Question

The length of a wire required to manufacture a solenoid of length l and self-induction L is (cross-sectional area is negligible)

• A. $\sqrt{\frac{2\pi Ll}{\mu_0}}$
• B. $\sqrt{\frac{\mu_0 Ll}{4\pi}}$
• C. $\sqrt{\frac{4\pi Ll}{\mu_0}}$
• D. $\sqrt{\frac{\mu_0 Ll}{2\pi}}$

Answer 1

Answer: (C)

$\sqrt{\frac{4\pi Ll}{\mu_0}}$

Answer 2

$L = \frac{\mu_0 N^2 A}{l}$ If $x$ is the length of the solenoid with $r$ as radius, then $x = 2\pi rN , A = \pi r^2$ $\therefore$ $L = \mu_0 \left( \frac{x^2}{4 \pi^2 r^2} \right) \frac{\pi r^2}{l} \left[ \therefore \, N = \frac{x}{2 \pi r} \right]$ $\therefore$ $x = \sqrt{\frac{4 \pi L l}{\mu_0}}$