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Electromagnetic Induction
Physics XII - Chapter 06: Complete NCERT Exercise Solutions
Master Electromagnetic Induction with NCERT solutions for Faraday's laws, Lenz's law, self and mutual inductance, and AC generators.
Question 6.1
Predict the direction of induced current in the situations described by the following Figs. 6.18(a) to (f).
Answer & Explanation:
Lenz's Law: Induced current opposes the change in magnetic flux.
(a) Clockwise
(b) Anticlockwise
(c) Clockwise
(d) Anticlockwise
(e) Clockwise
(f) Anticlockwise
(Directions are based on opposing the change in flux in each configuration.)
(a) Clockwise
(b) Anticlockwise
(c) Clockwise
(d) Anticlockwise
(e) Clockwise
(f) Anticlockwise
(Directions are based on opposing the change in flux in each configuration.)
Question 6.2
Use Lenz's law to determine the direction of induced current in the situations described by Fig. 6.19:
(a) A wire of irregular shape turning into a circular shape;
(b) A circular loop being deformed into a narrow straight wire.
(a) A wire of irregular shape turning into a circular shape;
(b) A circular loop being deformed into a narrow straight wire.
Answer & Explanation:
Lenz's Law: Induced current opposes the change in magnetic flux.
(a) Irregular to circular: Area increases → flux increases → induced current opposes increase → anticlockwise.
(b) Circular to straight wire: Area decreases → flux decreases → induced current opposes decrease → clockwise.
Answer: (a) Anticlockwise, (b) Clockwise.
(a) Irregular to circular: Area increases → flux increases → induced current opposes increase → anticlockwise.
(b) Circular to straight wire: Area decreases → flux decreases → induced current opposes decrease → clockwise.
Answer: (a) Anticlockwise, (b) Clockwise.
Question 6.3
A long solenoid with 15 turns per cm has a small loop of area 2.0 cm\(^2\) 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 & Explanation:
Formula: \( \varepsilon = -\frac{d\phi}{dt} \)
\( \phi = B A \), \( B = \mu_0 n I \)
Calculation:
\( n = 15 \, \text{turns/cm} = 1500 \, \text{turns/m} \)
\( \Delta I = 4 - 2 = 2 \, \text{A} \), \( \Delta t = 0.1 \, \text{s} \)
\( \frac{dB}{dt} = \mu_0 n \frac{dI}{dt} = 4\pi \times 10^{-7} \times 1500 \times \frac{2}{0.1} = 3.77 \times 10^{-2} \, \text{T/s} \)
\( A = 2 \times 10^{-4} \, \text{m}^2 \)
\( \varepsilon = A \frac{dB}{dt} = 2 \times 10^{-4} \times 3.77 \times 10^{-2} = 7.54 \times 10^{-6} \, \text{V} \)
Answer: \( 7.54 \times 10^{-6} \, \text{V} \)
\( \phi = B A \), \( B = \mu_0 n I \)
Calculation:
\( n = 15 \, \text{turns/cm} = 1500 \, \text{turns/m} \)
\( \Delta I = 4 - 2 = 2 \, \text{A} \), \( \Delta t = 0.1 \, \text{s} \)
\( \frac{dB}{dt} = \mu_0 n \frac{dI}{dt} = 4\pi \times 10^{-7} \times 1500 \times \frac{2}{0.1} = 3.77 \times 10^{-2} \, \text{T/s} \)
\( A = 2 \times 10^{-4} \, \text{m}^2 \)
\( \varepsilon = A \frac{dB}{dt} = 2 \times 10^{-4} \times 3.77 \times 10^{-2} = 7.54 \times 10^{-6} \, \text{V} \)
Answer: \( 7.54 \times 10^{-6} \, \text{V} \)
Question 6.4
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 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 & Explanation:
Formula: \( \varepsilon = B l v \)
(a) Velocity normal to longer side (8 cm):
\( l = 2 \, \text{cm} = 0.02 \, \text{m} \), \( v = 0.01 \, \text{m/s} \)
\( \varepsilon = 0.3 \times 0.02 \times 0.01 = 6 \times 10^{-5} \, \text{V} \)
Time = \( \frac{\text{length of field region}}{v} = \frac{0.08}{0.01} = 8 \, \text{s} \)
(b) Velocity normal to shorter side (2 cm):
\( l = 8 \, \text{cm} = 0.08 \, \text{m} \)
\( \varepsilon = 0.3 \times 0.08 \times 0.01 = 2.4 \times 10^{-4} \, \text{V} \)
Time = \( \frac{0.02}{0.01} = 2 \, \text{s} \)
Answer: (a) \( 6 \times 10^{-5} \, \text{V} \) for 8 s, (b) \( 2.4 \times 10^{-4} \, \text{V} \) for 2 s.
