WBCHSE Class 12 Physics Electromagnetic Induction Short Question And Answers

Electromagnetic Induction And Alternating Current

Electromagnetic Induction Short Questions And Answers

Question 1. Use Lenz’s law to determine the direction of induced current in the situations described:

1. A wire of irregular shape turns into a circular shape.
2. A circular loop is deformed into a narrow straight wire.

Answer:

1. In this case, fine magnetic flux linked with the coil increases with the change in shape of the loop. The direction of the: induced current should be such that it will oppose the increase, the direction of the induced current should produce a magnetic Field in a direction upwards on the plane of the paper. Applying the thumb rule it can be said that the induced current will be in the direction adeba.
2. In this case, the magnetic flux linked with the loop decreases. So the induced current will be along a’d’c’b’a’.

Question 2. A rectangular wire loop of sides 8 cm and 2 cm with a small cut is moving out of a region of uniform magnetic field of magnitude 0.3 T directed normally to the loop. What is the emf developed across the cut if the velocity of the loop is 1cm.s-1 in a direction normal to the

1. Longer side,
2. Shorter side, of the loop? For how long does the induced voltage last in each case?
3. Suppose in this case the loop is stationary but the current feeding the electromagnet that produces the magnetic field is gradual. reduced so that the field decreases from its initial value of 0.3 T at the rate of 0.02 T s^-1. If the cut is joined and the loop has a resistance of 1.611, how much power is dissipated by the loop as heat? What is the source of this power?

Answer:

1. e = Blv = 0.3 x 8 X 10^-2 X 10^-2 [in the longer side, l = 8 cm ]

= 2.4 X 10^-4 V

∴ Time, t1 = \(t_1=\frac{l^{\prime}}{v}=\frac{2}{l}=2 \mathrm{~s}\) [l’ = shorter side = 2 cm]

2. e = Bl’v = 0.3 x 2 x 10^-2 x 10^-2 = 0.6 x 10^-4 V

∴ Time, \(t_2=\frac{l}{v}=\frac{8}{1}=8 \mathrm{~s}\)

3. Induced emf, \(e=\frac{d \phi}{d t}=\frac{d B}{d t} \cdot A\)

= 0.02 x 8 x 2 x 10-4 = 3.2 x 10^-5 V

Induced current, \(I=\frac{e}{R}=\frac{3.2}{1.6} \times 10^{-5} \mathrm{~A}=2 \times 10^{-5} \mathrm{~A}\)

Power dissipated = e x I = 6.4 x 10^-10 W

The source of this power is the external agency which brings change in a magnetic field.

WBBSE Class 12 Electromagnetic Induction Short Q&A

Question 3. Indicate the direction of induced current in each case.

Answer:

1. According to Lenz’s law S-pole forms at q and N-pole forms at p. Therefore induced current will flow along qrp.
2. S-pole will be formed at both q And x. Therefore induced current will flow along prq in the first coil and along yzx in the second coil.
3. Along xyz.
4. Along zyx.
5. Along xry.
6. Since the lines of force are in the same plane as the coil, no emf will be induced. So there will be no induced current.

Question 4. A 1.0 m long metallic rod is rotated with an angular velocity 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:

∴ \(e=\frac{B l v}{2}=\frac{0.5 \times 1 \times 1 \times 400}{2}=100 \mathrm{~V}\)

Key Concepts in Electromagnetic Induction Short Answers

Question 5. A circular coil of radius 8.0 cm and 20 turns rotates about its vertical diameter with an angular speed of 50 rad.s^-1 in a uniform horizontal magnetic field of magnitude 3.0 x 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:

Flux through each turn of the coil,

∴ \(\phi=\pi r^2 B \cos (\omega t)\)

∴ \(e=-N \frac{d \phi}{d t}=-N \pi r^2 B \frac{d}{d t} \cos \omega t=N \pi r^2 B \omega \sin \omega t\)

∴ \(e_{\max }=N \pi r^2 B \omega\)

= 20 x 3.14 x (8 X 10^-2)² x 3 x 10^-2 x 50

= 0.603V

The average emf over a complete cycle is zero.

∴ \(I_{\max }=\frac{e_{\max }}{r}=\frac{0.603}{10}=0.0603 \mathrm{~A}\)

Average dissipated power = \(\frac{1}{2} e_{\max } \times I_{\max }\)

= \(\frac{1}{2} \times 0.603 \times 0.0603=0.018 \mathrm{~W}\)

The induced emf produces a torque, which opposes the motion of the coil. An external machine is to be used which will produce an equal and opposite torque to maintain the motion of the coil. This machine supplies the power apart which is being transferred to heat.

