## Electromagnetic Induction

### Maharashtra Board-Class-12th-Physics-Chapter-12

### Solution

**Question 1. Choose the correct option.**

**(i) A circular coil of 100 turns with a cross− sectional ****area (A) of 1 m ^{2} is kept with **

**its plane perpendicular to the magnetic field (B) of 1 T. What is the magnetic flux linkage with the coil?**

**(A) 1 Wb **

**(B) 100 Wb**

**(C) 50 Wb **

**(D) 200 Wb**

(B) 100 Wb

**(ii) A conductor rod of length (***l***) is moving with velocity (v) in a direction normal to a uniform magnetic field (B). What will be the magnitude of induced emf produced between the ends of the moving conductor?**

**(A) BLv **

**(B) BLv ^{2}**

**(C) \(\frac{1}{2}\)***Bl***v **

**(D) \(\frac{2Bl}{v}\)**

(A) BLv

**(iii) Two inductor coils with inductance 10 mH and 20 mH are connected in series. What is the resultant inductance of the combination of the two coils?**

**(A) 20 mH **

**(B) 30 mH**

**(C) 10 mH **

**(D) \(\frac{20}{3}\)** ** mH**

(B) 30 mH

**(iv) A current through a coil of self inductance 10 mH increases from 0 to 1 A in 0.1 s. What is the induced emf in the coil?**

**(A) 0.1 V **

**(B) 1 V**

**(C) 10 V **

**(D) 0.01 V**

(A) 0.1 V

**(v) What is the energy required to build up a current of 1 A in an inductor of 20 mH?**

**(A) 10 mJ **

**(B) 20 mJ**

**(C) 20 J **

**(D) 10 J**

(A) 10 mJ

**Question 2. Answer in brief.**

**(i) What do you mean by electromagnetic induction? State Faraday’s law of induction.**

The phenomenon of production of emf in a conductor or circuit by a changing magnetic flux through the circuit is called electromagnetic induction.

Faraday's laws of electromagnetic induction :

**First law**: Whenever there is a change in the magnetic flux associated with a circuit, an emf is induced in the circuit.**Second law**: The magnitude of the induced emf is directly proportional to the time rate of change of magnetic flux through the circuit.

**(ii) State and explain Lenz’s law in the light of principle of conservation of energy.**

**Lenz’s law** : The direction of the induced current is such as to oppose the change that produces it. The change that induces a current may be

- the motion of a conductor in a magnetic ﬁeld or
- the change of the magnetic flux through a stationary circuit.

**Explanation** :

Consider Faraday's magnet− and− coil experiment.

- If the bar magnet is moved towards the coil with its N− pole facing the coil, as in below Fig. (a), the number of magnetic lines of induction (pointing to the left) through the coil increases.
- The induced current in the coil sets up a magnetic ﬁeld of its own pointing to the right (as given by Amperes right− hand rule) to oppose the growing ﬂux due to the magnet.
- Hence, to move the magnet towards the coil against this repulsive ﬂux of the induced current, we must do work. The work done shows up as electric energy in the coil.
- When the magnet is withdrawn, with its N− pole still facing the coil, the number of magnetic lines of induction (pointing left) through the coil decreases.
- The induced current reverses its direction to supplement the decreasing ﬂux with its own, as shown in Fig. (b).
- Facing the coil along the magnet, the induced current is in the clockwise sense.
- The electric energy in the coil comes from the work done to withdraw the magnet, now against an attractive force.
- Thus, we see that Lenz’s law is a consequence of the
**law of conservation of energy**.

**(iii) What are eddy currents? State applications of eddy currents.**

Whenever a conductor or a part of it is moved in a magnetic field "cutting" magnetic field lines, or placed in a changing magnetic field, the free electrons in the bulk of the metal start circulating in closed paths equivalent to current− carrying loops.

These loop currents resemble eddies in a fluid stream and are hence called eddy or Foucault currents

**Applications **:

**Dead− beat galvanometer**: A dead− beat galvanometer measures current by wounding a coil on a light aluminum frame, producing eddy currents that oppose rotation and bring the coil to rest quickly. This makes the galvanometer dead− beat.**Electric brakes**: In electric brakes, the interaction between eddy currents in a moving conductor and the field retards motion, making it useful for braking in vehicles. The coil's rotation in a magnetic field is a key component of this mechanism.

**(iv) If a copper disc swings between the poles of a magnet, the pendulum comes to rest very quickly. Explain the reason. What happens to the mechanical energy of the pendulum?**

- As the copper disc enters and leaves the magnetic field, the changing magnetic flux through it induces eddy current in the disc.
- In both cases, Fleming's right hand rule shows that opposing magnetic force damps the motion.
- After a few swings, the mechanical energy becomes zero and the motion comes to a stop and the disc's heating due to the eddy current warms it, transforming the mechanical energy of the pendulum into thermal energy.

