Question

The plane of a circular loop of area \(150\ \text{cm}^2\) is perpendicular to a uniform magnetic field of \(0.5\) T. If the loop is turned such that its plane is in the direction of the field in \(0.5\) s, then the induced emf produced is

• A. $25$ mV
• B. $10$ mV
• C. $2.5$ mV
• D. $15$ mV
• E. $7.5$ mV

Hint:

For rotating loops: - Maximum flux when plane $\perp$ field - Zero flux when plane $\parallel$ field

Answer 1

The Correct Option is D

Concept: Magnetic flux: \[ \Phi = BA\cos\theta \] Induced emf: \[ \mathcal{E} = \frac{\Delta \Phi}{\Delta t} \]

Step 1: Convert area.
\[ A = 150\ \text{cm}^2 = 150 \times 10^{-4} = 1.5 \times 10^{-2}\ \text{m}^2 \]

Step 2: Initial flux.
Plane $\perp$ field $\Rightarrow \theta = 0^\circ$: \[ \Phi_1 = BA = 0.5 \times 1.5 \times 10^{-2} = 7.5 \times 10^{-3} \]

Step 3: Final flux.
Plane $\parallel$ field $\Rightarrow \theta = 90^\circ$: \[ \Phi_2 = 0 \]

Step 4: Change in flux.
\[ \Delta \Phi = 7.5 \times 10^{-3} \]

Step 5: Calculate emf.
\[ \mathcal{E} = \frac{7.5 \times 10^{-3}}{0.5} = 1.5 \times 10^{-2}\ \text{V} = 15\ \text{mV} \]