Question 6. A horizontal straight wire 10 m long extending from east to west is falling with a speed of 5 m s-1, at right angles to the horizontal component of earth’s magnetic field, 0.30 x 10^-4 Wb.m^-2.

1. What is the instantaneous value of the emf induced in the wire?
2. What is the direction of the emf?
3. Which end of the wire is at higher electrical potential?

Answer:

1. e = Bvl

∴ e = 0.30 X 10-4 X 5 X 10V

[∵ B = 0.30 x 10^-4 Wb m^-2, v = 5.0 m.s^-1, l = 10m]

⇒ 1.5 x 10^-3V

2. Applying Fleming’s right-hand rule.it is seen that the direction of the induced emf is from west to east.

3. The west end is at a higher potential.

Short Answer Questions on Faraday’s Law

Question 7. A square loop of side 12 cm with its sides parallel to the X and Y axes is moved with a velocity of 8 cm.s^-1 in the positive x-direction in an environment containing a magnetic field in the positive z-direction. The field is neither uniform in space nor constant in time. It has a gradient of 10^-3 T.cm^-1 along the negative x-direction and is decreasing in time at the rate of 10^-3 T.s^-1. Determine the direction and magnitude of induced current in the loop if its resistance is 4.50mil.

Answer:

Rate of change due to variation with time

= \(A \cdot \frac{d B}{d t}=\left(12 \times 10^{-2}\right)^2 \times 10^{-3}\)

= 1.44 x 10^-5 Wb.s^-1

Rate of change due to variation in space

= \(A \cdot \frac{d B}{d x} \cdot \frac{d x}{d t}=\left(12 \times 10^{-2}\right)^2 \times 10^{-3} \times 8 \times 10^{-2}\)

= 11.52 x l0^-5 Wb s^-1

Total rate of change of flux

= (1.44 + 11.52) x 10^-5 Wb s^-1

= 12.96 x 10^-5 Wb.s^-1

∴ e = 12.96 x 10^-5 V

∴ \(I=\frac{e}{R}=\frac{12.96 \times 10^{-5}}{4.5 \times 10^{-3}}=2.88 \times 10^{-2} \mathrm{~A}\)

Question 8.

1. Obtain an expression for the mutual inductance between a long straight wire and a square loop of side a as shown.
2. Now assume that the straight wire carries a current of 50 A and the loop is moved to the right with a constant velocity v = 10 m.s^-1. Calculate the induced emf in the loop at the instant when x = 0.2 m. Take a = 0.1 m and assume that the loop has a large resistance.

Answer:

1. Consider a small portion of the coil of thickness dt at a distance t from the current carrying wire as shown.

Then the magnetic field strength experienced by this portion,

∴ \(B=\frac{\mu_0 I}{2 \pi t}\)

Magnetic Flux linked with this portion,

⇒ \(d \phi=B \cdot d A=\frac{\mu_0 I}{2 \pi t} \cdot a d t[d A=a d t]\)

∴ Magnetic flux linked with the coil,

⇒ \(\phi=\int_x^{a+x} \frac{\mu_0 I a}{2 \pi t} d t \quad \text { or, } M I=\frac{\mu_0 I a}{2 \pi} \ln \left(\frac{a+x}{x}\right)\)

∴ \(M=\frac{\mu_0 a}{2 \pi} \ln \left(\frac{a+x}{x}\right)=\frac{\mu_0 a}{2 \pi} \ln \left(1+\frac{a}{x}\right)\)

2. Induced emf,

⇒ \(e=\frac{\mu_0}{2 \pi x} \cdot \frac{I a^2 v}{a+x}=\frac{2 \times 10^{-7} \times 50 \times(0.1)^2 \times 10}{0.2(0.1+0.2)}\).

∴ \(1.67 \times 10^{-5} \approx 1.7 \times 10^{-5} \mathrm{~V}\)

Common Short Questions on Lenz’s Law

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Question 9. A metal rod PQ is resting on the rails A’B’ and positioned between the poles of a permanent magnet. The rails, the rod and the magnetic field are in three mutually perpendicular directions. A galvanometer G connects the rails through a switch K. Length of the rod = 15 cm, B = 0.50 T, and the resistance of the closed loop containing the rod = 9.0 mΩ. Assume the field to be uniform.