(a) Velocity normal to longer side (8 cm):
\( l = 2 \, \text{cm} = 0.02 \, \text{m} \), \( v = 0.01 \, \text{m/s} \)
\( \varepsilon = 0.3 \times 0.02 \times 0.01 = 6 \times 10^{-5} \, \text{V} \)
Time = \( \frac{\text{length of field region}}{v} = \frac{0.08}{0.01} = 8 \, \text{s} \)
(b) Velocity normal to shorter side (2 cm):
\( l = 8 \, \text{cm} = 0.08 \, \text{m} \)
\( \varepsilon = 0.3 \times 0.08 \times 0.01 = 2.4 \times 10^{-4} \, \text{V} \)
Time = \( \frac{0.02}{0.01} = 2 \, \text{s} \)
Answer: (a) \( 6 \times 10^{-5} \, \text{V} \) for 8 s, (b) \( 2.4 \times 10^{-4} \, \text{V} \) for 2 s.
Question 6.5
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 & Explanation:
Formula: \( \varepsilon = \frac{1}{2} B \omega L^2 \) (for rod rotating about one end)
Calculation:
\( \varepsilon = \frac{1}{2} \times 0.5 \times 400 \times (1.0)^2 = 100 \, \text{V} \)
Answer: 100 V
Calculation:
\( \varepsilon = \frac{1}{2} \times 0.5 \times 400 \times (1.0)^2 = 100 \, \text{V} \)
Answer: 100 V
Question 6.6
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 \(3.0 \times 10^{-2}\) T. Obtain the maximum and average emf induced in the coil. If the coil forms a closed loop of resistance 10 \(\Omega\), 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 & Explanation:
Max emf: \( \varepsilon_{\text{max}} = N B A \omega \)
\( A = \pi r^2 = \pi (0.08)^2 = 0.0201 \, \text{m}^2 \)
\( \varepsilon_{\text{max}} = 20 \times 3.0 \times 10^{-2} \times 0.0201 \times 50 = 0.603 \, \text{V} \)
Average emf: Over half cycle, \( \varepsilon_{\text{avg}} = \frac{2}{\pi} \varepsilon_{\text{max}} \approx 0.384 \, \text{V} \)
Max current: \( I_{\text{max}} = \frac{\varepsilon_{\text{max}}}{R} = \frac{0.603}{10} = 0.0603 \, \text{A} \)
Average power: \( P_{\text{avg}} = \frac{\varepsilon_{\text{rms}}^2}{R} \), \( \varepsilon_{\text{rms}} = \frac{\varepsilon_{\text{max}}}{\sqrt{2}} \)
\( P_{\text{avg}} \approx 0.0182 \, \text{W} \)
Power source: Mechanical work done to rotate the coil.
Answer: Max emf = 0.603 V, Avg emf = 0.384 V, Max current = 0.0603 A, Avg power = 0.0182 W.
\( A = \pi r^2 = \pi (0.08)^2 = 0.0201 \, \text{m}^2 \)
\( \varepsilon_{\text{max}} = 20 \times 3.0 \times 10^{-2} \times 0.0201 \times 50 = 0.603 \, \text{V} \)
Average emf: Over half cycle, \( \varepsilon_{\text{avg}} = \frac{2}{\pi} \varepsilon_{\text{max}} \approx 0.384 \, \text{V} \)
Max current: \( I_{\text{max}} = \frac{\varepsilon_{\text{max}}}{R} = \frac{0.603}{10} = 0.0603 \, \text{A} \)
Average power: \( P_{\text{avg}} = \frac{\varepsilon_{\text{rms}}^2}{R} \), \( \varepsilon_{\text{rms}} = \frac{\varepsilon_{\text{max}}}{\sqrt{2}} \)
\( P_{\text{avg}} \approx 0.0182 \, \text{W} \)
Power source: Mechanical work done to rotate the coil.
