Question:
A circular coil of radius 8 cm, 400 turns and resistance \( 2\,\Omega \) is placed with its plane perpendicular to the horizontal component of the earth's magnetic field. It is rotated about its vertical diameter through \( 180^{\circ} \) in 0.30 sec. Horizontal component of the earth's magnetic field at the place is \( 3\times10^{-5} \) T. The magnitude of current induced in the coil is approximately:
A circular coil of radius 8 cm, 400 turns and resistance \( 2\,\Omega \) is placed with its plane perpendicular to the horizontal component of the earth's magnetic field. It is rotated about its vertical diameter through \( 180^{\circ} \) in 0.30 sec. Horizontal component of the earth's magnetic field at the place is \( 3\times10^{-5} \) T. The magnitude of current induced in the coil is approximately:
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Flipping an induction loop coil completely upside down (\( 180^\circ \)) inside a uniform magnetic field always doubles the total flux change (\( 2NBA \)), creating a highly predictable value structure.
Updated On: Jun 8, 2026
- \( 4\times10^{-2}~\text{A} \)
- \( 8\times10^{-4}~\text{A} \)
- \( 8\times10^{-2}~\text{A} \)
- \( 1.92\times10^{-3}~\text{A} \)
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The Correct Option is B
Solution and Explanation
Concept:
According to Faraday's Law of Induction, the induced electromotive force \( e \) is equal to the rate of change of magnetic flux linkage through the circuit:
\[
e = -\frac{\Delta \Phi}{\Delta t} = -\frac{N A B (\cos\theta_2 - \cos\theta_1)}{\Delta t}
\]
The induced current is then found using Ohm's Law: \( I = \frac{e}{R} \).
Step 1: Calculating the cross-sectional area \( A \) of the loop.
Given radius \( r = 8~\text{cm} = 0.08~\text{m} = 8\times10^{-2}~\text{m} \): \[ A = \pi r^2 = \pi \times (8\times10^{-2})^2 = 64\pi\times10^{-4}~\text{m}^2 \approx 2.01\times10^{-2}~\text{m}^2 \]
Step 2: Finding the net change in magnetic flux linkage.
The coil flips by \( 180^\circ \), so \( \theta_1 = 0^\circ \implies \cos(0^\circ) = 1 \), and \( \theta_2 = 180^\circ \implies \cos(180^\circ) = -1 \): \[ \Delta \Phi = NBA(-1 - 1) = -2NBA \] Plugging in the parameters \( N = 400 \) and \( B = 3\times10^{-5}~\text{T} \): \[ |e| = \frac{2NBA}{\Delta t} = \frac{2 \times 400 \times (3\times10^{-5}) \times (2.01\times10^{-2})}{0.30} \] \[ |e| = \frac{24\times10^{-3} \times 2.01\times10^{-2}}{0.30} = \frac{0.04824\times10^{-3}}{0.30} = 1.608\times10^{-3}~\text{V} \]
Step 3: Calculating the induced current \( I \).
Given internal resistance \( R = 2\,\Omega \): \[ I = \frac{|e|}{R} = \frac{1.608\times10^{-3}}{2} = 0.804\times10^{-3}~\text{A} \approx 8\times10^{-4}~\text{A} \] This matches option (B) perfectly.
Step 1: Calculating the cross-sectional area \( A \) of the loop.
Given radius \( r = 8~\text{cm} = 0.08~\text{m} = 8\times10^{-2}~\text{m} \): \[ A = \pi r^2 = \pi \times (8\times10^{-2})^2 = 64\pi\times10^{-4}~\text{m}^2 \approx 2.01\times10^{-2}~\text{m}^2 \]
Step 2: Finding the net change in magnetic flux linkage.
The coil flips by \( 180^\circ \), so \( \theta_1 = 0^\circ \implies \cos(0^\circ) = 1 \), and \( \theta_2 = 180^\circ \implies \cos(180^\circ) = -1 \): \[ \Delta \Phi = NBA(-1 - 1) = -2NBA \] Plugging in the parameters \( N = 400 \) and \( B = 3\times10^{-5}~\text{T} \): \[ |e| = \frac{2NBA}{\Delta t} = \frac{2 \times 400 \times (3\times10^{-5}) \times (2.01\times10^{-2})}{0.30} \] \[ |e| = \frac{24\times10^{-3} \times 2.01\times10^{-2}}{0.30} = \frac{0.04824\times10^{-3}}{0.30} = 1.608\times10^{-3}~\text{V} \]
Step 3: Calculating the induced current \( I \).
Given internal resistance \( R = 2\,\Omega \): \[ I = \frac{|e|}{R} = \frac{1.608\times10^{-3}}{2} = 0.804\times10^{-3}~\text{A} \approx 8\times10^{-4}~\text{A} \] This matches option (B) perfectly.
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