1. Suppose K is open and the rod is moved with a speed of 12 cm s-1 in the direction shown. Give the polarity and magnitude of the induced emf.
2. Is there an excess charge built up at the ends of the rods when K is open? What if K is closed?
3. With K open and the rod moving uniformly, there is no net force on the electrons in the rod PQ even though they do experience magnetic force due to the motion of the rod. Explain.
4. What is the retarding force on the rod when K is closed?
5. How much power is required by an external agent to keep the rod moving at the same speed (= 12 cm.s^-1 ) when K is closed? How much power is required when K is open?
6. How much power is dissipated as heat in the closed circuit? What is the source of this power?
7. What is the induced emf in the moving rod if the magnetic head is parallel to the rails instead of being perpendicular?

Answer:

1. Induced emf, e = Bvl

∴ e = 0.50 x 12 x 10^-2 x 15 x 10^-2 = 9 x 10^-3 V

[∵ B = 0.50 T, v = 12 x 10^-2 m.s^-1, l = 15 x 10^-2 m ]

The electrons of the rod will experience a force along PQ. Therefore, the P, end is positive and the Q end is negative.

2. Yes. When K is open, excess charge is developed at the end of the rod. When K is closed, the continuous flow of current maintains this excess charge.

3. The electric force developed due to excess charge of opposite nature at the ends of the rod cancels the magnetic force.

4. \(F=B I l=B \cdot \frac{e}{R} \cdot l=0.5 \times \frac{9 \times 10^{-3}}{9 \times 10^{-3}} \times 15 \times 10^{-2}\)

= 7.5 X 10^-2 N

5. Power = F x v

= 7.5 x 10^-2 x 12 x 10^-2

= 9 x 10^-3 W

No power is spent when K is open.

6. Power lost \(\frac{e^2}{R}=\frac{\left(9 \times 10^{-3}\right)^2}{9 \times 10^{-3}}\)

= 9 x 10^-3 W

7. Zero, because the motion of the rod does not cut or link with field lines.

Question 10. Which of the following is the unit of magnetic flux?

1. Tesla
2. Tesla x m²
3. Tesla/m²
4. Weber/m²

Answer: 3. Tesla/m²

The unit of magnetic flux is Tesla x m². The option 3 is correct

Question 11. A current flowing through a coil changes from +2 A to -2 A in 0.05 s and an emf of 8 V is induced in the coil. The value of self-inductance of the coil is

1. 0.8 H
2. 0.1 H
3. 0.2H
4. 0.4 H

Answer: 2. 0.1 H

We know, \(e=-L \frac{d I}{d t}\)

or, \(L=\frac{-e d t}{d I}=\frac{8 \times 0.05}{-2-(+2)}=\frac{0.4}{4}=0.1 \mathrm{H}\)

The option 2 is correct

Practice Short Questions on Induced EMF

Question 12. A metallic disc of radius 10 cm is rotating uniformly about a horizontal axis passing through its centre with angular v velocity 10 revolutions per second. A uniform magnetic field of intensity 10^-2 T acts along the axis of the disc. Find the potential difference induced between the centre and the rim of the disc.

Answer:

Radius of the metallic disc, R = 10 cm = 0.1 m

Now, magnetic field, B = l0^-2 T

Frequency, n = 10

Angular velocity, ω = 10 x 2 π rad = s^-1

The value of the induced potential difference between the endpoint and the centre of the metallic disc

= \(\frac{1}{2} B \omega R^2=\frac{1}{2} \times 10^{-2} \times 10 \times 2 \pi \times(0.1)^2\)

∴ 3.14 x 10^-3 V = 3.14 mV

Question 13. The dimension of magnetic flux is

1. \(M L^2 T^{-2} A^{-1}\)
2. \(M L T^{-1} A^{-2}\)
3. \(M L^{-1} \mathrm{TA}^{-1}\)
4. \(M L^{-1} A\)

Answer: 1. \(M L^2 T^{-2} A^{-1}\)

The option 1 is correct

Question 14. The magnetic flux through a coil varies according to the relation Φ = (4t² + 2t – 5)Wb, t measured in seconds. Calculate the induced current through the coil at t = 2s, if the resistance of the coil is 5 Ω.

Answer:

Φ = 4t² + 2t – 5

∴ \(\frac{d \phi}{d t}=8 t+2\)

When t = 2 s, \(\frac{d \phi}{d t}=8 \times 2+2=18\)

∴ Induced emf, e = 18V

Resistance of the coil, R = 5 Ω.