Answer: Max emf = 0.603 V, Avg emf = 0.384 V, Max current = 0.0603 A, Avg power = 0.0182 W.
Question 6.7
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, \(0.30 \times 10^{-4}\) 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?
(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 & Explanation:
(a) Induced emf: \( \varepsilon = B_H l v \)
\( \varepsilon = (0.30 \times 10^{-4}) \times 10 \times 5.0 = 1.5 \times 10^{-3} \, \text{V} \)
(b) Direction: By Fleming’s right‑hand rule: east to west.
(c) Higher potential: West end (since current flows from higher to lower potential).
Answer: (a) \( 1.5 \times 10^{-3} \, \text{V} \), (b) East to west, (c) West end.
\( \varepsilon = (0.30 \times 10^{-4}) \times 10 \times 5.0 = 1.5 \times 10^{-3} \, \text{V} \)
(b) Direction: By Fleming’s right‑hand rule: east to west.
(c) Higher potential: West end (since current flows from higher to lower potential).
Answer: (a) \( 1.5 \times 10^{-3} \, \text{V} \), (b) East to west, (c) West end.
Question 6.8
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 & Explanation:
Formula: \( \varepsilon = -L \frac{dI}{dt} \)
Calculation:
\( \frac{dI}{dt} = \frac{0 - 5.0}{0.1} = -50 \, \text{A/s} \)
\( L = \frac{\varepsilon}{dI/dt} = \frac{200}{50} = 4 \, \text{H} \)
Answer: 4 H
Calculation:
\( \frac{dI}{dt} = \frac{0 - 5.0}{0.1} = -50 \, \text{A/s} \)
\( L = \frac{\varepsilon}{dI/dt} = \frac{200}{50} = 4 \, \text{H} \)
Answer: 4 H
Question 6.9
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 & Explanation:
Formula: \( \Delta \phi = M \Delta I \)
Calculation:
\( \Delta I = 20 - 0 = 20 \, \text{A} \)
\( \Delta \phi = 1.5 \times 20 = 30 \, \text{Wb-turns} \)
Answer: 30 Wb-turns
Calculation:
\( \Delta I = 20 - 0 = 20 \, \text{A} \)
\( \Delta \phi = 1.5 \times 20 = 30 \, \text{Wb-turns} \)
Answer: 30 Wb-turns
Question 6.10
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 \(5 \times 10^{-4} \, \text{T}\) and the dip angle is \(30^\circ\).
Answer & Explanation:
Formula: \( \varepsilon = B_V l v \) where \( B_V = B \sin \theta \)
Calculation:
\( v = 1800 \, \text{km/h} = 500 \, \text{m/s} \)
\( B_V = 5 \times 10^{-4} \times \sin 30^\circ = 2.5 \times 10^{-4} \, \text{T} \)
\( \varepsilon = 2.5 \times 10^{-4} \times 25 \times 500 = 3.125 \, \text{V} \)
Answer: 3.125 V
Calculation:
\( v = 1800 \, \text{km/h} = 500 \, \text{m/s} \)
\( B_V = 5 \times 10^{-4} \times \sin 30^\circ = 2.5 \times 10^{-4} \, \text{T} \)
\( \varepsilon = 2.5 \times 10^{-4} \times 25 \times 500 = 3.125 \, \text{V} \)
Answer: 3.125 V
Question 6.11
Suppose the loop in Exercise 6.4 is stationary but the current feeding the electromagnet that produces the magnetic field is gradually 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 1.6 \(\Omega\), how much power is dissipated by the loop as heat? What is the source of this power?
Answer & Explanation:
Induced emf: \( \varepsilon = A \frac{dB}{dt} \)
Area \( A = 0.08 \times 0.02 = 1.6 \times 10^{-3} \, \text{m}^2 \)
\( \varepsilon = 1.6 \times 10^{-3} \times 0.02 = 3.2 \times 10^{-5} \, \text{V} \)
Current: \( I = \frac{\varepsilon}{R} = \frac{3.2 \times 10^{-5}}{1.6} = 2.0 \times 10^{-5} \, \text{A} \)
Power: \( P = I^2 R = (2.0 \times 10^{-5})^2 \times 1.6 = 6.4 \times 10^{-10} \, \text{W} \)
Source: Electrical energy from the electromagnet’s circuit.
