NCERT Solutions for Class 12 Physics Chapter 6 Electromagnetic Induction

NCERT solutions for class 12 Physics Chapter 6 Electromagnetic Induction is crucial for the students of 12th standard. NCERT physics class 12 chapter 6 PDF is provided here to help students understand the chapter in an easy and interesting way.

Class 12 Physics Chapter 6 Electromagnetic Induction NCERT solutions give comprehensive answers to the question provided in textbooks, previous year question papers and sample papers. In order to understand the topic thoroughly and to sort out class 12 Physics Chapter 6 Electromagnetic Induction NCERT notes students must practice the solutions regularly. The NCERT class 12 Physics solutions for all the chapters are created by subject experts according to the latest CBSE syllabus 2019-2020

Class 12 Physics NCERT solutions for Chapter 6 Electromagnetic Induction

The chapter discusses magnetism and electricity together. Faraday’s law should be well versed and understood. Faraday’s law states that any change in the magnetic field of a current carrying conductor results in voltage (emf) being induced in the conductor. Lenz’s law states that there is always a counter-force opposing the induced emf. Both the laws are important from exam point of view.

Numerical problems from these concepts are often asked. While solving the numerical problems please keep in mind the direction of induced emf and magnetic field. One can get a good understanding of these problems by solving practice problems of NCERT Class 12 Physics Electromagnetic Induction and by referring NCERT Solutions for Class 12 Chapter 6 of Physics to understand the correct method of solving problems.

Topics covered in Class 12 Physics Electromagnetic Induction are:

Section Number Topic
6.1 Introduction
6.2 The Experiments Of Faraday And Henry
6.3 Magnetic Flux
6.4 Faraday’s Law Of Induction
6.5 Lenz’s Law And Conservation Of Energy
6.6 Motional Electromotive Force
6.7 Energy Consideration: A Quantitative Study
6.8 Eddy Currents
6.9 Inductance
6.9.1 Mutual inductance
6.9.2 Self-inductance
6.10 Ac Generator

 

Class 12 Physics NCERT Solutions Electromagnetic Induction Important Questions


Q 1. Predict the direction of induced current in the situations described by the following figures

(i)

Class 12 Physics chapter 6 1

(ii)

Class 12 Physics chapter 6 2

(iii)

Class 12 Physics chapter 6 3

(iv)

Class 12 Physics chapter 6 4

(v)

Class 12 Physics chapter 6 5

(vi)

Class 12 Physics chapter 6 6

Answer:

Lenz’s law shows the direction of induced current in a closed loop. In the given two figures they shows the direction of induced current when the North pole of a bar magnet is moved towards and away from a closed loop respectively.

Class 12 Physics chapter 6 7

We can predict the direction induced current in different situation by using the Lenz’s rule.

(i) The direction of the induced current is along qrpq.

(ii) The direction of the induced current is along prqp.

(iii) The direction of the induced current is along yzxy.

(iv) The direction of the induced current is along zyxz.

(v) The direction of the induced current is along xryx.

(vi) No current is induced since the field lines are lying in the plane of the closed loop.

 

Q 2. We are rotating a 1 m long metallic rod with an angular frequency of 400 red s1s^{-1} with an axis normal to the rod passing through its one end. And on to the other end of the rod it is connected with a circular metallic ring. There exist an uniform magnetic field of 0.5 T which is parallel to the axis everywhere. Find out the emf induced between the centre and the ring .

Ans:

Length of the rod = 1m

Angular frequency,ω=400  rad/s\omega = 400 \; rad/s

Magnetic field strength, B = 0.5 T

At one of the end of the rod it have zero liner velocity, while on to its other end it have a linear velocity of Iω I \omega

Average linear velocity of the rod, v=Iω+02=Iω2v = \frac{I \omega + 0}{ 2 } = \frac{I \omega}{2}

Emf developed between the centre and ring.

e=Blv=Bl(iω2)=Bl2ω2=0.5×(1)2×4002=100V\\e = Blv = Bl\left ( \frac{i \omega }{2} \right ) = \frac{B l^{2 } \omega}{2} \\ \\ = \frac{0.5 \times \left ( 1 \right )^{2} \times 400}{2} = 100 V

Hence, the emf developed between the centre and the ring is 100 V.