∴ Induced current through the coil, \(I=\frac{e}{R}=\frac{18}{5}=3.6 \mathrm{~A}\)

Question 15. The mutual inductance of two coils can be increased by

1. Decreasing the number of turns on the coils
2. Increasing the number of turns on the coils
3. Winding the coils on the wooden core
4. None of these

Answer: 2. Increasing the number of turns on the coils

The option 2 is correct.

Important Definitions in Electromagnetic Induction

Question 16. If L and R denote inductance and resistance respectively, then the dimension of \(\frac{L}{R}\) is

1. \(\mathrm{M}^0 \mathrm{~L}^0 \mathrm{~T}^0\)
2. \(M^0 L^0 T^1\)
3. \(M^2 L^0 T^2\)
4. \(M^1 L^1 T^2\)

Answer: 2. \(M^0 L^0 T^1\)

The option 2 is correct.

Question 17.

1. A horizontal straight wire. 5 m long extending from east to west is falling with a speed of 10 m/s at right angles to the horizontal component of the earth’s magnetic field 0.40 x 10^-4 Wb. m^-2.

• Find the instantaneous value of emf induced in the wire.
• What is the direction of the emf?
• Which end of the wire will be at a higher potential? (Neglect acceleration due to gravity.)

2. Derive the expression for energy stored in an inductor of coefficient of self-inductance L carrying current i₀.

Answer:

1. The wire is in an east-west direction, the horizontal component of the earth’s magnetic field is directed from south to north and the direction of the wire’s velocity is vertically downwards. These three are perpendicular to each other.

Hence the instantaneous emf induced across the wire,

e = vBl = 10 x (0.40 x 10^-4) x 5

= 2 x l0^-3V = 2 mV

2. From the relation, \(\vec{F}_m=q \vec{v} \times \vec{B}\) since \(\vec{v}\) is directed downwards and \(\vec{B}\) is directed towards north, so any positive charge in the wire experiences a force from west to east and moves in that direction. The direction of the induced emf is in the direction in which the positive charge moves, i.e., from west to east.

3. Since the direction of the induced emf is from west to east, the west end of the wire will be at a higher potential.

Question 18. A coil of metallic wire is at rest in a non-uniform magnetic field. Would any electromotive force be induced in the

Answer:

As the coil of metallic wire is at rest in a non-uniform magnetic field, the magnetic flux linked with the coil will not change concerning time. Hence no emf is induced in the coil.

Question 19. A very small circular loop of radius a is initially (at t = 0) coplanar and concentric with a much larger fixed circular loop of radius b. A constant current I flows in the larger loop. The smaller loop is rotated with a constant angular speed a) about the common diameter. The emf induced in the smaller loop as a function of time t is

1. \(\frac{\pi a^2 \mu_0 I}{2 b} \omega \cos (\omega t)\)
2. \(\frac{\pi a^2 \cdot \mu_0 I}{2 b} \omega \sin \left(\omega^2 t^2\right)\)
3. \(\frac{\pi a^2 \mu_0 I}{2 b} \omega \sin (\omega t)\)
4. \(\frac{\pi a^2 \mu_0 I}{2 b} \omega \sin ^2(\omega t)\)

Answer:

= \(B=\frac{\mu_0 I}{2 b}, A=\pi a^2\)

=\(\phi=\vec{B} \cdot \vec{A}=B A \cos \omega t\)

= \(e=-\frac{d \phi}{d t}=-\frac{\mu_0 I}{2 b} \cdot \pi a^2(-\omega \sin \omega t)\)

= \(\frac{\mu_0 I \pi a^2}{2 b} \omega \sin \omega t\)

The option 3 is correct.

Question 20. A straight conductor 0.1 m long moves in a uniform magnetic field of 0.1 T. The velocity of the conductor is 15 m/s and is directed perpendicular to the field. The emf induced between the two ends of the conductor is

1. 0.10 V
2. 0.15 V
3. 1.50 V
4. 15.00 V

Answer: 2. 0.15 V

The emf induced between the two ends of the conductor

= Blv = 0.1 x 0.1 x 15 = 0.15 V

The option 2 is correct.

Question 21. A conducting loop in the form of a circle is placed in a uniform magnetic field with its plane perpendicular to the direction of the field. An emf will be induced in the loop if

1. It is translated parallel to itself
2. It is rotated about one of its diameters
3. It is rotated about its axis which is parallel to the field
4. The loop is deformed from the original shape

Answer:

If the conducting loop is translated parallel to itself or rotated about its axis, it won’t intersect any magnetic line of force.

The options 2 and 4 are correct.