Answer: \( 6.4 \times 10^{-10} \, \text{W} \)
Area \( A = 0.08 \times 0.02 = 1.6 \times 10^{-3} \, \text{m}^2 \)
\( \varepsilon = 1.6 \times 10^{-3} \times 0.02 = 3.2 \times 10^{-5} \, \text{V} \)
Current: \( I = \frac{\varepsilon}{R} = \frac{3.2 \times 10^{-5}}{1.6} = 2.0 \times 10^{-5} \, \text{A} \)
Power: \( P = I^2 R = (2.0 \times 10^{-5})^2 \times 1.6 = 6.4 \times 10^{-10} \, \text{W} \)
Source: Electrical energy from the electromagnet’s circuit.
Answer: \( 6.4 \times 10^{-10} \, \text{W} \)
Question 6.12
A square loop of side 12 cm with its sides parallel to 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 (that is it increases by \(10^{-3} \, T \, cm^{-1}\) as one moves in the negative x-direction), and it is decreasing in time at the rate of \(10^{-3} \, T \, s^{-1}\). Determine the direction and magnitude of the induced current in the loop if its resistance is 4.50 m\(\Omega\).
Answer & Explanation:
Total emf = motional + transformer emf.
Motional emf due to gradient: \( \varepsilon_1 = B l v \) but B varies → average B difference.
Transformer emf due to time variation: \( \varepsilon_2 = A \frac{dB}{dt} \).
Detailed calculation yields net emf ~ \( 1.44 \times 10^{-5} \, \text{V} \).
Current \( I = \frac{\varepsilon}{R} = \frac{1.44 \times 10^{-5}}{4.5 \times 10^{-3}} \approx 3.2 \times 10^{-3} \, \text{A} \).
Direction: Anticlockwise (Lenz’s law).
Answer: \( 3.2 \times 10^{-3} \, \text{A} \) anticlockwise.
Motional emf due to gradient: \( \varepsilon_1 = B l v \) but B varies → average B difference.
Transformer emf due to time variation: \( \varepsilon_2 = A \frac{dB}{dt} \).
Detailed calculation yields net emf ~ \( 1.44 \times 10^{-5} \, \text{V} \).
Current \( I = \frac{\varepsilon}{R} = \frac{1.44 \times 10^{-5}}{4.5 \times 10^{-3}} \approx 3.2 \times 10^{-3} \, \text{A} \).
Direction: Anticlockwise (Lenz’s law).
Answer: \( 3.2 \times 10^{-3} \, \text{A} \) anticlockwise.
Question 6.13
It is desired to measure the magnitude of field between the poles of a powerful loud speaker magnet. A small flat search coil of area 2 cm\(^2\) with 25 closely wound turns, is positioned normal to the field direction, and then quickly snatched out of the field region. Equivalently, one can give it a quick \(90^\circ\) turn to bring its plane parallel to the field direction). The total charge flown in the coil (measured by a ballistic galvanometer connected to coil) is 7.5 mC. The combined resistance of the coil and the galvanometer is 0.50 \(\Omega\). Estimate the field strength of magnet.
Answer & Explanation:
Formula: \( Q = \frac{N \Delta \phi}{R} = \frac{N B A}{R} \)
Calculation:
\( A = 2 \times 10^{-4} \, \text{m}^2 \), \( N = 25 \), \( R = 0.5 \, \Omega \), \( Q = 7.5 \times 10^{-3} \, \text{C} \)
\( B = \frac{Q R}{N A} = \frac{7.5 \times 10^{-3} \times 0.5}{25 \times 2 \times 10^{-4}} = 0.75 \, \text{T} \)
Answer: 0.75 T
Calculation:
\( A = 2 \times 10^{-4} \, \text{m}^2 \), \( N = 25 \), \( R = 0.5 \, \Omega \), \( Q = 7.5 \times 10^{-3} \, \text{C} \)
\( B = \frac{Q R}{N A} = \frac{7.5 \times 10^{-3} \times 0.5}{25 \times 2 \times 10^{-4}} = 0.75 \, \text{T} \)
Answer: 0.75 T
Question 6.14
Figure 6.20 shows a metal rod PQ resting on the smooth rails AB and positioned between the poles of a permanent magnet. The rails, the rod, and the magnetic field are in three mutual perpendicular directions. A galvanometer G connects the rails through a switch K. Length of the rod = 15 cm, \( B = 0.50 \, T \), resistance of the closed loop containing the rod = 9.0 m\(\Omega\). Assume the field to be uniform.
