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

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

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(B) The inductance $L$ of a solenoid is given by $L = \frac{\mu_0 N^2 A}{\ell}$,where $N$ is the number of turns,$A = \pi r^2$ is the cross-sectional area,and $\ell$ is the length of the solenoid.The total length of the wire $\ell_1$ used to make $N$ turns of radius $r$ is $\ell_1 = N(2 \pi r)$,which implies $r = \frac{\ell_1}{2 \pi N}$.Substituting $A = \pi \left(\frac{\ell_1}{2 \pi N}\right)^2 = \frac{\ell_1^2}{4 \pi N^2}$ into the inductance formula:$L = \frac{\mu_0 N^2}{\ell} \cdot \frac{\ell_1^2}{4 \pi N^2} = \frac{\mu_0 \ell_1^2}{4 \pi \ell}$.Rearranging for $\ell_1$:$\ell_1^2 = \frac{4 \pi L \ell}{\mu_0}$.Therefore,$\ell_1 = \sqrt{\frac{4 \pi L \ell}{\mu_0}}$.

The length of thin wire required to manufacture a solenoid of inductance $L$ and length $\ell$ (if cross-sectional diameter is considered much less than its length) is:

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