Question 22. Two coils of self-inductances 6mH and 8mH are connected in series and are adjusted for the highest coefficient of coupling. Equivalent self-inductance L for the assembly is approximately

1. 50 mH
2. 36 mH
3. 28 mH
4. 18 mH

Answer: 3. 28 mH

If the self-inductance of two coils are respectively L1 and L2 and their mutual inductance is M, then

⇒ \(L=L_1+L_2+2 M=L_1+L_2+2 k \sqrt{L_1 L_2}\) [k: = coefficient of coupling of the coils]

⇒ 6 + 8 + 2 x l x \(\sqrt{6 \times 8}\) [∵ L₁ = 6mH, L₁ = BmH and highest value of k = 1 ]

≈ 28mH

The option 3 is correct.

Question 23. As shown in the figure, a rectangular loop of a conducting wire is moving away with a constant velocity v in a perpendicular direction from a very long straight conductor carrying a steady current I. When the breadth of the rectangular loop is very small compared to its distance from the straight conductor, how does the emf E induced in the loop vary with time t?

1. \(E \propto \frac{1}{t^2}\)
2. \(E \propto \frac{1}{t}\)
3. \(E \propto-\ln (t)\)
4. \(E \propto \frac{1}{t^3}\)

Answer: 1. \(E \propto \frac{1}{t^2}\)

Electromotive force,

⇒ \(E=-\frac{d \phi}{d t}=-\frac{d}{d t}(\vec{B} \cdot \vec{A})\)

⇒ \(-A \frac{d B}{d t}\) [∵ the direction of A and B are same]

⇒ \(-A \frac{d}{d t}\left(\frac{\mu_0 I}{2 \pi v t}\right)\)

[Since for an infinity long conductor, magnetic field at a distance r is, \(B=\frac{\mu_0 I}{2 \pi r}. \text { Now, } r=v \times t\)]

⇒ \(-\frac{A \mu_0 I}{2 \pi \nu} \frac{d}{d t}\left(t^{-1}\right)=\frac{A \mu_0 I}{2 \pi \nu} t^{-2}\)

∴ \(E \propto \frac{1}{t^2}\)

The option 1 is correct

Question 24. In a coil of resistance 100 Ω, a current is induced by changing the magnetic flux through it as shown in the figure. The magnitude of change in flux through the coil is

1. 200 Wb
2. 225 Wb
3. 250 Wb
4. 275 Wb

Answer: 3. 250 Wb

Induced emf,

⇒ \(|e|=\frac{d \phi}{d t} \quad \text { or, } t R=\frac{d \phi}{d t}\)

or, \(\int d \phi=R \int i d t\)

or, Φ = R x area enclosed by the graph and axes

= 100 x \(\frac{1}{2}\) x 0.5 x 10 = 250 Wb

The option 3 is correct.

Examples of Applications of Electromagnetic Induction

Question 25. A thin semicircular conducting ring (PQR) of radius falling with its plane vertical in a horizontal magnetic field B, as shown. The potential difference developed across the ring when its speed is v is

1. Zero
2. \(\frac{B v \pi r^2}{2}\) and P is at higher potential
3. πrBv and R is at higher potential
4. 2rBv and R have a higher potential

Answer: 4. 2rBv and R have a higher potential

⇒ \(|e|=\frac{d \phi}{d t}=\frac{d}{d t}(B A)=B \frac{d A}{d t}\)

⇒ B x diameter x velocity = B x 2r x w = 2rvB

According to Fleming’s right-hand rule, if P and R are connected to a closed external circuit, the direction of the induced current is from P to R. In the external circuit, the current is from R to P so the potential of R is greater than that of P.

The option 4 is correct.

Question 26. A conducting square frame of side a and a long straight wire carrying current I are located in the same plane as shown. The frame moves to the right with a constant velocity v. The emf induced in the frame will be proportional to

1. \(\frac{1}{x^2}\)
2. \(\frac{1}{(2 x-a)^2}\)
3. \(\frac{1}{(2 x+a)^2}\)
4. \(\frac{1}{(2 x-a)(2 x+a)}\)

Answer: 4. \(\frac{1}{(2 x-a)(2 x+a)}\)

The electromotive force and the current induced due to the motion of the frame are in the clockwise direction. Hence, the emf and the current will be oppositely directed in the left and right arms of the frame. If the magnetic field in those two arms due to current 1 is B1 and B2 respectively, then the emf induced will be directly proportional to (B₁-B₂).