(a) 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.
(b) Is there an excess charge built up at the ends of the rods when K is open? What if K is closed?
(c) With K open and the rod moving uniformly, there is no net force on the electrons in the rod PQ even though they experience magnetic force due to the motion of the rod. Explain.
(d) What is the retarding force on the rod when K is closed?
(e) 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?
(f) How much power is dissipated as heat in the closed circuit? What is the source of this power?
(g) What is the induced emf in the moving rod if the magnetic field is parallel to the rails instead of being perpendicular?
(a) 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.
(b) Is there an excess charge built up at the ends of the rods when K is open? What if K is closed?
(c) With K open and the rod moving uniformly, there is no net force on the electrons in the rod PQ even though they experience magnetic force due to the motion of the rod. Explain.
(d) What is the retarding force on the rod when K is closed?
(e) 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?
(f) How much power is dissipated as heat in the closed circuit? What is the source of this power?
(g) What is the induced emf in the moving rod if the magnetic field is parallel to the rails instead of being perpendicular?
Answer & Explanation:
(a) \( \varepsilon = B l v = 0.5 \times 0.15 \times 0.12 = 9 \times 10^{-3} \, \text{V} \). Polarity: P positive, Q negative.
(b) Open: Yes, charges accumulate. Closed: No, current flows.
(c) Electric field due to accumulated charges balances magnetic force → no net force.
(d) Retarding force \( F = \frac{B^2 l^2 v}{R} = \frac{(0.5)^2 (0.15)^2 (0.12)}{9 \times 10^{-3}} = 0.075 \, \text{N} \).
(e) Closed: Power = \( F v = 0.075 \times 0.12 = 9 \times 10^{-3} \, \text{W} \). Open: Zero.
(f) Heat power = \( I^2 R = \left( \frac{\varepsilon}{R} \right)^2 R = \frac{\varepsilon^2}{R} = 9 \times 10^{-3} \, \text{W} \). Source: External agent.
(g) Zero (since motion is parallel to field, no flux change).
Answer: As above.
(b) Open: Yes, charges accumulate. Closed: No, current flows.
(c) Electric field due to accumulated charges balances magnetic force → no net force.
(d) Retarding force \( F = \frac{B^2 l^2 v}{R} = \frac{(0.5)^2 (0.15)^2 (0.12)}{9 \times 10^{-3}} = 0.075 \, \text{N} \).
(e) Closed: Power = \( F v = 0.075 \times 0.12 = 9 \times 10^{-3} \, \text{W} \). Open: Zero.
(f) Heat power = \( I^2 R = \left( \frac{\varepsilon}{R} \right)^2 R = \frac{\varepsilon^2}{R} = 9 \times 10^{-3} \, \text{W} \). Source: External agent.
(g) Zero (since motion is parallel to field, no flux change).
Answer: As above.
Question 6.15
An air-cored solenoid with length 30 cm, area of cross-section 25 cm\(^2\) and number of turns 500, carries a current of 2.5 A. The current is suddenly switched off in a brief time of 10\(^{-3}\) s. How much is the average back emf induced across the ends of the open switch in the circuit? Ignore the variation in magnetic field near the ends of the solenoid.
Answer & Explanation:
Formula: \( \varepsilon = -L \frac{dI}{dt} \), \( L = \frac{\mu_0 N^2 A}{l} \)
Calculation:
\( L = \frac{4\pi \times 10^{-7} \times (500)^2 \times 25 \times 10^{-4}}{0.3} \approx 2.62 \times 10^{-3} \, \text{H} \)
\( \frac{dI}{dt} = \frac{0 - 2.5}{10^{-3}} = -2500 \, \text{A/s} \)
\( \varepsilon = 2.62 \times 10^{-3} \times 2500 \approx 6.55 \, \text{V} \)
Answer: ~6.55 V
Calculation:
\( L = \frac{4\pi \times 10^{-7} \times (500)^2 \times 25 \times 10^{-4}}{0.3} \approx 2.62 \times 10^{-3} \, \text{H} \)
\( \frac{dI}{dt} = \frac{0 - 2.5}{10^{-3}} = -2500 \, \text{A/s} \)
\( \varepsilon = 2.62 \times 10^{-3} \times 2500 \approx 6.55 \, \text{V} \)
Answer: ~6.55 V
Question 6.16
(a) Obtain an expression for the mutual inductance between a long straight wire and a square loop of side \( a \) as shown in Fig. 6.21.
