Question

A coil of an AC generator, having 100 turns and area 0.1 m² each, rotates at half a rotation per second in a magnetic field of 0.02 T. The maximum emf generated in the coil is:

• A. 0.31 V
• B. 0.20 V
• C. 0.63 V
• D. 0.10 V

Hint:

The maximum induced emf in a rotating coil depends on the number of turns, the area of the coil, the strength of the magnetic field, and the angular velocity of the coil.

Answer 1

The Correct Option is A

Step 1: Formula

The maximum induced electromotive force (emf) in a coil rotating within a magnetic field is determined by the formula:

\[ \text{emf}_{\text{max}} = N A B \omega \]

In this equation:

• \( N \) represents the number of turns in the coil.
• \( A \) denotes the area of the coil.
• \( B \) signifies the strength of the magnetic field.
• \( \omega \) indicates the angular velocity.

Step 2: Calculation

The provided values are:

• Number of turns, \( N = 100 \).
• Coil area, \( A = 0.1 \, \text{m}^2 \).
• Magnetic field strength, \( B = 0.02 \, \text{T} \).
• Angular velocity, \( \omega = 2\pi \times 0.5 = \pi \, \text{rad/s} \).

Substituting these values into the formula yields:

\[ \text{emf}_{\text{max}} = 100 \times 0.1 \times 0.02 \times \pi = 0.31 \, \text{V} \]

Outcome:

The calculated maximum emf induced in the coil is \( 0.31 \, \text{V} \). This result aligns with the correct option.