Question

NCERT Solutions Class 12 Physics Chapter 6 Electromagnetic Induction is perfect study material for the preparation for the board exams and last-minute revision of difficult concepts. NCERT Solutions is based on the topic of electromagnetic induction. NCERT Solutions Class 12 is discussed in a detailed manner.

• Electromagnetic Induction – the electromagnetic induction electromotive force inside the electrical conductor in a variable magnetic field. The theory of magnetic induction was discovered by Michael Faraday in 1831. Faraday’s law of induction is mathematically described by James Clerk Maxwell.
• The Experiments of Faraday and Henry – in the experiment conducted by Henry and Faraday a coil was connected to a galvanometer and a bar magnet is inserted in the coil. The north pole of the magnet is pointed towards the coil. When the bar magnet experienced a translation it showed a deflection in the galvanometer.
• Magnetic Flux – the total magnetic field which passes through a given area is called the magnetic flux. It helps us describe the effects of the magnetic force on some occupying area. The measurement of magnetic flux is calculated in a particular area only.
• Faraday’s Law of Induction – the electromagnetic flux experienced in a circuit is proportional to the rate of change in the magnetic flux which passes across the circuit.
• Lenz’s Law and Conservation of Energy – the law of conservation of energy is related to Lenz's law which deals with the conservation of energy. The Lenz law shows that the induced current tends to resist the reason which produces it. The magnetic flux changes periodically.
• Motional Electromotive Force – the motional electromotive force is termed as the emf induced by the motion of the conductor around the given magnetic field.  Motional electromotive force is explained mathematically by E = -MLB.

Students must refer to our NCERT Solutions Class 12 Physics for clear tough concepts of physics. This solution is articulated by the mentors of the subject matter.

Answer 1

NCERT Solutions Class 12 Physics Chapter 6 Electromagnetic Induction

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

Answer:

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

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

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

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

(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.

2. We are rotating a 1 m long metallic rod with an angular frequency of 400 red s⁻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 a 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.

Answer:

Length of the rod = 1m

Angular frequency, ω = 400 rad/s

Magnetic field strength, B = 0.5 T

At one of the ends of the rod, it has zero liner velocity, while on to its other end it has a linear velocity of Iω

Average linear velocity of the rod, \(v = \frac{I \omega + 0}{ 2 } = \frac{I \omega}{2}\) 

Emf developed between the centre and ring.

\(e = Blv = Bl\left ( \frac{I \omega }{2} \right ) = \frac{B l^{2 } \omega}{2} \)

\(= \frac{0.5 \times \left ( 1 \right )^{2} \times 400}{2} = 100\; V\)

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

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?

Answer:

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

The 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:

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

Where,

ϕ = Induced flux through the small loop

= BA       .......(2)

B = Magnetic field

= μ₀​ni

μ_0​ = Permeability of free space

= 4π × 10⁻⁷ H/m

Hence, equation (1) can be reduced to:

\(e = \frac{d}{dt}\left ( BA \right )\)

\(e = A\, \mu _{0}\, n \times \left (\frac{di}{dt} \right )\)

\(= 2 \times 10^{-4} \times 4 \pi \times 10 ^{-7} \times 1500 \times \frac{2}{0.1}\)

\(= 7.54 \times 10^{-6}\; V\)

Hence, the induced voltage in the loop is 7.54 × 10⁻⁶ V

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?

Answer:

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 a rectangle, area of the wired loop,

A = lb

= 0.08 × 0.02

= 16 × 10⁻⁴ m²

Strength of magnetic field, B = 0.3 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 × 10⁻⁴V

Time taken to travel along the width ,t = \(\frac{Distance \; travelled}{Velocity} = \frac{b}{v}\)

\(= \frac{0.02}{0.01} = 2s\)

Hence, the induced voltage is 2.4 × 10⁻⁴ V which lasts for 2 s.