(b) Now assume that the straight wire carries a current of 50 A and the loop is moved to the right with a constant velocity, \( v = 10 \, \text{m/s} \). Calculate the induced emf in the loop at the instant when \( x = 0.2 \, \text{m} \). Take \( a = 0.1 \, \text{m} \) and assume that the loop has a large resistance.
(b) Now assume that the straight wire carries a current of 50 A and the loop is moved to the right with a constant velocity, \( v = 10 \, \text{m/s} \). Calculate the induced emf in the loop at the instant when \( x = 0.2 \, \text{m} \). Take \( a = 0.1 \, \text{m} \) and assume that the loop has a large resistance.
Answer & Explanation:
(a) \( M = \frac{\mu_0 a}{2\pi} \ln\left(1 + \frac{a}{x}\right) \)
(b) \( \varepsilon = M \frac{dI}{dt} \) but here I constant, loop moves → flux change due to motion.
\( \varepsilon = \frac{d\phi}{dt} = \frac{d}{dt} \left( \frac{\mu_0 I a}{2\pi} \ln\left(1 + \frac{a}{x}\right) \right) \)
At \( x=0.2, a=0.1 \), \( v = \frac{dx}{dt} = 10 \, \text{m/s} \), calculation yields \( \varepsilon \approx 1.25 \times 10^{-5} \, \text{V} \).
Answer: (a) \( M = \frac{\mu_0 a}{2\pi} \ln\left(1 + \frac{a}{x}\right) \), (b) \( 1.25 \times 10^{-5} \, \text{V} \).
(b) \( \varepsilon = M \frac{dI}{dt} \) but here I constant, loop moves → flux change due to motion.
\( \varepsilon = \frac{d\phi}{dt} = \frac{d}{dt} \left( \frac{\mu_0 I a}{2\pi} \ln\left(1 + \frac{a}{x}\right) \right) \)
At \( x=0.2, a=0.1 \), \( v = \frac{dx}{dt} = 10 \, \text{m/s} \), calculation yields \( \varepsilon \approx 1.25 \times 10^{-5} \, \text{V} \).
Answer: (a) \( M = \frac{\mu_0 a}{2\pi} \ln\left(1 + \frac{a}{x}\right) \), (b) \( 1.25 \times 10^{-5} \, \text{V} \).
Question 6.17
A line charge \( \lambda \) 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,
\( B = -B_0 \mathbf{k} \quad (r \leq a; \quad a < R) \)
\( = 0 \quad (\text{otherwise}) \)
What is the angular velocity of the wheel after the field is suddenly switched off?
\( B = -B_0 \mathbf{k} \quad (r \leq a; \quad a < R) \)
\( = 0 \quad (\text{otherwise}) \)
What is the angular velocity of the wheel after the field is suddenly switched off?
Answer & Explanation:
Angular impulse = change in angular momentum.
Magnetic torque when field switched off induces rotation.
Final angular velocity \( \omega = \frac{\lambda B_0 a^2}{M R} \).
Derivation: Force on element → torque → angular impulse = \( I \omega \).
Answer: \( \omega = \frac{\lambda B_0 a^2}{M R} \)
Magnetic torque when field switched off induces rotation.
Final angular velocity \( \omega = \frac{\lambda B_0 a^2}{M R} \).
Derivation: Force on element → torque → angular impulse = \( I \omega \).
Answer: \( \omega = \frac{\lambda B_0 a^2}{M R} \)
📘 Exam Preparation Tip:
These exercise questions will help you understand induced electromotive forces and electromagnetic applications. You'll learn to apply Faraday's and Lenz's laws to calculate induced EMF, determine self and mutual inductance for coils, analyze energy stored in inductors, and understand working principles of AC generators and transformers. Crucial for power generation and electrical engineering concepts.