The length of a wire required to make a solenoid of length $l$ and self-induction $L$ is
The length of a wire required to make a solenoid of length $l$ and self-induction $L$ is
- $\sqrt{\frac{4 \pi L l}{\mu_0}}$
- $\sqrt{\frac{L I}{4 \pi \mu_0}}$
- $\sqrt{\frac{2 \pi L I}{\mu_0}}$
- $\sqrt{\frac{\mu_0 L I}{4 \pi}}$
Solution
Length of wire in solenoid $=2 \pi r \times N$
$
=2 \pi r \sqrt{\frac{L l}{\mu_0 \pi r^2}}\left\{\begin{array}{c}
\therefore L=\frac{\mu_0 N^2 A}{l} \\
A=\pi r^2
\end{array}\right\}=\sqrt{\frac{4 \pi L l}{\mu_0}}
$
Asked in: AP EAMCET 2018 (22 Apr Shift 2)
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