n = N /l so.B = μ0 (N/l )I Case #1:

If the length is double and N remains the same . B = μ0 (N/2l )I = 1/2 μ0 (N/l )I = 1/2 B

Which means magnetic field reduce to half Case #2:

If the number of turns (N) is double and l remains the same. B = μ0 (2N/l )I = 2 μ0 (N/l )I = 2 B

Which means magnetic field increases to two times?

Q#4At a given instant the proton moves in the x direction in a region

where there is magnetic field in the negative z direction. What is direction of the magnetic force? Will the proton continue to move the positive x direction ? ExplainAns# 4:

If proton moves along the positive x –axis in a region of magnetic field is along negative z –axis .the magnetic force is given by R.H. rule ,Fb = q(v x B) , this force(Fb) is along the positive Y- axis and it will be start to move in circular path in xy plane around z –axis and it will not continue to move in the positive x – direction.

Q#5


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Two charged particles are projected into a region where there is a magnetic field perpendicular to their velocities. If the charges are reflected in opposite directions, what can you say about them?Ans#5:

Two charge particles are projected in a magnetic field perpendicular to its their velocity and they are deflected in opposite directions. As we know that in magnetic field particles are deflected so they are oppositely charged .So, one particle is positively charged and other is negatively charged.Q#6

Suppose that a charge q is moving in a uniform magnetic field with a velocity v. Why is there no work done by the magnetic force that acts on the charge q?-Ans#6:

A charge q is moving in uniform magnetic field B, with a velocity V in a Circular path due to magnetic force. The angle between magnetic force and velocity will be 900 .work done is given by , W = F.d = Fd cos 900 =0 (as cos of 900 is zero). So work is being done .

Q#7

If a charged particle moves in a straight line through some region of space, can you say that the magnetic field in the region is zero?Ans#7:

No, when a particle moves in straight line in the direction of magnetic field. B

. q o v

The force acting on charge will be zero as angle is 00 charge q and magnetic field is parallel so, F= qvBsin00 = qvB(0) = 0Therefore we can not say magnetic field in this zero.Q#8

Why does the picture on TV screen become distorted when a magnet is brought near the screen?Ans#:8


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90o

F


As the picture is formed on the TV screen due to the electrons , when a magnetic is brought is near the TV screen ,the beam of electron is deflected due the magnetic force which is given by, F= evBsinθ0 .on the each electron.Q#9

Is it possible to orient a current loop in a uniform magnetic field cush that the loop will tend to rotate? Explain.Ans#9

The torque acting on a current carrying loop in the uniform magnetic field is given by,τ(torque )=NIABcosα ,Where α is the angle between B and plane of loop .if α= 900 then cos of 900= 0 so torque will be zero here this shows that if plane of loop is oreintted at 900 with B .Then loop will not tend to rotate. Q#10

How can a current loop be used to determine the presence of a magnetic field in a given region of space?Ans#10:

When current carrying loop is placed in the presence of a magnetic field with its plane makes an angle α with B .The torque acting on it will be τ(torque )=NIBAcosα if current loop deflects , filed is present other wise not.

Q#11How can you use a magnetic field to separate isotopes of chemical

element?Ans#11:

When isotopes of chemical element are projected in a magnetic field at right angle. then magnetic force ,Fb= evBsin90 = evB Provide centripetal force , Fc = mv2/r Bev = mv2/r r= mv/Be Isotopes has same charge , but due to different masses they adopt different paths due to which fall at different places and are separated . Q 12:

What should be the orientation of a current carrying coil in a magnetic field so that torque acting upon the coil (a)maximum (b) minimumAns#12:


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When the current carrying wire placed in the uniform magnetic field . the torque acting on it is given by,τ(torque )=NIBAcosα where α is the angle between and plane of the coil

(a) when plane of the coil is oriented parallel to to B .thenα=0 and cos0 = 1 so, τ(torque )=NIBA(1) = NIAB(max) α=90 and cos 90 = 0 τ(torque )=NIBA(0) =0 at this the torque will be minimum.Torque will be maximum (b) when plane of coil is oriented at right angle to B then angle will be 90 so, Q#13

A loop of wire is suspended between the poles of a magnet with its plane parallel to the pole faces. What happens if a direct current uis put through the coil? What happens if an alternating current is used instead?Ans#13:

Torque on loop wire is given by , τ(torque )=IBAcos αWhen the loop of wire is suspended between the poles of a magnet with its plan parallel to pole faces then α =90 and cos 90 = 0 then we get the result τ(torque )=IBAcos90 =0 if direct current or alternating current is passed through the coil .these current have no effect on the loop Q#14

Why the resistance of an ammeter should be very low?Ans#14:

Ammeter is always connected in series the magnitude of current decreases through through the circuit with the presence of ammeter .to measure

I Rmaximum current the resistance of ammeter should be very small as compared to the resistance of circuit. Ans#15:

Why the voltmeter should have a very high resistance?Ans#15:

Voltmeter is always connected I in parallel with two points where potential is to be measured. the potential across the points decreases with presences of voltmeter because it draws some current Iv of the circuit . to reduce the current through voltmeter ,the resistance of voltmeter should be large as compared to the resistance of circuit .Prepared By:

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A IR R B

CHAPTER # 15 Q#1Does the induced emf in a circuit depend upon the resistance of the circuit?Does the induced emf depends upon the resistance of the circuit ?Ans#According to Faradays law .emf is given by ε =-N ∆Ф/∆ tThis relation shows that emf induced in coil depend upon the rate of change of magnetic flux and does not depends on the resistance of the coil .While in case of ohm law induced current flowing through coil depends upon the resistance R, of the coilI= ε / RQ#2A square loop of wire is moving through a uniform magnetic field. The normal to the loop is oriented parallel to the magnetic field. Is emf induced in the loop ? Give a reason for your answer?

Ans#2. B

. A v

E= VBLsinθ0 If the vector area is parallel to B . then according to faradays law the time rate of change of flux becomes zero. ∆Ф/∆ t=0 for N=1ε =-N ∆Ф/∆ t = -1 X 0=0Hence there is no emf produced in a loop Q#3A light metallic ring is released from above into a vertical bar magnet. Viewed for above, does the current flow clockwise or anticlockwise in the ring? Ans#3


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According to lens’s law the direction of induced current will oppose the motion of ring towards the bar magnet .for this purpose, the induced current should flow in the clock wise direction the ring.Q#4What is the direction of the current through resistor R in fig.? when switch S is

(a) closed(b) opened

Ans#4a) As the switch is closed, according to lens’s law, the direction of current

induced in R will be in anti clockwise direction.b) A switch S is opened according to lens’s law the direction of current

induced in R will be in the clock wise direction Q#5Does the induced emf always act to decrease the magnetic flux through a circuit?Ans#5No, induced emf does not always act to decrease the magnetic flux through the circuit .according to lens’s law ,when bar magnet with N pole is moved towards a loop flux decreases and when it taken away from a loop flux increases which case emf induced.Q#6When the switch in the circuit is closed a current is established in the coil and the metal ring jumps upward (fig.) why?Describe what would happen to the ring if the battery polarity were reversed?Ans#6When switch S is closed, the changing flux of coil causes to induce current in the metal ring which produces flux opposite to that of coil and ring is replaced and jumps upward. Same thing will be happen if polarity of the battery is reversed. Q#7The fig. shows a coil of wire in the xy plane with a magnetic field directed along the y – axis.Around which of the three coordinate axes the coil should be rotated in order to generate an emf and a current in the coil?Ans#7The coil must be rotated about about x-axis to get change of magnetic flux and an induced current through it.The emf induced in the coil is given byPrepared By:


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ε =NwABsinθ0 .only in this case emf is induced.If the coil rotated about z-axis then no change of magnetic flux takes place .If coil rotated through y-axis then the flux through it will be zero.b/c plane of the coil is at all times parallel to the lines of magnetic field Q#8How would you position a flat loop of wire in a changing magnetic field so that there is no emf induced in the loop?Ans#8Induced emf is given by ,ε =NwABsinθ0 N=1 so, ε =wABsinθ0

As flate loop of wire is parallel to changing magnetic field then θ=00

ε =wABsin00 = 0 this shows that nom flux change through it and hence no emf is induced.Q#9In a certain region the earth’s magnetic field vertically down .when a plane files due to north, which wingtip is positively charged? Ans#9The magnetic force on a moving electric charge is given by ,F = -e (V X B).this relation shows that F acting at right angle to v is towards north B is vertically downward ,the electron moves towards left N direction of F is directed towards west .

