Question

A coil of 100 turns and 0.10 m\(^2\) area, making two rotations per second is placed in a 0.01 T uniform magnetic field perpendicular to its axis of rotation. The maximum voltage generated in the coil is:

• A. 0.1 V
• B. 12.56 V
• C. 1.256 V
• D. 0.628 V

Hint:

To find the maximum induced emf in a rotating coil, remember the formula involves the number of turns, area, magnetic field, and angular velocity. Angular velocity \( \omega \) is calculated as \( \omega = 2 \pi \times \text{frequency} \).

Answer 1

The Correct Option is C

The formula for the maximum induced emf in a coil rotating in a magnetic field is given by: \[ \mathcal{E}_{\text{max}} = N A B \omega \] Where: - \( \mathcal{E}_{\text{max}} \) is the maximum induced emf,
- \( N \) is the number of turns of the coil,
- \( A \) is the area of the coil,
- \( B \) is the magnetic field strength,
- \( \omega \) is the angular velocity of the coil.
Given: - \( N = 100 \) turns,
- \( A = 0.10 \, \text{m}^2 \),
- \( B = 0.01 \, \text{T} \),
- The coil makes 2 rotations per second, so \( \omega = 2\pi \times 2 = 4\pi \, \text{rad/s} \).
Substitute these values into the formula: \[ \mathcal{E}_{\text{max}} = 100 \times 0.10 \times 0.01 \times 4\pi \] \[ \mathcal{E}_{\text{max}} = 100 \times 0.10 \times 0.01 \times 12.566 \] \[ \mathcal{E}_{\text{max}} = 1.256 \, \text{V} \] Thus, the correct answer is 1.256 V.