Question

Two co-axial solenoids are made by a pipe of cross sectional area $10 \mathrm{~cm}^{2}\( and length \)20 \mathrm{~cm}$ If one of the solenoid has 300 turns and the other 400 turns, their mutual inductance is...... (a) \(4.8 \pi \times 10^{-4} \mathrm{H}\) (b) \(4.8 \pi \times 10^{-5} \mathrm{H}\) (c) \(2.4 \pi \times 10^{-4} \mathrm{H}\) (d) \(4.8 \pi \times 10^{4} \mathrm{H}\)

Short answer

The mutual inductance between the two co-axial solenoids is (b) \(4.8π × 10^{-5} \, \mathrm{H}\).

Step-by-step solution

Calculate the self-inductance of each solenoid

First, we need to find the self-inductance L1 and L2 of the solenoids using the formula: \[L = \frac{μ₀ * A * n^2}{l}\] where μ₀ = permeability of free space = \(4π × 10^{-7} \, \mathrm{T \, m/A}\), A is the cross-sectional area of the pipe, n is the number of turns per unit length and l is the length of the solenoid. For solenoid 1 \(n_1 = \frac{300}{0.2} \, \mathrm{turns/m}\) For solenoid 2 \(n_2 = \frac{400}{0.2} \, \mathrm{turns/m}\)

Calculate the mutual inductance between the solenoids

To find the mutual inductance M, we use the following formula: \[M = \frac{μ₀ * A * n_1 * n_2}{l}\] Insert the values and perform the calculation: \(M = \frac{4π × 10^{-7} * 10^{-2} * (\frac{300}{0.2}) * (\frac{400}{0.2})}{0.2}\)

Calculate the answer

Now, simplify and solve the expression for M: \(M = 4.8π × 10^{-5} \, \mathrm{H}\) So the answer is (b) \(4.8π × 10^{-5} \, \mathrm{H}\).