w E SQ#10Show that ε and ∆Фe/∆t have same units?Ans#10ε = W/q as unit of work is joule and charge is coulomb so,ε = J/C = JC-1 = volt a)Hence the unit of Є is volt.Unit of ∆Ф/∆ t is ---------b)∆Ф = B∆A = NA-1m-1 m2 ∆ t = secThen b ) becomes ,∆Ф/∆ t = NA-1m-1 m2/sPrepared By:


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∆Ф/∆ t = N m/A s As N m =J, A s = coulomb = C∆Ф/∆ t = J/C = voltQ#11When an electric motor, such as an electrical drill, is being used, does it also act as a generator? if so what is consequences of this? Ans#11When an electrical motor is running, its armature is rotating in a magnetic field. a torque acts on the armature and at the same time magnetic flux is changing through is changing which produced an induced emf .the induced emf opposes the rotation of armature .this means that motor also act as generator when it is running .when motor rotate without load it will rotate quite easy,then large back emf is induced .but in case of when motor is over loaded the induced back emf in the coil produces high current which cause to burn the motor out . Q#12Can a D.C motor be turned into a D.C generator? What changes are required to be done?Ans#12Yes, D.C motor and D.C generator are same in construction but the difference is that D.C is mechanically driven instrument while D.C motor is electrically driven.If voltage is supplied by connecting a battery to slip ring of D.C generator, then D.C generator acts as D.C motor.When current passes through armature of motor a torque is produced which rotates the armature .by converting electrical energy into mechanical energy .thus a D.C motor can be converted into D.C generator.There are two changes occurs to convert D.C motor into D.C generator

1) The magnetic field must be applied by permanent magnet and not by electromagnet.

2) An arrangement to rotate the coil armature should be provided. Q#13Is it possible to change both area of the loop and the magnetic field passing through the field and still not have an induced emf in the loop?Ans#13Yes, it it is possible in the magnetic flux takes due to the change in the area of the loop and magnetic field, no emf will be induced in the loop.Magnetic field is given by .∆Ф = B . A =B∆cosθ if the vector area ∆A is parallel to magnetic field B.θ = 00 so,Prepared By:


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∆Ф =B∆Acos00 cos0=1∆Ф =B∆Aif b B and A are constant then , ∆Ф =0Є =-N∆Ф/∆t Є =-N X 0/ ∆t =0Hence no induce emf in the loop will be produced Q#14Can an electric motor be used to driven an electric generator with the out put from the generator being used to operate the motor?Ans#14No, it is impossible .if it is possible it will be self operating system without getting energy .which is against the law of conservation of energy.Moreover the efficiency is always less than 100.Hence a combination of motor and generator can never produce more energy out than its energy.Q#15A suspended magnet is oscillating freely in a horizontal plane. The oscillations are strongly damped when a metal plate is placed under the magnet. Explain why this occurs? Ans#15The oscillating magnet produces change of magnetic flux close to it. The metal plate below it experience a the change of magnetic flux .As a result eddy current produced inside the metal .according to lens’s law ,Eddy currents oppose the cause which produces it. Therefore ,the oscillations of magnet are strongly damped. Q#16Four unmarked wires emerge from a transformer. What steps would you take to determine the turn’s ratio?Ans#16Four wires coming out of transformer are the ends of two coils (p&s) of transformer .each coil has negligible resistance for D.C .the coils are separated as primary and secondary. an alternating voltage of known value vp to the primary coil the output voltage across the secondary coil. The turn ratio of the coil is determined as,Vs/Vp =Ns/Np

Q#17a) Can a step up transformer increases the power level? b) In a transformer, there is no transfer of charge from the primary to

secondary .How is then the power transferred?Prepared By:


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Ans#17a) No, a step up transformer can not increase the power level. the power of

out put is always less than input. in ideal case input is equal to the out put.

b) Two coils of transformer are magnetically linked, the change of flux through one coil is linked with the other coil and emf is produced.

Q#18 When the primary of a transformer is connected to a.c mains the current in it,

a) is very small if the secondary circuit is open ,but,b) Increases when the secondary circuit is closed .explain these facts.

Ans#18a) if the secondary circuit is open, then out power will be zero .because out

put power always les than input power .therefore very small value of current is being drawn by a primary coil of transformer a.c mains

b) When the secondary circuit is closed, the out power will be increase .therefore the transformer will draw a large current from the a.c mains to increase the primary power.


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Solution of Book Answers 2nd Physics