A 10 cm long air core solenoid of cross-sectional area 1.0 cm² contains 300 turns. Its inductance is

$36π × 10^{-6} H$

The correct answer is Option (1) → $36π × 10^{-6} H$

Inductance of a solenoid:

$L = \mu_0 \frac{N^2 A}{l}$

Given: $N = 300$, $l = 0.1\ \text{m}$, $A = 1 \times 10^{-4}\ \text{m}^2$, $\mu_0 = 4\pi \times 10^{-7}\ \text{H/m}$

Substitute values directly:

$L = (4 \pi \times 10^{-7}) \cdot \frac{300^2 \cdot 1 \times 10^{-4}}{0.1} = 36 \pi \times 10^{-6}\ \text{H}$

Final Answer: $L = 36 \pi \times 10^{-6}\ \text{H}$