(ii) Emf developed, e = Bbv

= 0.3 × 0.02 × 0.01 = 0.6 × 10⁻⁴ V

Time taken to travel along the length, t = \(\frac{Distance \; travelled}{Velocity} = \frac{l}{v}\)

\(= \frac{0.08}{0.01} = 8s\)

Hence, the induced voltage is 0.6 × 10⁻⁴ V which lasts for 8 s.

5. A 1.0 m long metallic rod is rotated with an angular frequency of 400 rad s^-1 about an axis normal to the rod passing through its one end. The other end of the rod is in contact with a circular metallic ring. A constant and uniform magnetic field of 0.5 T parallel to the axis exists everywhere. Calculate the emf developed between the centre and the ring.

Answer:

Length of the rod, l = 1 m

Angular frequency, \(\omega\) = 400 rad/s

Magnetic field strength, B = 0.5 T

One end of the rod has zero linear velocity, while the other end has a linear velocity of Ιω.

Average linear velocity of the rod, \(v = \frac{l\omega + 0}{2} = \frac{l\omega}{2}\) 

Emf devloped between the centre and the ring,

\(e = Blv = Bl \left(\frac{l\omega}2\right) = \frac{Bl^2\omega}{2}\)

\( = \frac{0.5\times(1)^2\times400}{2} = 100\,V\)

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

Answer 2

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?

Answer:

Maximum emf induced = 0.603 V

Average emf induced = 0 V

Maximum current in the coil = 0.0603 A

Power loss (average) = 0.018 W

(Power which is coming from external rotor)

Circular coil radius, r = 8 cm = 0.08 m

Area of the coil, A = πr² = π × (0.08)² m²

Number of turns on the coil, N = 20

Angular speed, ω = 50 rad/s

Strength of magnetic, B = 3 × 10⁻² T

Total resistance produced by the loop, R = 10Ω

Maximum emf induced is given as:

e = Nω AB

= 20 × 50 × π × (0.08)² × 3 × 10⁻²

= 0.603V

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:

\(I = \frac{e}{R} \)

\(= \frac{0.603}{10} = 0.0603\; A\)

Average power because of the Joule heating:

\(P = \frac{el}{2} \)

\(= \frac{0.603 \times 0.0603}{2} = 0.018\; W\)

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.

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?

Answer:

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 × 10⁻⁴ Wb m⁻²

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

= 0.3 × 10⁻⁴ × 5 × 10

= 1.5 × 10⁻³ V

(b) We can determine the direction of the induced current by using 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

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.

Answer:

Current at initial point, I₁​ :  5.0 A

Current at final pint, I₂ ​: 0.0 A

Therefore, change in current is, dI = I_1​ – I₂​ = 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 :

\(e = L \frac{di}{dt}  \)

\(L = \frac{e}{\left (\frac{di}{dt} \right )}  \)

\(= \cfrac{200}{\frac{5}{0.1}} = 4 H\)

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

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?

Answer:

Given,

Mutual inductance, μ = 1.5H

Current at initial point, I₁​ :  0 A

Current at final point, I₂ ​: 20 A

Therefore, change in current is, dI = I₂​ – I₁​ = 20 – 0 = 20 A

Time taken for the change, t = 0.5s

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

dϕ = change in the flux linkage with the coil.

Relation of emf and inductance is:

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

On equating both the equation, we get

\(\frac{d\phi }{dt} = \mu \frac{dl}{dt}\)

\(\frac{d\phi }{dt} = 1.5 \times \left ( 20 \right ) \)

= 30 Wb

Therefore, the change in flux linkage is 30 Wb.

10. A jet plane is travelling towards the 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°.

Answer:

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 × 10⁻⁴ T

Dip angle,δ = 30^∘

Vertical component of Earth’s magnetic field,

B_v ​= Bsinδ

= 5 × 10⁻⁴ sin30^∘

= 2.5 × 10⁻⁴ T

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

e = (B_v​) × l × v

= 2.5 × 10⁻⁴ × 25 × 500

= 3.125V

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

11. Let us assume that the loop in 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 1.6Ω. Calculate how much power is lost in the form of heat? What is the source of this power?

