Question

Two coaxial solenoids are made by winding thin insulated wire over a pipe of cross-sectional area \( A = 10 \, \text{cm}^2 \) and length 20 cm. If one of the solenoids has 300 turns and the other 400 turns, their mutual inductance is

• A. \( 2.4 \pi \times 10^{-5} \, \text{H} \)
• B. \( 2.4 \times 10^{-7} \, \text{H} \)
• C. \( 4.8 \pi \times 10^{-4} \, \text{H} \)
• D. \( 4.8 \pi \times 10^{-5} \, \text{H} \)

Hint:

The mutual inductance between two solenoids depends on their number of turns, cross-sectional area, and the length of the solenoids.

Answer 1

The Correct Option is D

Step 1: Use the formula for mutual inductance.
The mutual inductance \( M \) of two coaxial solenoids is given by the formula: \[ M = \frac{\mu_0 N_1 N_2 A}{l} \] where \( N_1 \) and \( N_2 \) are the number of turns in the solenoids, \( A \) is the cross-sectional area, \( l \) is the length of the solenoids, and \( \mu_0 \) is the permeability of free space.
Step 2: Calculate the mutual inductance.
Substitute the given values to find the mutual inductance as \( 4.8 \pi \times 10^{-5} \, \text{H} \). Final Answer: \[ \boxed{4.8 \pi \times 10^{-5} \, \text{H}} \]