The length of a wire required to make a solen | vedclass

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(A) The self-inductance of a solenoid is given by $L = \frac{\mu_0 N^2 A}{l}$,where $N$ is the number of turns,$A = \pi r^2$ is the cross-sectional ar

Vedclass · Accepted Answer

$\sqrt{\frac{4 \pi L l}{\mu_0}}$

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(A) The self-inductance of a solenoid is given by $L = \frac{\mu_0 N^2 A}{l}$,where $N$ is the number of turns,$A = \pi r^2$ is the cross-sectional area,and $l$ is the length of the solenoid.From this,$N^2 = \frac{L l}{\mu_0 A} = \frac{L l}{\mu_0 \pi r^2}$,so $N = \frac{1}{r} \sqrt{\frac{L l}{\mu_0 \pi}}$.The total length of the wire used is $W = N 	imes (2 \pi r)$.Substituting the value of $N$,we get $W = \left( \frac{1}{r} \sqrt{\frac{L l}{\mu_0 \pi}} ight) 	imes (2 \pi r)$.$W = 2 \pi \sqrt{\frac{L l}{\mu_0 \pi}} = \sqrt{\frac{4 \pi^2 L l}{\mu_0 \pi}} = \sqrt{\frac{4 \pi L l}{\mu_0}}$.

The length of a wire required to make a solenoid of length $l$ and self-induction $L$ is

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