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The self-inductance of a long solenoid of cross-sectional area $\mathrm{A}$, length $l$ and $\mathrm{n}$ turns'per unit length is given by
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Verified Answer
The correct answer is:
$\mu_0 n^2 \mathrm{Al}$
Magnetic flux, $\phi_{\mathrm{B}}=\mathrm{B} . \mathrm{A}$
for solenoid, $\mathrm{B}=\mu_0 \mathrm{n}$ I
$$
\phi_{\mathrm{B}}=\mu_0 \mathrm{nIA}
$$
The total magnetic flux associates with the solenoid
$$
\begin{aligned}
& \phi_{\mathrm{B}}=\mu_0 \mathrm{nIA} \times \mathrm{N} \\
& \mathrm{N}=\mathrm{nL} \\
& =\mu \mathrm{n}^2 \mathrm{IAL}
\end{aligned}
$$
for solenoid, $\mathrm{B}=\mu_0 \mathrm{n}$ I
$$
\phi_{\mathrm{B}}=\mu_0 \mathrm{nIA}
$$
The total magnetic flux associates with the solenoid
$$
\begin{aligned}
& \phi_{\mathrm{B}}=\mu_0 \mathrm{nIA} \times \mathrm{N} \\
& \mathrm{N}=\mathrm{nL} \\
& =\mu \mathrm{n}^2 \mathrm{IAL}
\end{aligned}
$$
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