 

 

Q 3. A long solenoid with 15 turns per cm has a small loop of area 2.0 cm2 placed inside the solenoid normal to its axis. If the current carried by the solenoid changes steadily from 2.0 A to 4.0 A in 0.1 s, what is the induced emf in the loop while the current is changing?

Ans:

Number of turns on the solenoid – 15 turn / cm = 1500 turn / m

Number of turns per unit length, n = 1500 turns

The solenoid has a small loop of area, A = 2.0 cm2 = 2 × 10−4 m2

Current carried by the solenoid changes from 2 A to 4 A.

Therefore, Change in current in the solenoid, di = 4 – 2 = 2 A

Change in time, dt = 0.1 s

According to Faraday’s law, induced emf in the solenoid is given by:

e=dϕdt                  ...(1)e = \frac{d\phi }{dt} \;\;\;\;\;\;\;\;\; . . . (1)

Where,

ϕ\phi = Induced flux through thee small loop

= BA                                         . . . (2)

B = Magnetic field

=μ0ni\mu _{0} ni μ0\mu _{0} = Permeability of free space

= 4n×107  H/m4n \times 10 ^{-7} \; H/m

Hence, equation (1) can be reduced to:

e=ddt(BA)e = \frac{d}{dt}\left ( BA \right ) e=Aμ0n×(didt)=2×104×4π×107×1500×20.1=7.54×106  Ve = A\, \mu _{0}\, n \times \left (\frac{di}{dt} \right ) \\ \\ = 2 \times 10^{-4} \times 4 \pi \times 10 ^{-7} \times 1500 \times \frac{2}{0.1}\\ \\ = 7.54 \times 10^{-6}\; V

Hence, the induced voltage in the loop is 7.54×106  V7.54 \times 10^{-6}\; V

 

Q 4. A rectangular wire loop of sides 8 cm and 2 cm with a small cut is moving out of a region of the uniform magnetic field of magnitude 0.3 T directed normal to the loop. What is the emf developed across the cut if the velocity of the loop is 1 cm s–1 in a direction normal to the (a) longer side, (b) shorter side of the loop? For how long does the induced voltage last in each case?

 

Ans:

Length of the wired loop, l = 8 cm = 0.08 m

Width of the wired loop, b = 2 cm = 0.02 m

Since, the loop is rectangle, area of the wired loop,

A = lb

= 0.08 × 0.02

= 16×104m216 \times 10 ^{-4} \, m^{2}

Strength of magnetic field, B = 0.2 T

Velocity of the loop, v = 1 cm / s = 0.01 m / s

 

(i)Emf developed in the loop is given as:

e = Blv

= 0.3 × 0.08 × 0.01 =2.4×104  V2.4\times 10^{-4} \; V

Time taken to travel along the width ,t=Distance  travelledVelocity=bv=0.020.01=2st = \frac{Distance \; travelled}{Velocity} = \frac{b}{v} \\ \\ = \frac{0.02}{0.01} = 2s

Hence, the induced voltage is 2.4×104  V2.4\times 10^{-4} \; V which lasts for 2 s.

 

(ii) Emf developed, e = Bbv

= 0.3 × 0.02 × 0.01 =0.6×104V0.6 \times 10^{-4} \, V

Time taken to travel along the length, t=Distance  travelledVelocity=lv=0.080.01=8st = \frac{Distance \; travelled}{Velocity} = \frac{l}{v} \\ \\ = \frac{0.08}{0.01} = 8s

Hence, the induced voltage is 0.6×104V0.6 \times 10^{-4} \, V which lasts for 8 s.