Here, \(B_1-B_2=\frac{\mu_0}{4 \pi} \frac{2 I}{x-a / 2}-\frac{\mu_0}{4 \pi} \frac{2 I}{x+a / 2}\)

= \(\frac{\mu_0 I}{\pi}\left(\frac{1}{2 x-a}-\frac{1}{2 x+a}\right)\)

= \(\frac{\mu_0 I}{\pi} \frac{2 a}{(2 x-a)(2 x+a)}\)

∴ \(B_1-B_2 \propto \frac{1}{(2 x-a)(2 x+a)}\)

The option 4 is correct.

Question 27. A long solenoid has 1000 turns. When a current of 4 A flows through it, the magnetic flux linked with each turn of the solenoid is 4 x l0^-3 Wb. The self-inductance of the solenoid is

1. 3H
2. 2H
3. 1H
4. 4H

Answer: 3. 1H

If the number of turns of the solenoid is N, the current through is I and the magnetic flux linked with each turn of the solenoid is Φ, then its self-inductance,

∴ \(L=\frac{N \phi}{I}=\frac{1000 \times\left(4 \times 10^{-3}\right)}{4}=1 \mathrm{H}\)

The option 3 is correct.

Question 28. A circular coil of radius 10 cm, 500 turns and resistance 2 Ω is placed with its plane, perpendicular to the horizontal component of the earth’s magnetic field. It is rotated about its vertical diameter through 180° in 0.25 s. The induced emf in the coil is (take H_E = 3.0 x 10^-5 T).

1. 6.6 x 10^-4 V
2. 1.4 x 10^-2 V
3. 2.6 X 10^-2 V
4. 3.8 x 10^-3 V

Answer: 4. 3.8 x 10^-3 V

A = π x 10² = 3.14 x l0²cm² = 3.14 x 10^-2 m²

Considering the surroundings as vacuum or air,

B = 3.0 x 10^-5T; N = 500

Initially, magnetic flux linked with the coil,

Φ₁ = NBA

By rotating the coll through 180°, magnetic flux linked with the coil, Φ₂ = -NBA

∴ Induced emf,

⇒ \(e=-\frac{\Delta \phi}{\Delta t}=-\frac{\phi_2-\phi_1}{\Delta t}=\frac{2 N B A}{\Delta t} .\)

⇒ \(\frac{2 \times 500 \times\left(3 \times 10^{-5}\right) \times\left(3.14 \times 10^{-2}\right)}{0.25}\)

≈ 3.8 x 10^-3 V

The option 4 is correct.

Question 29. Inside a parallel plate capacitor, the electric field E varies with time as t². The variation of the induced magnetic field with time is given by

1. t²
2. No variation
3. t³
4. t

Answer: 4. t

A varying electric field induces a magnetic field and this induced magnetic field is proportional to the rate of change in an electric field. According to the question, the electric field is proportional to \(t^2 \text { and } \frac{d}{d t}\left(t^2\right)=2 t\), so the induced magnetic field is proportional to t.

The option 4 is correct.

Question 30. The magnetic potential energy stored in a certain inductor is 25 mJ, when the current in the inductor is 60 mA. This inductor is of inductance

1. 1.389 H
2. 138.88 H
3. 0.138 H
4. 13.89 H

Answer: 4. 13.89 H

Energy stored in the inductor,

⇒ \(U=\frac{1}{2} L i^2\)

or, \(25 \times 10^{-3}=\frac{1}{2} \times L \times\left(60 \times 10^{-3}\right)^2\)

or, \(L =\frac{2 \times 25 \times 10^{-3}}{\left(60 \times 10^{-3}\right)^2}\)

or, L = 13.888 = 13.89 H

The option 4 is correct.

Question 31. A bar magnet is moved in the direction indicated by the arrow between two coils PQ and CD. Predict the directions of induced current in each coil.

Answer:

As per Lenz’s law, an S-pole is induced at the C-side of the coil CD. So the induced current in CD is such that it is clockwise when seen from the position of NS. Similarly, another S-pole is induced at the Q-side of PQ. So, the induced current in PQ is clockwise when seen from the position of NS.

Question 32. The motion of a copper plate is damped when it is allowed to oscillate between the two poles of a magnet. What is the cause of this damping?

Answer:

The eddy current generated in the copper plate is in such a direction, as per Lenz’s law, that its oscillation is damped.

Question 33. A rectangular conductor LMNO is placed in a uniform magnetic field of 0.5 T. The field is directed perpendicular to the plane of the conductor. When the arm MN of the length of 20 cm is moved towards the left with a velocity of 10 m.s^-1, calculate the emf induced in the arm. Given the resistance of the arm to be 5 Ω, (assuming that the other arms are of negligible resistance) find the value of the current in the arm.

