Question

A circular coil of area \(3 \times 10^{-2} \text{ m}^2\), 900 turns and a resistance of 1.8 Ω is placed with its plane perpendicular to a uniform magnetic field of \(3.5 \times 10^{-5}\) T. The current induced in the coil when it is rotated through 180° in half a second is

• A. 2.1 mA
• B. 1.8 mA
• C. 1.5 mA
• D. 2.7 mA

Hint:

When a coil is rotated by 180° from a position where its plane is perpendicular to the field, the change in flux is always \(2BA\). When it's rotated by 90°, the change is \(BA\). Memorizing these common cases can save time.

Answer 1

The Correct Option is A

Step 1: Calculate change in magnetic flux. Initial angle: The plane is perpendicular to the field, so the area vector is parallel to the field (\(\theta_1 = 0^\circ\)). Initial Flux \(\phi_i = NBA \cos(0^\circ) = NBA\). Final angle: Rotated by \(180^\circ\) (\(\theta_2 = 180^\circ\)). Final Flux \(\phi_f = NBA \cos(180^\circ) = -NBA\). Change in flux \(\Delta \phi = \phi_f - \phi_i = -NBA - NBA = -2NBA\). Magnitude \(|\Delta \phi| = 2NBA\).
Step 2: Calculate Induced EMF. \[ \epsilon = \frac{|\Delta \phi|}{\Delta t} = \frac{2NBA}{\Delta t} \]
Step 3: Calculate Induced Current. \[ I = \frac{\epsilon}{R} = \frac{2NBA}{R \Delta t} \] Given: \(N = 900\) \(B = 3.5 \times 10^{-5} \, \text{T}\) \(A = 3 \times 10^{-2} \, \text{m}^2\) \(R = 1.8 \, \Omega\) \(\Delta t = 0.5 \, \text{s}\) Substitute values: \[ I = \frac{2 \times 900 \times 3.5 \times 10^{-5} \times 3 \times 10^{-2}}{1.8 \times 0.5} \] Numerator: \( 2 \times 900 \times 3.5 \times 3 \times 10^{-7} = 18900 \times 10^{-7} = 1.89 \times 10^{-3} \) Denominator: \( 0.9 \) \[ I = \frac{1.89 \times 10^{-3}}{0.9} = 2.1 \times 10^{-3} \, \text{A} = 2.1 \, \text{mA} \]