 

Q 6. A circular coil of radius 8.0 cm and 20 turns is rotated about its vertical diameter with an angular speed of 50 rad s–1 in a uniform horizontal magnetic field of magnitude 3.0 × 10–2 T. Obtain the maximum and average emf induced in the coil. If the coil forms a closed loop of resistance 10 Ω, calculate the maximum value of current in the coil. Calculate the average power loss due to Joule heating. Where does this power come from?

Ans:

Maxm emf induced = 0.603 V

Avg emf induced = 0 V

Maxm current in the coil = 0.0603 A

Power loss (average) = 0.018 W

(Power which is coming from external rotor)

Circular coilradius, r = 8 cm = 0.08 m

Area of the coil,A=nr2=n×(0.08)2m2A = nr^{2} = n \times \left ( 0.08 \right )^{2}m^{2}

Number of turns on the coil, N = 20

Angular speed,ω=50  rad/s\omega = 50 \; rad/s

Strength of magnetic,B=3×102  TB = 3 \times 10 ^{-2} \; T

Total resistance produced by the loop,R=10ΩR = 10 \Omega

Maxm emf induced is given as:

e=Nω  AB=20×50×n×(0.08)2×3×102=0.603  Ve = N\omega \; AB\\ = 20 \times 50 \times n \times \left ( 0.08 \right )^{2} \times 3 \times 10^{-2} \\ = 0.603 \; V

The maximum emf induced in the coil is 0.603 V.

Over a full cycle, the average emf induced in the coil is zero.

Maximum current is given as:

I=eR=0.60310=0.0603  AI = \frac{e}{R} \\ = \frac{0.603}{10} = 0.0603\; A

Average power because of the joule heating:

P=el2=0.603×0.06032=0.018  WP = \frac{el}{2} \\ = \frac{0.603 \times 0.0603}{2} = 0.018\; W

The torque produced by the current induced in the coil is opposing the normal rotation of the coil. To keep the rotation of the coil continuously, we must find a source of torque which opposes the torque by the emf, so here the rotor works as an external agent. Hence, dissipated power comes from the external rotor.

 

 

Q 7. A horizontal straight wire 10 m long extending from east to west is falling with a speed of 5.0 m s–1, at right angles to the horizontal component of the earth’s magnetic field, 0.30 × 10–4 Wb m–2.
(a) What is the instantaneous value of the emf induced in the wire?
(b) What is the direction of the emf?
(c) Which end of the wire is at the higher electrical potential?

Ans:

Wire’s Length, l = 10 m

Speed of the wire with which it is falling, v = 5.0 m/s

Strength of magnetic field, B = 0.3×104Wbm20.3 \times 10 ^{-4} Wb \, m^{-2}

 

(a) EMF induced in the wire, e = Blv

=0.3×104×5×10=1.5×103V\\ =0.3 \times 10 ^{-4} \times 5 \times 10 \\ = 1.5 \times 10 ^{-3} \, V

 

(b) We can determine the direction of the induced current by using the Fleming’s right hand thumb rule, here the current is flowing in the direction from West to East.

 

(c) In this case the eastern end of the wire will be having higher potential

 

 

Q 8. Current in a circuit falls from 5.0 A to 0.0 A in 0.1 s. If an average emf of 200 V induced, give an estimate of the self-inductance of the circuit.

Ans:

Current at initial point, I1I_{1}:  5.0 A

Current at final pint, I2I_{2}: 0.0 A

Therefore, change in current is, dI = I1I2I_{1} – I_{2} = 5 A

Total time taken, t = 0.1 s

Average EMF, e = 200 V

We have the relation, for self – inductance (L) and average emf of the coil :

e=LdidtL=e(didt)=2005=4H\\e = L \frac{di}{dt} \\ \\ L = \frac{e}{\left (\frac{di}{dt} \right )} \\ \\ = \frac{200}{5} = 4 H

Hence, the self – induction of the coil is 4 H.

 

Q 9. A pair of adjacent coils has a mutual inductance of 1.5 H. If the current in one coil changes from 0 to 20 A in 0.5 s, what is the change of flux linkage with the other coil?

