Question

You are required to design an air-filled solenoid of inductance \( 0.016 \, \text{H} \) having a length \( 0.81 \, \text{m} \) and radius \( 0.02 \, \text{m} \). The number of turns in the solenoid should be:

• A. 2592
• B. 2866
• C. 2976
• D. 3140

Hint:

The inductance of a solenoid depends on the permeability of the material inside it, the number of turns, the cross-sectional area, and the length of the solenoid.

Answer 1

The Correct Option is B

To determine the number of turns in the solenoid, we can use the formula for the inductance \( L \) of a solenoid: \[ L = \frac{\mu_0 N^2 A}{l} \] where \( L = 0.016 \, \text{H} \) (inductance), \( \mu_0 = 4\pi \times 10^{-7} \, \text{T}\cdot\text{m/A} \) (permeability of free space), \( N \) is the number of turns, \( A = \pi r^2 \) is the cross-sectional area, and \( l = 0.81 \, \text{m} \) is the length of the solenoid. 

First, calculate the cross-sectional area \( A \): \[ A = \pi r^2 = \pi (0.02)^2 = 4\pi \times 10^{-4} \, \text{m}^2 \] Substitute the known values into the inductance formula: \[ 0.016 = \frac{4\pi \times 10^{-7} \times N^2 \times 4\pi \times 10^{-4}}{0.81} \] Simplify to solve for \( N^2 \): \[ N^2 = \frac{0.016 \times 0.81}{4\pi \times 10^{-7} \times 4\pi \times 10^{-4}} \] \[ N^2 = \frac{0.01296}{16\pi^2 \times 10^{-11}} \] \[ N^2 = \frac{0.01296}{1.57896 \times 10^{-9}} \] \[ N^2 \approx 8219479.5 \] \[ N \approx \sqrt{8219479.5} \] \[ N \approx 2866 \]

Thus, the number of turns in the solenoid is approximately 2866.