Answer:

In unit time, the arm MN describes an area Iv. Sb, the rate of change of magnetic flux linked with MN is

⇒ \(\frac{d \phi}{d t}=\frac{d}{d t}(B \times \text { area })=B \frac{d}{d t}(\text { area })=B l v\)

∴ The magnitude of emf Induced in the arm (omitting the negative sign) is

⇒ \(|e|=\frac{d \phi}{d t}=B l v=0.5 \times\left(20 \times 10^{-2}\right) \times 10=1 \mathrm{~V}\)

∴ Induced current in the arm MN,

∴ \(i=\frac{1}{5}=0.2 \mathrm{~A}\).

Question 34. A wheel with 8 metallic spokes each 50 cm long is rotated with a speed of 120 rev.min^-1 in a plane normal to the horizontal component of the Earth’s magnetic field. The Earth’s magnetic field at the place is 0.4 G and the angle of dip is 60°. Calculate the emf induced between the axle and the rim of the wheel. How will the value of emf be affected if the number of spokes were increased?

Answer:

Length of each spoke, L = 50 cm = 0.5 m;

⇒ \(\omega=\frac{2 \pi \times 120}{60}=4 \pi \mathrm{rad} \cdot \mathrm{s}^{-1}\)

Horizontal component of earth’s magnetic field,

B = 0.4 cos 60° = 0.2G = 0.2 x 10^-4 T

∴ The emf induced between the two ends of a spoke,

= \(e=\frac{1}{2} B \omega L^2\)

= \(\frac{1}{2} \times 0.2 \times 10^{-4} \times 4 \pi \times(0.5)^2\)

= 3.14 x 10^-5 V

Each metal spoke behaves like a cell of emf e. All such identical cells are connected in parallel. So the total emf does not depend on the number of spokes.

Question 35. How does the mutual inductance of a pair of coils change when

1. Distance between the coils is increased and
2. The number of turns in the coils is increased.

Answer:

1. The flux linkage decreases in this case; so the mutual inductance decreases.
2. The flux linkage increases in this case, so the mutual inductance increases.

Question 36. A light metal disc on the top of an electromagnet is thrown up as the current is switched on. Why? Give reason.

Answer:

Let at the top end of the electromagnet, the current in the coil is clockwise when observed from above. As per Lenz’s law, eddy current is induced, in the metal disc in such a direction that it would tend to neutralise the current, at this instant the current is switched on. So the eddy current will be anticlockwise. As unlike parallel currents repel each other, the metal disc is thrown up.

Question 37. The motion of copper plates is damped when it is allowed to oscillate between the two poles of a magnet. If slots are cut in the plate, how will the damping be affected?

Answer:

The damping is due to eddy currents induced in the copper plate. If slots are cut in the plate, the eddy current loops become much shorter as a result, the net value of the eddy current decreases. Then the damping becomes less.

Question 38. A conducting loop is held above a current-carrying wire PQ as shown in the figure. Depict the direction of the current induced in the loop when the current in the wire PQ is constantly increasing.

Answer:

The current in the wire PQ is increasing constantly. So the magnetic field connected with the conducting loop will also increase constantly. As per Lenz’s law, the direction of the current induced in the loop will be such that the magnetic flux will tend to decrease.

Then this induced current will be anticlockwise [Use Max-well’s corkscrew rule to determine the directions of the magnetic fields due to the current in PQ, and due to the induced current in the loop].

Real-Life Scenarios in Electromagnetic Induction

Question 39. How does one understand this motional emf by invoking the Lorentz force acting on the free charge carriers of the conductor? Explain.

Answer:

Let \(\vec{B}=B(-\hat{k})\) (in the negative z-direction)

The rod PQ is moved towards the right, i.e., in the y-direction. So, the f charge carriers (electrons) in the rod moves also in the y-direction Q with a velocity

∴ \(\vec{v}=v \hat{j}\)

As the charge of an electron is -e, the Lorentz magnetic force acting on it is

∴ \(\vec{F}=-e \vec{v} \times \vec{B}=-e[v \hat{j} \times B(-\hat{k})]=e v B \hat{i}\)

So, the electron drift in the rod PQ is in the x-direction, i.e., from P to Q. Then, as per convention, the motional emf in PQ will be directed from Q to P. This is the direction obtained by applying Flemings’s right-hand rule.

Question 40. Two loops, one rectangular of dimensions 10 cm x 2.5 cm and the second of square shape of side 5 cm are moved out of a uniform magnetic field \(\vec{B}\) perpendicular to the planes of the loops with equal velocity v as is shown.