 Ans:

Given,

Mutual inductance, μ=1.5  H\mu = 1.5 \; H

Current at initial point, I1I_{1}:  0 A

Current at final point, I2I_{2}: 20 A

Therefore, change in current is, dI = I1I2I_{1} – I_{2} = 20 – 0 = 20 A

Time taken for the change, t = 0.5s

Emf induced, e=dϕdte = \frac{d\phi }{dt}                                            . . . (1)

dϕd\phi = change in the flux linkage with the coil.

Relation of emf and inductance is:

e=μdldte = \mu \frac{dl}{dt}                                                                 . . . (2)

On equating both the equation, we get

dϕdt=μdldt\frac{d\phi }{dt} = \mu \frac{dl}{dt} dϕdt=1.5×(20)=30  Wb\frac{d\phi }{dt} = 1.5 \times \left ( 20 \right ) \\ \\ = 30 \;Wb

Therefore, the change in flux linkage os 30 Wb.

 

 

Q 10. A jet plane is travelling towards west at a speed of 1800 km/h. What is the voltage difference developed between the ends of the wing having a span of 25 m, if the Earth’s magnetic field at the location
has a magnitude of 5 × 10–4 T and the dip angle is 30°.

 

Ans:

Speed of the plane with which it is moving,  v = 1800 km/h = 500 m/s

Wing span of the jet, l = 25 m

Magnetic field strength by earth, B = 5×104  T5 \times 10 ^{-4} \; T

Dip angle, δ=30\delta = 30^{\circ}

Vertical component of Earth’s magnetic field,

Bv=Bsinδ=5×104sin30=2.5×104T\\B_{v} = B \sin \delta \\ = 5 \times 10^{-4} \sin 30^{\circ} \\ = 2.5 \times 10^{-4} T

Difference in voltage between both the ends can be calculated as:

e=(Bv)×l×v=2.5××104×25×500=3.123V\\e = \left (B_{v} \right ) \times l \times v \\ \\ = 2.5 \times \times 10^{-4} \times 25 \times 500 \\ \\ = 3.123 V

Hence, the voltage difference developed between the ends of the wings is 3.125 V.

 

 

 

Additional Exercises

Q 11.Let us assume that the loop in the question number 4 is stationary or constant but the current source which is feeding the electromagnet which is producing the magnetic field is slowly decreased. It was having an initial value of 0.3 T and the rate of reducing the field is 0.02 T / sec. If the cut is joined to form the loop having a resistance of 1.6Ω1.6 \Omega. Calculate how much power is lost in the form of heat? What is the source of this power?

 

Ans:

 Rectangular loop are having sides as 8 cm and 2 cm.

Therefore, the area of the loop will be, A = L × B

= 8 cm × 2 cm

= 16cm216 cm^{2}

= 16×104cm216 \times 10^{-4} cm^{2}

Value of magnetic field at initial phase, B’ = 0.3 T

Magnetic fields decreasing rate, dBdt=0.02T/s\frac{dB}{dt} = 0.02 T/s

Emf induced in the loop is:

e=dϕdte = \frac{d\phi }{dt} dϕd\phi = change in flux in the loop area

= AB

e=d(AB)dt=AdBdt=16×104×0.02=0.32×104V∴ e = \frac{d\left ( AB \right )}{dt} = \frac{AdB}{dt} \\ \\ = 16 \times 10^{-4} \times 0.02 = 0.32 \times 10^{-4} V

Resistance in the loop will be, R = 1.6Ω1.6 \Omega

The current developed in the loop will be:

i=eR=032×1041.6=2×105Ai = \frac{e}{R} \\ \\ = \frac{032 \times 10 ^{-4}}{1.6} = 2 \times 10^{-5} A

Power loss in the loop in the form of the heat is :

P=i2R=(2×105)2×1.6=6.4×1010WP = i^{2}R \\ \\ = \left ( 2 \times 10^{-5} \right )^{2} \times 1.6 \\ \\ = 6.4 \times 10^{-10} W

An external agent is the source for this heat loss, which is responsible for the change in the magnetic field with time.

