Question

A conducting ring of radius 1 m kept in a uniform magnetic field B of 0.01 T, rotates uniformly with an angular velocity 100 rad s\(^{-1}\) with its axis of rotation perpendicular to B. The maximum induced emf in it is

• A. 1.5\(\pi\)V
• B. \(\pi\)V
• C. 2\(\pi\)V
• D. 0.5\(\pi\)V
• E. 4\(\pi\)V

Hint:

The formula \( \epsilon = NBA\omega \sin(\omega t) \) is the principle behind AC generators. Here, since it's a single ring, the number of turns \( N \) is 1.

Answer 1

The Correct Option is B

Concept: When a conducting loop rotates in a uniform magnetic field, the magnetic flux (\( \Phi \)) through it changes continuously, inducing an AC EMF.
• Flux Equation: \( \Phi = B A \cos(\omega t) \).
• Induced EMF: \( \epsilon = - \frac{d\Phi}{dt} = B A \omega \sin(\omega t) \).
• Maximum EMF (\( \epsilon_{max} \)): Occurs when \( \sin(\omega t) = 1 \), so \( \epsilon_{max} = B A \omega \).

Step 1: Calculate the area of the ring.
Given radius \( r = 1 \text{ m} \): \[ A = \pi r^2 = \pi (1)^2 = \pi \text{ m}^2 \]

Step 2: Calculate the maximum induced EMF.
Substitute the given values: \( B = 0.01 \text{ T} \), \( A = \pi \text{ m}^2 \), and \( \omega = 100 \text{ rad/s} \). \[ \epsilon_{max} = B \cdot A \cdot \omega \] \[ \epsilon_{max} = 0.01 \cdot \pi \cdot 100 \] \[ \epsilon_{max} = \pi \text{ V} \]