Question

A solenoid has the length \(1\ \text{m}\) and the area of cross section \(0.02\ \text{m}^2\). If the number of turns in the solenoid is \(5000\), then the self inductance of the solenoid is

• A. \(0.2\pi\ \text{henry}\)
• B. \(0.4\pi\ \text{henry}\)
• C. \(0.02\pi\ \text{henry}\)
• D. \(0.04\pi\ \text{henry}\)

Hint:

For a long air-core solenoid, \[ L=\frac{\mu_0N^2A}{l}. \] Self inductance is directly proportional to the square of the number of turns and the cross-sectional area, and inversely proportional to the length of the solenoid.

Answer 1

The Correct Option is A

Step 1: Write the self-inductance formula.
\( L = \mu_0 \frac{N^2 A}{l} \). Given: \( N = 5000 \), \( A = 0.02\,\text{m}^2 \), \( l = 1\,\text{m} \), \( \mu_0 = 4\pi\times10^{-7}\,\text{H m}^{-1} \).

Step 2: Substitute and simplify.
\[ L = 4\pi\times10^{-7}\times\frac{25\times10^6\times0.02}{1} = 4\pi\times10^{-7}\times5\times10^5 = 0.2\pi\,\text{H} \] \[ \boxed{0.2\pi\,\text{henry}} \]