  

 Q 12. We have a square loop having side as 12 cm and its sides are parallel top x and y axis is moved with a velocity of 8 cm /s in the positive x direction in a region which have a magnetic field in the direction of positive z axis.  The field is not uniform whether in case of its space or in the case of time. It has a gradient of 10−3 T cm−1 along the negative x – direction(i.e its value increases by 103  T  cm110 ^{-3} \; T \; cm^{-1} as we move from positive to negative direction ), and it is reducing in the case of time with the rate of  103  T  s110 ^{-3} \; T \; s^{-1}. Calculate the magnitude and direction of induced current in the loop (Given : Resistance = 4.50  mΩ4.50 \; m \Omega).

 

Ans:

Side of the Square loop, s = 12cm = 0.12m

Area of the loop, A = s × s = 0.12 × 0.12 = 0.0144 m2m^{2}

Velocity of the lop, v = 8 cm  s1 cm \;s^{-1} = 0.08 m  s1 m \;s^{-1}

Gradient of the magnetic field along negative x-direction,

dBdx=103  Tcm1=101  m1\frac{dB}{dx} = 10^{-3} \; T\,cm^{-1} = 10^{-1} \; m^{-1}

And, the rate of decrease of the magnetic field,

dBdt=103  Ts1\frac{dB}{dt} = 10^{-3} \; T \, s^{-1}

Resistance, R = 4.50  mΩ=4.5×103  Ω4.50 \; m \Omega = 4.5 \times 10 ^{-3} \; \Omega

Rate of change of the magnetic flux due to the motion of the loop in a non-uniform magnetic field is given as:

dBdt=A×dBdx×v144×104m2×101×0.08=11.52×105  Tm2s1\frac{dB}{dt} = A \times \frac{dB}{dx} \times v \\ \\ 144 \times 10 ^{-4} m ^{2} \times10^{-1} \times 0.08 \\ \\ = 11.52 \times 10 ^{-5} \; Tm^{2} s^{-1}

Rate of change of the flux due to explicit time variation in field B is given as:

dϕdt=A×dBdt=144×104×103=1.44×105Tm2s1\frac{d\phi }{dt} = A \times \frac{dB}{dt} \\ \\ = 144 \times 10^{-4} \times 10 ^{-3} \\ \\ = 1.44 \times 10^{-5} T\,m^{2}s^{-1}

Since the rate of change of the flux is the induced emf, the total induced emf in the loop can be calculated as:

e=1.44×105+11.52×105=12.96×105VInduced  current,  i=eR=12.96×1054.5×103i=2.88×102  Ae = 1.44 \times 10^{-5} + 11.52 \times 10^{-5}\\ \\ = 12.96 \times 10^{-5} V \\ \\ ∴ Induced \; current, \; i = \frac{e}{R}\\ \\ = \frac{12.96 \times 10 ^{-5}}{4.5 \times 10 ^{-3}} \\ \\ i = 2.88 \times 10^{-2}\; A

Hence, the direction of the induced current is such that there is an increase in the flux through the loop along positive z-direction.

  

Q 13. We have a powerful loud speaker magnet, and have to measure the magnitude of field between the poles of the speaker. And  a small search coil is placed normal to the field direction and then quickly removed out of the field region, the coil is of 2cm22 \, cm^{2} area and have 25 closely wound turns. Similarly, we can give the coil a quick 9090 ^{\circ} turn to bring its plane parallel to the field direction. We have measured the total charge flown in the coil by using a ballistic galvanometer and it comes to 7.5 mC. Total resistance after combining the coil and the galvanometer is 0.50  Ω0.50 \; \Omega. Estimate the field strength of magnet.

