Electromagnetic Induction - NEET Physics Questions
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Electromagnetic Induction

Question 31: difficult

Two resistors of 10 W and 20 W and an ideal inductor of 10 H are connected to a 2 V battery
as shown. The key K is inserted at time t = 0. The initial (t = 0) and final (t →∞) currents
through battery are

1. \[\frac{1}{15} A,\frac{1}{10} A\]
2. \[\frac{1}{10} A,\frac{1}{15} A\]
3. \[\frac{2}{15} A,\frac{1}{10} A\]
4. \[\frac{1}{15} A,\frac{2}{25} A\]
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Question 32: moderate

The network shown in the figure is part of a complete circuit. If at a certain instant, the current
I is 5A and it is decreasing at a rate of

\[10^{3} As^{-1} then V_{B}-A_{A} equals\]

1. 20 V
2. 15 V
3. 10 V
4. 5 V
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Question 33: moderate

For the circuit shown in figure R = 10Ω, L = 5H, E = 20 V, i = 2 A. This current is decreasing at
a rate of 1.0 A /s Find Vab at this instant.

1. 45 V
2. 35 V
3. 25 V
4. 40 V
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Question 34: moderate

Using Lenz’s Law, determine the direction of current flow in the loop for each of the two
situations shown below.

1. anticlockwise for (a) and Clockwise for (b)
2. clockwise for (a) and Clockwise for (b)
3. anticlockwise for (a) and Anticlockwise for (b)
4. clockwise for (a) and Anticlockwise for (b)
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Question 35: moderate

Select the correct alternative. A thin semicircular conducting ring of radius R is falling with its plane vertical in a horizontal magnetic induction \[\overrightarrow{B}\] . At the position MNQ the speed of the ring is v & the potential difference developed across the ring is :

1. zero
2. BvπR²/2 & M is at higher potential
3. π RBV & Q is at higher potential
4. 2 RBV & Q is at higher potential
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Question 36: difficult

A conducting rod AC of length 4l is rotated about a point O in a uniform magnetic field \[\overrightarrow{B}\] directed into the paper. AO = l and OC = 3l. Then

1. \[V_{A}-V_{0}=\frac{B\omega l^{2}}{2}\]
2. \[V_{0}-V_{C}=\frac{9}{2}B\omega l^{2}\]
3. \[V_{A}-V_{C}=8B\omega l^{2}\]
4. \[V_{C}-V_{0}=\frac{9}{2}B\omega l^{2}\]
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Question 37: easy

A conducting rod is moved with a constant velocity v in a magnetic field. A potential difference appears across the two ends

1. \[if \overrightarrow{v} \parallel \overrightarrow{l}\]
2. \[if \overrightarrow{v} \parallel \overrightarrow{B}\]
3. \[if \overrightarrow{l} \parallel \overrightarrow{B}\]
4. none of these
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Question 38: moderate

A square metal loop of side 10 cm and resistance 1 Ω is moved with a constant velocity partly inside a magnetic field of 2 Wbm–², directed into the paper, as shown in the figure. This loop is connected to a network of five resistors each of value 3 Ω. If a steady current of 1 mA flows in the loop, then the speed of the loop is :

1. \[0.5 cms^{-1}\]
2. \[1 cms^{-1}\]
3. \[2 cms^{-1}\]
4. \[4 cms^{-1}\]
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Question 39: easy

A conducting wire frame is placed in a magnetic field which is directed into the paper. The magnetic field is increasing at a constant rate. The directions of induced current in wires AB and CD are :

1. B to A and D to C
2. A to B and C to D
3. A to B and D to C
4. B to A and C to D
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in a circuit consisting of two loops connected in series (like a figure-eight), the larger loop effectively dictates the direction of the current for the entire circuit.

Question 40: difficult

The figure shows three circuits with identical batteries, inductors, and resistors. Rank the circuits according to the current through the battery (i) just after the switch is closed and (ii) a long time later, greatest first :

1. (i) i2 > i3 > i1 (i1 = 0) (ii) i2 > i3 > i1
2. (i) i2 < i3 > i1 (i1 ≠ 0) (ii) i2 > i3 > i1
3. (i) i2 = i3 = i1 (i1 = 0) (ii) i2 > i3 > i1
4. (i) i2 = i3 > i1 (i1 ≠ 0) (ii) i2 > i3 > i1
View Answer

(i) Just after the switch is closed (t = 0): An inductor opposes any sudden change in current. Initially, it acts like an infinite resistance (open circuit). No current can flow through any branch containing the inductor at this instant.

(ii) A long time later (t =infinity): Once the current reaches a steady state, the inductor no longer opposes the flow. It acts like an ideal wire (short circuit) with zero resistance. You can then rank the circuits by calculating the total equivalent resistance of the remaining resistors.