Answer:

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

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

= 8 cm × 2 cm

= 16cm²

= 16 × 10⁻⁴ cm²

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

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

Emf induced in the loop is:

\(e = \frac{d\phi }{dt}\)

dϕ = change in flux in the loop area

= AB

\(∴ e = \frac{d\left ( AB \right )}{dt} = \frac{AdB}{dt}\)

= 16 × 10⁻⁴ × 0.02 = 0.32 × 10⁻⁴ V

Resistance in the loop will be, R = 1.6Ω

The current developed in the loop will be:

\(i = \frac{e}{R} \)

\(= \frac{032 \times 10 ^{-4}}{1.6} = 2 \times 10^{-5} A\) 

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

P = i²R

= (2 × 10⁻⁵)² × 1.6

= 6.4 × 10⁻¹⁰ W

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

Answer 3

12. We have a square loop having side as 12 cm and its sides are parallel to 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 10⁻³ T cm⁻¹ as we move from positive to negative direction), and it is reducing in the case of time with the rate of  10⁻³ Ts⁻¹. Calculate the magnitude and direction of induced current in the loop (Given: Resistance = 4.50 mΩ).

Answer:

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

Area of the loop, A = s × s = 0.12 × 0.12 = 0.0144 m²

Velocity of the loop, v = 8 s⁻¹ = 0.08 ms⁻¹

Gradient of the magnetic field along negative x-direction,

\(\frac{dB}{dx} = 10^{-3} \; T\,cm^{-1} = 10^{-1} \; m^{-1}\)

And, the rate of decrease of the magnetic field,

\(\frac{dB}{dt} = 10^{-3} \; T \, s^{-1}\)

Resistance, R = 4.50 mΩ = 4.5 × 10⁻³Ω

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

\(\frac{dB}{dt} = A \times \frac{dB}{dx} \times v \)

\(=144 \times 10 ^{-4} m ^{2} \times10^{-1} \times 0.08\)

\(= 11.52 \times 10 ^{-5} \; Tm^{2} s^{-1}\)

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

\(\frac{d\phi }{dt} = A \times \frac{dB}{dt} \)

\(= 144 \times 10^{-4} \times 10 ^{-3} \)

\(= 1.44 \times 10^{-5} T\,m^{2}s^{-1}\) 

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

\(e = 1.44 \times 10^{-5} + 11.52 \times 10^{-5}\)

\(= 12.96 \times 10^{-5} V\)

\(∴ Induced \; current, \; i = \frac{e}R\)

\(= \frac{12.96 \times 10 ^{-5}}{4.5 \times 10 ^{-3}}\)

\(i = 2.88 \times 10^{-2}\; A\)

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

13. We have a powerful loudspeaker magnet and have to measure the magnitude of the 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 2 cm² area and has 25 closely wound turns. Similarly, we can give the coil a quick 90^∘ 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 0.50Ω. Estimate the field strength of the magnet.

Answer:

Given,

Coil’s Area, A = 2cm² = 2 × 10⁻⁴ m²

Number of turns on the coil, N = 25

Total Charge in the coil, Q = 7.5 mC = 7.5 × 10⁻³ C

Total resistance produced by the combo of coil and galvanometer, R = 0.50Ω

Current generated in the coil,

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

EMF induced is shown as:

\(e = -N\frac{d\Phi}{dt}\)​              .......(2)

Where,

dϕ = Change in flux

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

\(I = -\cfrac{N\frac{d\Phi }{dt}}{R} \)

\(Idt = -\frac{N}{R}d \phi\)             ...(3)

Flux through the coil at initial phase, ϕ_i​ = BA

Where B = Strength of the magnetic field

Flux through the coil at final phase, ϕf​ = 0

After integrating equation (3) on both of the sides, we get

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

Total Charge, \(Q=\int Idt\)

\(∴ Q = \frac{-N}{R}\left ( \phi_{f} – \phi_{i} \right )= \frac{-N}{R}\left ( – \phi_{i} \right ) = + \frac{N \phi_{i}}{R} \)

\(Q = \frac{NBA}{R}\)

\(∴ B = \frac {QR}{NA}\)

\(= \frac {7.5 \times 10^{-3} \times 0.5}{25 \times 2 \times 10^{-4}} = 0.75 \; T\)

Hence, the field strength is 0.75 T.