 

Ans:

Given,

Coil’s Area, A = 2cm2=2×104m22 \, cm^{2} = 2 \times 10^{-4} m^{2}

Number of turns on the coil, N = 25

Total Charge in the coil, Q = 7.5 mC = 7.5×103  C7.5 \times 10 ^{-3}\; C

Total resistance produced by the combo of coil and galvanometer, R=0.50  ΩR = 0.50 \; \Omega

Current generated in the coil,

I=Induced  emf  (e)R                ...(1)I = \frac{Induced \; emf\; \left ( e \right )}{R}\;\;\;\;\;\;\;\; . . . (1)

EMF induced is shown as:

e=Ndϕdt                  ...(2)e = -N \frac{d\phi }{dt} \;\;\;\;\;\;\;\;\; . . . (2)

Where,

dϕd \phi = Change in flux

From equation (1) and (2), we have

I=NdϕdtIdt=NRdϕ                ...(3)I = – N \frac{d \phi}{dt} \\ \\ Idt = -\frac{N}{R}d \phi \;\;\;\;\;\;\;\; . . . (3)

Flux through the coil at initial phase, ϕi=BA\phi _{i} = BA

Where, B = Strength of the magnetic field

Flux through the coil at final phase, ϕf=0\phi _{f} = 0

After integrating eq (3) on both of the side, we get

Idt=NRϕiϕfdϕ\int Idt = \frac{-N}{R}\int_{ \phi_{i} }^{ \phi_{f} } d \phi

Total Charge, Q=IdtQ = \int Idt Q=NR(ϕfϕi)=NR(ϕi)=+NϕRQ=NBARB=QRNA=7.5×103×0.525×2×104=0.75  T\\∴ Q = \frac{-N}{R}\left ( \phi_{f} – \phi_{i} \right )= \frac{-N}{R}\left ( – \phi_{i} \right ) = + \frac{N \phi}{R} \\ \\ Q = \frac{NBA}{R}\\ \\ ∴ B = \frac {QR}{NA} \\ \\ = \frac {7.5 \times 10^{-3} \times 0.5}{25 \times 2 \times 10^{-4}} = 0.75 \; T

Hence, the field strength is 0.75 T.

 

 

Q 14. In the given figure we have a metal rod PQ which is put on the smooth rails AB and these are kept in between the two poles of a permanent magnets. All these three (rod, rails and the magnetic field ) are in mutual perpendicular direction. There is a galvanometer ‘G’ connected through the rails by using a switch ‘K’.Given, Rod’s length = 15 cm , Magnetic field strength, B = 0.50 T, Resistance produced by the closed loop = 9.0  mΩ9.0\; m\Omega. Let’s consider the field is uniform.

8

(i) Determine the polarity and the magnitude of the induced emf if we will keep the K open and the rod will be moved with the speed of 12 cm/s in the direction shown in the figure.

 (ii) When the K was open is there any excess charge built up? Assume that K is closed then what will happen after it?

 (iii) When the rod were moving uniformly and the K was open, then on the electron in the rod PQ there were no net force even though they did not experienced any magnetic field because of the motion of the rod. Explain.

 (iv) After closing the K, calculate the retarding force.

 (v) When the K will be closed calculate the total external power which will be required to keep moving the rod with the same speed ( 12 cm/s)? and also calculate the power required when K will be closed.

 (vi)What would be the power loss ( in form of heat) when the circuit is closed? What would be the source of this power?

 (vii) Calculate the emf induced in the moving rod if the direction of magnetic field is changed from perpendicular to parallel to the rails?

 

Ans:

Length of the rod, l  = 15 cm = 0.15 m

Strength of the magnetic field, B = 0.50 T

Resistance produced by the closed loop, R = 9.0  mΩ=9×103Ω9.0\; m\Omega = 9 \times 10^{-3} \Omega

(i) emf induced = 9 mV,

Polarity of the emf induced is in such a way that its P end is showing positive which the other end .ie. Q is showing negative.

Since, speed, v = 12cm/s = 0.12 m/s

Emf induced is: e = Bvl

= 0.5 × 0.12 × 0.15

= 9×103  v9 \times 10 ^{-3}\; v

= 9 mVs

Here, the polarity of the emf induced is a way that P end shows +ve and Q end shows -ve.

