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 $\pi^2 = 10$)

$8.0 × 10^{-4} H$

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, $L = 7.9\times10^{-4}\ \text{H}$