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

Electromagnetic Induction

Solution:

$L = \frac{\mu_0 N^2 A}{l}$
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 $ $x = \sqrt{\frac{4 \pi L l}{\mu_0}}$