 

(ii) Yes, when the key K was opened then at both the end there was excess charge built up.

And excess charge were also built up when the key K was closed, and that charge was maintained by the continuous flow of current.

 

(iii) Because of the electric charge set up there were excess charge of opposite nature at both of the ends of the rod. Because of that the Magnetic force was is cancelled up.

When the key K is opened then there were no net force on the electrons in the rod PQ, and the rod was moving uniformly. It is because of the cancelled magnetic field on the rod.

 

(iv) Regarding force exerted on the rod, F = IBl

Where,

I = current flowing through the rod

=eR=9×1039×103=1  AF=1×0.5×0.15=75×103  N= \frac {e}{R} = \frac{9 \times 10^{-3}}{9 \times 10^{-3}} = 1 \; A\\ \\ ∴F = 1 \times 0.5 \times 0.15\\ \\ = 75 \times 10^{-3} \; N

 

(v) 9 mW,

No power will be expended when the key K will be opened.

Speed of the rod, v = 12 cm/s = 0.12 m /s

Hence,

Power, P = Fv

=75×103×0.12=9×103  W=9  mW\\= 75 \times 10^{-3} \times 0.12\\ \\ = 9 \times 10^{-3} \; W \\ \\ = 9 \; mW

When the key K is opened no power is expended.

 

(vi) 9mW,

Power is provided by an external agent.

Power loss in the form of heat = I2RI^{2}R 12×9×1031^{2} \times 9 \times 10^{-3}

= 9 mW

 

(vii) Zero (0)

There would be no emf induced in the coil. As the emf induces if the motion of the rod cuts the field lines. But in this case motion of the rod does not cut across the field lines.

 

Q 15. We have an air – cored solenoid having a length of 30 cm, whose area is 25  cm225 \; cm^{2} and number of turns is 500. And the solenoid is carried a current of 2.5 A. Suddenly the current in turned off and the time taken for it is 10310 ^{-3}s. What would be the average value of the induced back emf by the ends of the open switch in the circuit? (Neglect the variation in the magnetic fields near the ends of the solenoid.)

 

Ans:

Given,

Length of the solenoid, l = 30 cm = 0.3 m

Area of the solenoid, A =25  cm2=25×104  m225 \; cm^{2} = 25 \times 10^{-4} \; m^{2}

Number of turns on the solenoid, N = 500

Current in the solenoid, I = 2.5 A

Time duration for the current flow, t = 10310^{-3}

Average back emf, e=dϕdt            ...(1)Where,dϕ=e = \frac{d \phi}{dt} \;\;\;\;\;\; . . . (1)\\ \\Where, \\ \\ d \phi = change in flux

= NAB                                                  . . . . (2)

Where,

B = Strength of magnetic field

=μ0NIl                ...(3)= \mu_{0} \frac{NI}{l} \;\;\;\;\;\;\;\; . . . (3)

Where,

μ0\mu _{0} = Permeability of free space =  4n×107  T  m  A14 n \times 10 ^{-7} \; T\; m \; A^{-1}

Using equations (2) and (3) in equation (1), we get

e=μ0N2IAlt=4π×107×(500)2×25×1040.3×103=6.5V\\e = \frac{\mu _{0}N ^{2} I A}{lt} \\ \\ = \frac{4 \pi \times 10^{-7} \times \left ( 500 \right )^{2} \times 25 \times 10^{-4}}{0.3 \times 10^{-3}} = 6.5 V

Hence, the average back emf induced in the solenoid is 6.5 V.

 

 Q 16. (i) We are given a long straight wire and a square loop of given size (refer to figure). Find out an expression for the mutual inductance between both.

(ii) Now, consider that we passed an electric current through the straight wire of 50 A, and the loop is then moved to the right with constant velocity, v = 10 m/s.Find the emf induced in the loop at an instant where x = 0.2 m. Take a = 0.01 m and assume that the loop has a large resistance.

9 

Ans:

(i) Take a small element dy in the loop at a distance y from the long straight wire(as shown in the given figure).