1. In which case will the emf induced be more?
2. In which case will the current flowing through the two loops be less?

Answer:

1. When the loop is completely inside the magnetic field there is no change in the magnetic flux.

∴ |e| = 0

When the loop is moving out of the field,

Φ = BA = Blx

∴ \(|e|=\frac{d \phi}{d t}=B l \frac{d x}{d t}=B l v\)

So, \(\left|e_{\text {rec }}\right|=(B \times 2.5 \times v) \mathrm{V}\)

∴ \(\left|e_{\mathrm{sq}}\right|=(B \times 5 \times v) \mathrm{V}\)

When the loop moves out of the field, |e| = 0 Hence, the emf induced in the square loop is more.

2. Current, I = \(I=\frac{|e|}{R}\)

∵ |e| is less for the rectangular loop, the current induced in the rectangular loop will be less.

Question 41. Sketch the change in flux, emf and force when a conducting rod PQ of resistance R and length L moves freely and fro between A and C with speed v on a rectangular conductor placed in the uniform magnetic field as shown.

Answer:

The flux enclosed by the rod is,

⇒ Φ = Blx for 0 ≤ x ≤ b = Blb for b ≤ x ≤ 2b

Now, the magnitude of induced emf is,

⇒ \(e=-\frac{d \phi}{d t}=-B L \frac{d x}{d t}=-B L v\) for 0 ≤ x ≤ b

⇒ \(-B L \frac{d b}{d t}=0\) for b ≤ x ≤ 2b

Now, the magnitude of induced current when induced emf is non-zero,

⇒ \(I=\frac{e}{R}=\frac{B L v}{R}\)

The force required to keep the conductor in motion is,

⇒ \(F=B I L=B \frac{B L v}{R} L=\frac{B^2 L^2 v}{R}\)

∴ \(F=\frac{B^2 L^2 v}{R}\) for 0 ≤ x ≤ b = 0 for b ≤ x ≤ 2b

Therefore, the variation of flux, emf and force are shown respectively.

Question 42. A long straight current-carrying wire passes normally through the centre of a circular loop. If the current through the wire increases, will there be an induced emf in the loop? justify.

Answer:

We know that, induced emf, \(e=-\frac{d \phi}{d t}=-\frac{d}{d t}(\vec{B} \cdot \vec{A})\)

In this case, the magnetic lines of force due to the current-carrying wire are parallel to the plane of the loop. So, the angle between the magnetic field and normal of the normal plane of the circular loop is 90°.

∴ e = 0 \( [∵ \vec{B} \cdot \vec{A}=0]\)

Hence, there will be no induced emf in the loop.

Question 43. Predict the polarity of the capacitor in the situation described below:

Answer:

According to Lenz’s law, the direction of induced emf is such that, this induced emf can oppose the motion of the tire magnet in both cases. So polarity of plate A will be positive concerning the plate B of the capacitor.

Question 44. What is the direction of induced currents in metal rings 1 and 2 when current I in the wire is increasing steadily?

Answer:

The current in the wire is increasing steadily. So the magnetic flux linked with the metal rings 1, and 2 will also increase steadily. According to Lenz’s law, the direction of the current induced in the rings 1 and 2 will be such that the magnetic flux will tend to decrease.

So, the induced current in rings 1 and 2 will be clockwise and anticlockwise respectively.

Question 45. A horizontal conducting rod 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, 0.3 x 10^-4 Wb m^-2. Find the instantaneous value of the emf induced in the rod.

Answer:

The instantaneous value of the induced emf in the rod,

e = Blvsind = 0.3 x 10^-4 x 10 x 5 x sin90°

= 15 X 10^-4 V = 1.5 mV

Question 46. An aeroplane is flying horizontally from west to east with a velocity of 900 km/h. Calculate the potential difference developed between the ends of its wings having a span of 20 m. The horizontal component of the Earth’s magnetic field is 5 x 10^-4T and the angle of dip is 30°.

Answer:

A potential difference between the wings of the aeroplane

e = B sin30° .l.v

= \(l=20 \mathrm{~m}, v=900 \mathrm{~km} / \mathrm{h}=900 \times \frac{5}{18}=250 \mathrm{~m} \cdot \mathrm{s}^{-1}\)

= H = B sin 30 or, \(B=\frac{H}{\cos 30^{\circ}}\)

∴ e = H tan 30. l.v = \(5 \times 10^{-4} \times \frac{1}{\sqrt{3}} \times 20 \times 250\)

= 1. 44 V.