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Inductance per unit length near the middle of a long solenoid is $\left(\mu_0=\right.$ permeability of free space, $\mathrm{n}=$ number of turns per unit length, $\mathrm{d}=$ the diameter of the solenoid)
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Verified Answer
The correct answer is:
$\mu_0 \pi\left(\frac{\mathrm{nd}}{2}\right)^2$
The inductance long solenoid is
$$
\begin{aligned}
& \mathrm{L}=\frac{\mu_0 \mathrm{~N}^2 \mathrm{~A}}{l} \\
& \mathrm{~L}=\mu_0\left(\frac{\mathrm{N}}{l}\right)^2 \times \pi \times \frac{\mathrm{d}^2}{4}
\end{aligned}
$$
$\therefore \quad$ The inductance per unit length near the middle of a long solenoid is:
$$
\frac{\mathrm{L}}{l}=\mu_0 \pi\left(\frac{\mathrm{nd}}{2}\right)^2 \quad \ldots .\left(\because \frac{\mathrm{N}}{l}=\mathrm{n}\right)
$$
$$
\begin{aligned}
& \mathrm{L}=\frac{\mu_0 \mathrm{~N}^2 \mathrm{~A}}{l} \\
& \mathrm{~L}=\mu_0\left(\frac{\mathrm{N}}{l}\right)^2 \times \pi \times \frac{\mathrm{d}^2}{4}
\end{aligned}
$$
$\therefore \quad$ The inductance per unit length near the middle of a long solenoid is:
$$
\frac{\mathrm{L}}{l}=\mu_0 \pi\left(\frac{\mathrm{nd}}{2}\right)^2 \quad \ldots .\left(\because \frac{\mathrm{N}}{l}=\mathrm{n}\right)
$$
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