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)

$L=\frac{{\mu }_0{N}^{2}A}{l}$
If $x$ is the length of the solenoid with $r$ as radius, then $x=2\pi rN,A=\pi r^2$
$\therefore$ $L={\mu }_0\left(\frac{x^2}{4{\pi }^2{r}^2}\right)\frac{\pi r^2}{l}\left[\therefore N=\frac{x}{2\pi r}\right]$
$\therefore$ $x=\sqrt{\frac{4\pi Ll}{{\mu }_0}}$