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 number of turns, the cross-sectional area, and the length of the solenoid. Use the formula \( L = \mu_0 \frac{N^2 A}{l} \) to solve for the unknowns.

Answer 1

The Correct Option is B

To find the number of turns in an air-filled solenoid with a given inductance, length, and radius, we use the formula for the inductance \(L\) of a solenoid: 

\(L = \dfrac{\mu_0 N^2 A}{l}\)

Where:

• \(L = 0.016 \, \text{H}\) (inductance)
• \(\mu_0 = 4\pi \times 10^{-7} \, \text{T m/A}\) (permeability of free space)
• \(A = \pi r^2 = \pi (0.02 \, \text{m})^2\) (cross-sectional area)
• \(l = 0.81 \, \text{m}\) (length of the solenoid)
• \(N\) is the number of turns we are solving for.

First, calculate the cross-sectional area \(A\):

\(A = \pi (0.02)^2 = 1.25664 \times 10^{-3} \, \text{m}^2\)

Rearrange the formula to solve for \(N\):

\(N = \sqrt{\dfrac{L \cdot l}{\mu_0 \cdot A}}\)

Substitute the known values into the equation:

\(N = \sqrt{\dfrac{0.016 \times 0.81}{4\pi \times 10^{-7} \times 1.25664 \times 10^{-3}}}\)

Calculate step-by-step:

\(N = \sqrt{\dfrac{0.01296}{1.577924 \times 10^{-9}}}\)

\(N \approx \sqrt{8212273}\)

\(N \approx 2866\)

Therefore, the number of turns in the solenoid should be \(2866\).