Answer 4

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 permanent magnets. All these three (rod, rails and the magnetic field ) are in mutually 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 mΩ. Let’s consider the field is uniform.

(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 was moving uniformly and the K was open, then on the electron in the rod PQ there was no net force even though they did not experience 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 the magnetic field is changed from perpendicular to parallel to the rails?

Answer:

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 × 10⁻³Ω

(i) The 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 × 10⁻³ V

= 9 mV

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 was 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 was 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 was 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

\(= \frac {e}{R} = \frac{9 \times 10^{-3}}{9 \times 10^{-3}} = 1 \; A\)

\( ∴F = 1 \times 0.5 \times 0.15\)

\( = 75 \times 10^{-3} \; N\)

(v) 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

\(= 75 \times 10^{-3} \times 0.12\)

\(= 9 \times 10^{-3} \; W\)

= 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 is given as follows:

= I²R

= 1² × 9 × 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 movement of the rod does not cut across the field lines.

15. We have an air-cored solenoid having a length of 30 cm, whose area is 25 cm² and a number of turns are 500. And the solenoid has carried a current of 2.5 A. Suddenly the current is turned off and the time is taken for it is 10⁻³ 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.)

Answer:

Given,

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

Area of the solenoid, A = 25cm² = 25 × 10⁻⁴ m²

Number of turns on the solenoid, N = 500

Current in the solenoid, I = 2.5 A

Time duration for the current flow, t = 10⁻³

Average back emf,

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

Where,

dϕ= change in flux

= NAB   ..........(2)

Where,

B = Strength of the magnetic field

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

Where,

μ₀​ = Permeability of free space =  4π × 10⁻⁷ Tm A⁻¹

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

\(e = \frac{\mu _{0}N ^{2} I A}{lt}\)

\(= \frac{4 \pi \times 10^{-7} \times \left ( 500 \right )^{2} \times 2.5 \times 25 \times 10^{-4}}{0.3 \times 10^{-3}} = 6.5 V\)

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

Answer 5

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.

Answer:

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

Magnetic flux associated with element dy, dϕ =a dy

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

I = Current in the wire

μ₀​ = Permeability of free space = 4π × 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 to 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:

\(\phi = MI\)

\(∴ MI = \frac {\mu _{0} I a}{2 \pi} \log _{e}\left ( \frac{a}{x} + 1 \right )\)

\(M = \frac {\mu _{0} a}{2 \pi} \log _{e}\left ( \frac{a}{x} + 1 \right )\) 

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

Given, I = 50 A

x = 0.2 m

a = 0.1 m

v = 10 m/s

\(e = \frac {4 \pi \times 10 ^{-7} \times 50 \times 0.1 \times 10}{2 \pi \times 0.2}\)

\(e = 5 \times 10^{-5}\; V\)

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 in fig.  A uniform magnetic field extends over a circular region within the rim. It is given by,

B = −B₀​k (r ≤ a; a < R) = 0 (otherwise) 

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

Answer:

Line charge per unit length = \(\lambda = \frac{Total \; charge}{Length} = \frac{Q}{2 \pi r}\)

Where,

r = Distance of the point within the wheel

Mass of the wheel = M

The radius of the wheel = R

Magnetic field, \( \vec{B} = -B_{0} \dot{k}\) 

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

Where,

v = linear velocity of the wheel

\(∴ B2 \pi r \lambda = \frac {Mv}{r}\)

\(v = \frac{B2 \pi \lambda r^{2}}{M}\)

∴ Angular Velocity, \(\omega = \frac{v}{R}\)

\(= \frac{B2 \pi \lambda r^{2}}{M R}\)

For r \(\leq\) a and a < R, we get

\(\omega =\, – \frac{2 \pi B_{0} a^{2} \lambda}{M R} \hat{k}\)