Question

A square loop of side 0.50 m is placed in a uniform magnetic field of 0.4 T perpendicular to the plane of the loop. The loop is rotated through an angle of $60^\circ$ in 0.2 s. The value of emf induced in the loop will be:

• A. 5 V
• B. 3.5 V
• C. 2.5 V
• D. Zero V

Hint:

When calculating induced emf using Faraday's Law, always use the rate of change of magnetic flux. For rotational motion, remember to consider the angle change during the motion.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Flux through the loop depends on the angle between the field and the area vector. Rotating the loop changes this angle, so the flux changes and an emf is induced.
The average emf is the change in flux divided by the time taken.

Step 2: Key Formula or Approach:
\( \Phi = BA\cos\theta \) and \( \varepsilon_{avg} = \left|\dfrac{\Delta\Phi}{\Delta t}\right| \).

Step 3: Detailed Explanation:
1. Area \( A = (0.50)^2 = 0.25\ \text{m}^2 \).
2. Initially the field is perpendicular to the plane, so the area vector is parallel to \( \vec{B} \): \( \theta_i = 0^\circ \).
3. Initial flux \( \Phi_i = 0.4 \times 0.25 \times \cos 0^\circ = 0.10\ \text{Wb} \).
4. After rotation by \( 60^\circ \): \( \Phi_f = 0.4 \times 0.25 \times \cos 60^\circ = 0.05\ \text{Wb} \).
5. Change in flux: \( |\Delta\Phi| = 0.10 - 0.05 = 0.05\ \text{Wb} \).
6. Average emf: \[ \varepsilon = \frac{0.05}{0.2} = 0.25\ \text{V} \]
7. Option (D) is wrong because the flux changes. Option (A) 5 V and option (B) 3.5 V are far too large.
8. The printed data give 0.25 V. The only option of the same digits is (C) 2.5 V, which is exactly 10 times larger (it would need B = 4 T).

Step 4: Final Answer:
Calculated emf for the printed data \( = 0.25\ \text{V} \). The nearest listed option is (C) 2.5 V, which is taken as the intended answer.