Question

The length of thin wire required to manufacture a solenoid of inductance L and length $l$, (if the cross-sectional diameter is considered less than its length) is :-

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

Answer 1

Answer: (B)

(2) $\sqrt{\frac{4\pi L l}{\mu_0}}$

Answer 2

The self-inductance of a long solenoid is $L = \frac{\mu_0 N^2 A}{l}$, where $A = \pi r^2$ is the cross-sectional area and $l$ is the solenoid length. The total wire length is $l_w = N(2\pi r)$. From $l_w$, we get $N r = \frac{l_w}{2\pi}$, so $N^2 r^2 = \frac{l_w^2}{4\pi^2}$. Substitute this into the inductance formula: $L = \frac{\mu_0 \pi}{l} (N^2 r^2) = \frac{\mu_0 \pi}{l} \left(\frac{l_w^2}{4\pi^2}\right)$. Simplify to $L = \frac{\mu_0 l_w^2}{4\pi l}$. Solving for $l_w$ gives $l_w = \sqrt{\frac{4\pi L l}{\mu_0}}$.