The self-inductance of a solenoid is L, which | vedclass

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(D) Let $r$ be the radius of the solenoid and $N$ be the number of turns.The length of the wire $l_w$ is given by $l_w = N(2\pi r)$, so $N = \frac{l_w

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$\frac{\mu_0 l_w^2}{4\pi L}$

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(D) Let $r$ be the radius of the solenoid and $N$ be the number of turns.The length of the wire $l_w$ is given by $l_w = N(2\pi r)$, so $N = \frac{l_w}{2\pi r}$.The self-inductance $L$ of a solenoid of length $\ell$ is $L = \frac{\mu_0 N^2 A}{\ell} = \frac{\mu_0 N^2 \pi r^2}{\ell}$.Substitute $N = \frac{l_w}{2\pi r}$ into the formula for $L$:$L = \frac{\mu_0 (l_w / 2\pi r)^2 \pi r^2}{\ell} = \frac{\mu_0 l_w^2 \pi r^2}{4\pi^2 r^2 \ell} = \frac{\mu_0 l_w^2}{4\pi \ell}$.Solving for the length of the solenoid $\ell$:$\ell = \frac{\mu_0 l_w^2}{4\pi L}$.

The self-inductance of a solenoid is $L$, which is made by a wire of length $l_w$. What is the length of the solenoid?

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