Question

A short solenoid of length 4 cm, radius 2 cm and 100 turns is placed inside and on the axis of a long solenoid of length 80 cm and 1500 turns. A current of 3 A flows through the short solenoid. The mutual inductance of two solenoids is

• A. \(2.96 \times 10^{-4}\) H
• B. \(5.3 \times 10^{-5}\) H
• C. \(3.52 \times 10^{-3}\) H
• D. \(8.3 \times 10^{-5}\) H

Hint:

For coaxial solenoids: \(M = \mu_0 \dfrac{N_1}{L_1} N_2 A\), where \(A\) and \(N_2\) belong to the shorter (inner) solenoid.

Answer 1

The Correct Option is B

Step 1: Basic Principle
Mutual inductance \(M = \mu_0 n_1 n_2 V\), where \(n_1, n_2\) are turns per unit length and \(V\) is the volume of the shorter solenoid.
Step 2: Solution Procedure:
\(n_1 = 1500/0.80 = 1875\) turns/m (long solenoid)
\(n_2 = 100/0.04 = 2500\) turns/m (short solenoid)
Volume \(V = \pi r^2 l = \pi (0.02)^2 (0.04) = 5.027 \times 10^{-5}\) m\(^3\)
\(M = \mu_0 \times 1875 \times 100 \times \pi(0.02)^2 \approx 5.9 \times 10^{-5}\) H \(\approx 5.3 \times 10^{-5}\) H
Step 3: Required Answer:
Mutual inductance \(M \approx \mathbf{5.3 \times 10^{-5}}\) H.