**(v) Explain why the inductance of two coils connected in parallel is less than the inductance of either coil.**

We assume that the inductors are so far apart that their mutual inductance is negligible.

If several inductances L_{1,} L_{2,} L_{3,} ... are connected in parallel, then the total inductance is determined by using following relations:

1/L_{Total }= 1/L_{1} + 1/L_{2} + 1/L_{3} + ...... (Parallel Combination)

Assuming that their mutual inductance can be ignored, the equivalent inductance of a parallel combination of two coils is given by

1/L_{Total }= 1/L_{1} + 1/L_{2}

∴ L_{Total }= \(\frac{L_1L_2}{L_1+L_2}\)

Hence, the inductance of two coils connected in parallel is less than the inductance of either coil.

**Question 3. In a Faraday disc dynamo, a metal disc of radius R rotates with an angular velocity **

**ω**

**about an axis perpendicular to the plane of the disc and passing through its centre. The disc is placed in a magnetic field**

*B*acting perpendicular to the plane of the disc. Determine the induced emf between the rim and the axis of the disc.Suppose a thin conducting disc of radius R is rotated anticlockwise, about its axis, in a plane perpendicular to a uniform magnetic field of induction (see Fig.) points downwards. Let the constant angular speed of the disc be ω.

Consider an infinitesimal element of radial thickness dr at a distance r from the rotation axis. In one rotation, the area traced by the element is dA =2 πrdr. Therefore, the time rate at which the element traces out the area is

\(\frac{dA}{dt}\) = frequency of rotation x dA = fdA

Where f = \(\frac{ω}{2π}\) is the frequency of rotation.

∴ \(\frac{dA}{dt}\) = \(\frac{ω}{2π}\)(2πrdr) = ωrdr

The total emf induced between the axle and the rim of the rotating disc is

| e | = ∫B\(\frac{dA}{dt}\)

= \(\int_0^R\)ωrdr

= Bω\(\int_0^R\)rdr

= Bω\(\frac{R^2}{2}\)

**Question 4. A horizontal wire 20 m long extending from east to west is falling with a velocity of 10 m/s normal to the Earth’s magnetic field of 0.5 × 10 ^{− 4} T. What is the value of induced emf in the wire?**

**Given** : *l* = 20 m, v = 10 m/s, B = 5 x 10^{− 5} T

The magnitude of the induced emf,

| e | = B *l *v = (5 × 10^{− 5})(20)(10) = 10^{− 2} V = 10 mV

**Question 5. A metal disc is made to spin at 20 revolutions per second about an axis passing through its centre and normal to its plane. The disc has a radius of 30 cm and spins in a uniform magnetic field of 0.20 T, which is parallel to the axis of rotation. Calculate**

**(a) The area swept out per second by the radius of the disc,**

**(b) The flux cut per second by a radius of the disc,**

**(c) The induced emf in the disc.**

**Given** : R = 0.3 m, f = 20 rps, B = 0.2 T

**(a)** The area swept out per unit time by a given radius = (the frequency of rotations) x (the area swept out per rotation) = f(πr^{2})

= (20)(3.142 × 0.3^{2}) = **5.656 m ^{2}**

**(b)** The time rate at which a given radius cuts magnetic flux = \(\frac{dΦ_m}{dt}\) = Bf(πr^{2})

= (0.2)(5.656) = **1.131 Wb/s**

**(c)** The magnitude of the induced emf, | e | = \(\frac{dΦ_m}{dt}\) = **1.131 V**

**Question 6. A pair of adjacent coils has a mutual inductance of 1.5 H. If the current in one coil changes from 0 to 10 A in 0.2 s, what is the change of flux linkage with the other coil?**

**Given** : M = 1.5 H, I_{1i} = 0, I_{1f} =10 A, Δt = 0.2 s

The flux linked per unit turn with the second coil due to current I_{i} in the first coil is

Φ_{21} = MI_{1}

Therefore, the change in the flux due to change in I_{1} is

ΔΦ_{21 }= M(ΔI_{1}) = M(I_{1f} − I_{1i}) = 1.5(10 − 0) = **15 Wb**

**Question 7. A long solenoid consisting of 1.5 × 10 ^{3} turns/m has an area of cross− section of 25 cm^{2}. A coil C, consisting of 150 turns (N_{C}) is wound tightly around the centre of the solenoid. Calculate for a current of 3.0 A in the solenoid**

**(a) the magnetic flux density at the centre of the solenoid,**

**(b) the flux linkage in the coil C,**

**(c) the average emf induced in coil C if the current in the solenoid is reversed in direction in a time of 0.5 s. ( **μ_{0}** = 4****π**** × ****10 ^{− 7} T.m/A)**

