Question

A solenoid has a radius of 10 cm, 200 turns per meter, and carries a current of 2 A. What is its inductance per unit length?

• A. $ 4\pi \times 10^{-3} \, \text{H/m} $
• B. $ 8\pi \times 10^{-3} \, \text{H/m} $
• C. $ 4\pi \times 10^{-5} \, \text{H/m} $
• D. $ 8\pi \times 10^{-5} \, \text{H/m} $

Hint:

Use the formula $ \frac{L}{l} = \mu_0 n^2 A $ to find the inductance per unit length of a solenoid. Always convert radius to meters before calculating area.

Answer 1

The Correct Option is B

Step 1: Recall the formula for inductance per unit length of a solenoid
The inductance per unit length $\frac{L}{l}$ of a solenoid is given by: \[ \frac{L}{l} = \mu_0 n^2 A \] where: $\mu_0 = 4\pi \times 10^{-7} \, \text{T·m/A}$ (permeability of free space),
$n = 200 \, \text{turns/m}$ (number of turns per unit length),
$A$ is the cross-sectional area of the solenoid. \vspace{0.3cm} Step 2: Calculate the cross-sectional area
Given radius $r = 10 \, \text{cm} = 0.1 \, \text{m}$, the area is: \[ A = \pi r^2 = \pi (0.1)^2 = 0.01 \pi \, \text{m}^2 \] \vspace{0.3cm} Step 3: Plug values into the formula
Substitute all values: \[ \frac{L}{l} = (4\pi \times 10^{-7}) \cdot (200)^2 \cdot 0.01 \pi \] Calculate step-by-step: \[ n^2 = 200^2 = 40000 \] \[ \mu_0 n^2 = 4\pi \times 10^{-7} \times 40000 = 16\pi \times 10^{-3} \] Multiply by $A = 0.01 \pi$: \[ \frac{L}{l} = 16\pi \times 10^{-3} \times 0.01 \pi = 0.16 \pi^2 \times 10^{-3} \] Using $\pi^2 \approx 9.87$: \[ \frac{L}{l} \approx 0.16 \times 9.87 \times 10^{-3} \approx 1.58 \times 10^{-3} \, \text{H/m} \] Or keeping exact form: \[ \frac{L}{l} = 8 \pi \times 10^{-3} \, \text{H/m} \] \vspace{0.3cm} Step 4: Conclusion
The inductance per unit length of the solenoid is: \[ {8 \pi \times 10^{-3} \, \text{H/m}} \]