Question

A solenoid has 2000 turns wound over a length of 0.30 m. The area of its cross-section is \( 1.2 \times 10^{-4} \, \text{m}^2 \). Around its central section, a coil of 300 turns is wound. If an initial current of 2 A in the solenoid is reversed in 0.25 s, then the emf induced in the coil is:

• A. \( 6 \times 10^4 \, \text{V} \)
• B. \( 48 \times 10^{-3} \, \text{V} \)
• C. \( 6 \times 10^2 \, \text{V} \)
• D. \( 48 \, \text{mV} \)

Hint:

The induced emf depends on the number of turns in the coil and the rate of change of magnetic flux.

Answer 1

The Correct Option is D

Step 1: Use Faraday’s Law to calculate emf.
The induced emf is given by: \[ \mathcal{E} = - N \frac{d\Phi}{dt} \] where \( N \) is the number of turns, and \( \Phi = B A \) is the magnetic flux.
Step 2: Explanation.
After applying the formula, we find the induced emf in the coil to be \( 48 \, \text{mV} \).
Final Answer: \[ \boxed{48 \, \text{mV}} \]