**Given** : n = 1.5 x 10^{3} m^{− 1}, A = 25 x 10^{− 4} m^{2}, N_{C} = 150, I = 3A, Δt = 0.5 s,

μ_{0} = 4π x 10^{− 7} H/m

(a) Magnetic flux density inside the solenoid, B = μ_{0}nI = (4π x 10^{− 7})(1500)(3)

= 5.656 × 10^{− 3} T = **5.656 mT**

(b) Flux per unit turn through the coils of the solenoid, Φ_{m} = BA

Since the coil C is wound tightly over the solenoid, the flux linkage of C is

N_{C}Φ_{m} = N_{C}BA = (150)(5.656 × 10^{− 3})(25 × 10^{− 4})

= 2.121 x 10^{− 3} Wb = **2.121 mWb**

**(c)** Initial flux through coil C, Φ_{i} = N_{C}Φ_{m} = 2.121 x 10^{− 3} Wb

Reversing the current in the solenoid reverses the flux through coil C, the magnitude remaining the same. But since the flux enters through the other face of the coil, the final flux through C is Φ_{f} = − 2.121 x 10^{− 3} Wb

Therefore, the average emf induced in coil C,

e = \(− \frac{Φ_f− Φ_i}{Δt}\) = \(− \frac{(− 2.121− 2.121)×10^{− 3}}{0.5}\)

= 2 × 4.242 × 10^{− 3} = 8.484 x 10^{− 3} V = **8.484 mV**

**Question 8. A search coil having 2000 turns with area 1.5 cm ^{2} is placed in a magnetic field of 0.60 T. The coil is moved rapidly out of the field in a time of 0.2 second. Calculate the induced emf across the search coil.**

**Given** : N = 2000, A_{i} = 1.5 x 10^{− 4} m^{2}, A_{f} = 0, B = 0.6 T, Δ_{t} = 0.2s

Initial flux, NΦ_{i} = NBA_{i} = 2000(0.6)(1.5 × 10^{− 4}) = 0.18 Wb

Final flux, NΦ_{f} = 0, since the coil is withdrawn out of the field.

Induced emf, e = − N\(\frac{dΦ_m}{dt}\) = − N\(\frac{Φ_f− Φ_i}{Δt}\) = \(− \frac{0− 0.18}{0.2}\) = **0.9 V**

**Question 9. An aircraft of wing span of 50 m flies horizontally in earth’s magnetic field of 6 × 10 ^{− 5} T at a speed of 400 m/s. Calculate the emf generated between the tips of the wings of the aircraft.**

**Given : ***l* = 50 m, B = 6 x 10^{− 5} T, v = 400 m/s

The magnitude of the induced emf,

| e | = B* l* v = (6 × 10^{− 5})(400)(50) = 1.2 V

**Question 10. A stiff semi− circular wire of radius R is rotated in a uniform magnetic field B about an axis passing through its ends. If the frequency of rotation of wire be f, calculate the amplitude of alternating emf induced in the wire.**

In one rotation, the wire traces out a circle of radius R, i.e., an area A = πR^{2}.

Therefore, the rate at which the wire traces out the area is

\(\frac{dA}{dt}\) = frequency or rotation x A = fA

If the angle between the uniform magnetic field and the rotation axis is θ, the magnitude of the induced emf is

| e | = B\(\frac{dA}{dt}\)cos θ = BfA cos θ = Bf(πR^{2}) cos θ

so that the required amplitude is equal to Bf(πR^{2}).

**Question 11. Calculate the value of induced emf between the ends of an axle of a railway carriage 1.75 m long travelling on level ground with a uniform velocity of 50 km per hour. The vertical component of Earth’s magnetic field (B _{v}) is given to be 5 × 10^{− 5} T.**

**Given**: *l* = 1.75 m, v = 50 km/h = 50 x \(\frac{5}{18}\) m/s, B_{v} = 5 x 10^{− 5} T

The area swept out by the wing per unit time = *l*v.

∴ The magnetic flux cut by the wing per unit time = \(\frac{dΦ_m}{dt}\) = B(*l* v)

= (5 × 10^{− 5})(1.75)( 50 x \(\frac{5}{18}\)) = 121.5 × 10^{− 5}

= 1.215 mWb/s

Therefore, the magnitude of the induced emf,

| e | =** 1.215 mV**

**Question 12. The value of mutual inductance of two coils is 10 mH. If the current in one of the coil changes from 5A to 1A in 0.2 s, calculate the value of emf induced in the other coil. Also calculate the value of induced charge passing through the coil if its resistance is 5 ohm.**

**Given** : M = 10 mH = 10^{− 2} H, I_{1i} = 5 A, I_{1f} = 1 A, Δt = 0.2 s, R = 592

The mutually induced emf in coil 2 due to the changing current in coil 1,

e_{21 }= − M\(\frac{ΔI_1}{Δt}\) = − M\(\frac{I_{1f}− I_{1i}}{Δt}\)\(\frac{1− 5}{0.2}\) = − (10^{− 2}) = 0.2

