A 50 cm long air core solenoid has a 2 cm radius and 500 turns. The self-inductance of the solenoid would be
(given $μ_0 = 4 × 10^{-7} T m\, A^{-1}$ and $\pi^2 = 10$)
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → $8.0 × 10^{-4} H$
Given:
Length $l = 50\ \text{cm} = 0.5\ \text{m}$
Radius $r = 2\ \text{cm} = 0.02\ \text{m}$
Number of turns $N = 500$
Permeability of air $\mu_0 = 4\pi\times10^{-7}\ \text{H/m}$
Formula:
$L = \frac{\mu_0 N^2 A}{l}$
where $A = \pi r^2$
Substitute values:
$A = \pi (0.02)^2 = 1.2566\times10^{-3}\ \text{m}^2$
$L = \frac{4\pi\times10^{-7}\times(500)^2\times1.2566\times10^{-3}}{0.5}$
$L = \frac{4\pi\times10^{-7}\times2.5\times10^{5}\times1.2566\times10^{-3}}{0.5}$
$L = \frac{3.9478\times10^{-4}}{0.5}=7.8956\times10^{-4}\ \text{H}$
Therefore, $L = 7.9\times10^{-4}\ \text{H}$