Question

A long solenoid having 100 turns per cm carries a current of \(\frac{4}{\pi}\) A. At the centre of it is placed a coil of 200 turns of cross sectional area 25 cm\(^2\) having its axis parallel to the field produced by the solenoid. When the direction of the current in the solenoid is reversed within 0.04 s, the induced emf in the coil is

• A. 0.2 V
• B. 0.04 V
• C. 0.002 V
• D. 0.016 V

Hint:

Use \(\mathcal{E} = N A \frac{\Delta B}{\Delta t}\) when magnetic field changes through a coil.

Answer 1

The Correct Option is B

Magnetic field of solenoid: \[ B = \mu_0 n I = 4\pi \times 10^{-7} \cdot 10^4 \cdot \frac{4}{\pi} = 1.6 \times 10^{-2}\ \text{T} \] EMF: \[ \mathcal{E} = \frac{N \cdot A \cdot \Delta B}{\Delta t} = \frac{200 \cdot 25 \times 10^{-4} \cdot 2 \cdot 1.6 \times 10^{-2}}{0.04} = 0.04\ \text{V} \]