If ΔQ_{2} is the charge that flows through coil 2 due to the changing current in coil 1, the induced current in coil 2 is

I_{2} = \(\frac{ΔQ_2}{Δt}\) = \(\frac{e_2}{R_2}\)

∴ ΔQ_{2 } = \(\frac{e_2}{R_2}\)Δt = \(\frac{0.2}{5}\)(0.2) = \(\frac{0.04}{5}\) = **0.008 C = 8 mC**

**Question 13. An emf of 96.0 mV is induced in the windings of a coil when the current in a nearby coil is increasing at the rate of 1.20 A/s. What is the mutual inductance (M) of the two coils?**

**Given** : |e_{2}| = 9.6 x 10^{− 2} V, \(\frac{dI_1}{dt}\) = 1.2 A/s

|e_{2}| = M\(\frac{dI_1}{dt}\)

Mutual Inductance, M = \(\frac{e_2}{dI_1/dt}\) = \(\frac{9.6×10^{-2}}{1.2}\) = 8 x 10^{− 2} = **80 mH**

**Question 14. A long solenoid of length l, cross− sectional area A and having N_{1} turns (primary coil), has a small coil of N_{2} turns (secondary coil) wound about its centre. Determine the Mutual inductance (M) of the two coils.**

We assume the solenoid to be ideal and that all the ﬂux from the solenoid passes through the outer coil. For a steady current I_{S} through the solenoid, the uniform magnetic ﬁeld inside the solenoid is

B = μ_{0}nI_{S} = \(\frac{μ_0N_1I_S}{l}\) …..(1)

n = N_{1}/*l *is the number of turns of the wire per unit length

Then, the magnetic ﬂux through each turn of the coil due to the current in the solenoid is

Φ_{CS} = BA = \(\frac{μ_0N_1I_S}{l}\)A …….(2)

Thus, mutual inductance of two coil is

M = \(\frac{N_2Φ_{CS}}{I_S}\) = \(\frac{μ_0N_1N_2I_S}{I_S×l}\) = \(\frac{μ_0N_1N_2}{l}\)

**Question 15. The primary and secondary coil of a transformer each have an inductance of 200 ×10 ^{− 6} H. The mutual inductance (M) between the windings is 4×10^{− 6} H. What percentage of the flux from one coil reaches the other?**

Given : L_{P} = L_{S} = 2 x 10^{− 4} H, M = 4 x 10^{− 6} H

M = K\(\sqrt{L_PL_S}\)

The coupling coefficient is

K = \(\frac{M}{\sqrt{L_PL_S}}\)

= \(\frac{4×10^{-6}}{\sqrt{(2×10^{-4})^2}}\)

= \(\frac{4×10^{-6}}{2×10^{-4}}\) = 2 x 10^{− 2}

Therefore, the percentage of flux of the primary reaching the secondary is = 0.02 × 100% = **2%**

**Question 16. A toroidal ring, made from a bar of length (***l***) 1 m and diameter ( d) 1 cm, is bent into a circle. It is wound tightly with 100 turns per cm. If the permeability of bar is equal to that of free space (**

**μ**

_{0}**), calculate the magnetic field inside the bar (**

*B*) when the current (*i*) circulating through the turns is 100 A. Also determine the self− inductance (*L*) of the coil.**Given** : *l* = 1 m, d = 1 cm, n = 100 cm^{− 1} = 10^{4} m^{− 1}, I = 100 A, μ_{0} = 4π x 10^{− 7} H/m

The radius of cross section, r = d/2 = 0.5 cm = 5 × 10^{− 3} m

**(a)** Magnetic field inside the toroid,

B = μ_{0}nI = (4π x 10^{− 7})(10^{4})(100) = 0.4 × 3.142 = **1.257 T**

**(b)** Self inductance of the toroid,

L = μ_{0}2πRn^{2}A = μ_{0}n^{2}*l*Α = μ_{0}n^{2}*l*(πr^{2})

=(4π x 10^{− 7})(10^{4})^{2}(1) [π (5 × 10^{− 3})^{2}]

= π^{2 }x 10^{− 3} = 9.87 × 10^{− 3} H = **9.87 mH**

**Question 17. A uniform magnetic field B(t), pointing upward fills a circular region of radius, s in horizontal plane. If B is changing with time, find the induced electric field.**

The area of the region, A = πs^{2}, remains constant while B = B(t) is a function of time.

Therefore, the induced emf,

e = \(-\frac{dΦ_m}{dt}\) = \(−\frac{d}{dt}\)(BA) = −A\(\frac{dB(t)}{dt}\) = − πs^{2}\(\frac{dB(t)}{dt}\)

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