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Magnetic flux associated with element dy,dϕ=adydy, \, d \phi = a \, dy

B = Magnetic field at distance y=μ0I2πyy = \frac{\mu _{0}I}{2 \pi y}

I = Current in the wire

μ0\mu _{0} = Permeability of free space = 4n×107dϕ=μ0Ia2πdyyϕ=μ0Ia2πdyyy  tends  from  x  a+xϕ=μ0Ia2πxa+xdyy=μ0Ia2π[logey]xa+x=μ0Ia2πlogea+xx4n \times 10^{-7}\\ \\ ∴ d\phi = \frac{\mu _{0}Ia}{2 \pi}\frac{dy}{y}\\ \\ \phi = \frac{\mu _{0}Ia}{2 \pi} \int \frac{dy}{y} \\ \\ y \; tends\; from \; x \; a + x\\ \\ ∴ \phi = \frac{\mu_{0}Ia}{2 \pi} \int_{x}^{a + x} \frac{dy}{y} \\ \\ = \frac{\mu_{0}Ia}{2 \pi} \left [ \log_{e} y \right ]_{x}^{a + x} \\ \\ = \frac{\mu_{0}Ia}{2 \pi} \log_{e} \frac{a + x}{x}

For mutual inductance M, the flux is given as:

ϕ=MIMI=μ0Ia2πloge(ax+1)    M=μ0a2πloge(ax+1)\phi = MI \\ \\∴ MI = \frac {\mu _{0} I a}{2 \pi} \log _{e}\left ( \frac{a}{x} + 1 \right ) \\ \\ \; \;M = \frac {\mu _{0} a}{2 \pi} \log _{e}\left ( \frac{a}{x} + 1 \right )

 

(ii) EMF induced in the loop, e = B’av=μ0I2πxav= \frac {\mu _{0} I}{2 \pi x} av

Given, I = 50 A

x = 0.2 m

a = 0.1 m

v = 10 m/s

e=4π×107×50×0.1×102π×0.2e=5×105  Ve = \frac {4 \pi \times 10 ^{-7} \times 50 \times 0.1 \times 10}{2 \pi \times 0.2} \\ \\ e = 5 \times 10^{-5}\; V

 

 

Q 17.A line charge λ per unit length is lodged uniformly onto the rim of a wheel of mass M and radius R. The wheel has light non-conducting spokes and is free to rotate without friction about its axis (Fig. 6.22). A uniform magnetic field extends over a circular region within the rim. It is given by,

B=B0k(ra;a<R)B = -B _{0} \, k \left ( r \leq a ; a < R \right )

= 0 (otherwise)

What is the angular velocity of the wheel after the field is suddenly switched off?

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Ans:

Line charge per unit length = λ=Total  chargeLength=Q2πr\lambda = \frac{Total \; charge}{Length} = \frac{Q}{2 \pi r}

Where,

r = Distance of the point within the wheel

Mass of the wheel = M

Radius of the wheel = R

Magnetic field, B=B0k˙\vec{B} = -B_{0} \dot{k}

At distance r, the magnetic force is balanced by the centripetal force i.e.,BQv=Mv2rBQv = \frac{Mv^{2}}{r}

Where,

v = linear velocity of the wheel

B2πrλ=Mvrv=B2πλr2MAngular  Velocity,ω=vR=Mvrv=B2πλr2MRFor  ra  and  a<R,we  getω=2πB0a2λMRk^\\∴ B2 \pi r \lambda = \frac {Mv}{r} \\ \\ v = \frac{B2 \pi \lambda r^{2}}{M}\\ \\ ∴ Angular \; Velocity, \omega = \frac{v}{R} = \frac {Mv}{r} \\ \\ v = \frac{B2 \pi \lambda r^{2}}{M R} \\ \\ For\; r \leq a \; and \; a < R, we\; get \\ \\ \omega = – \frac{2 \pi B_{0} a^{2} \lambda}{M R} \hat{k}

 

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1 Comment

  1. very good

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