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The self induction (L) produced by solenoid of length ' $l$ ' having ' $N$ ' number of turns and cross sectional area ' $\mathrm{A}$ ' is given by the formula ( $\phi=$ magnetic flux, $\mu_0,=$ permeability of vacuum)
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The correct answer is:
$\mathrm{L}=\frac{\mu_0 \mathrm{~N}^2 \mathrm{~A}}{l}$
Magnetic field inside the solenoid is, $\mathrm{B}=\frac{\mu_0 \mathrm{NI}}{l}$ Flux inside the coil, $\phi=\mathrm{N}(\mathrm{BA})$
$\therefore \quad \phi=\frac{\mu_0 \mathrm{~N}^2 \mathrm{IA}}{l}$
Also, self inductance, $\mathrm{L}=\frac{\phi}{\mathrm{I}}=\frac{\frac{\mu_0 \mathrm{~N}^2 \mathrm{IA}}{l}}{\mathrm{I}}$ $\therefore \quad \mathrm{L}=\frac{\mu_0 \mathrm{~N}^2 \mathrm{~A}}{l}$
$\therefore \quad \phi=\frac{\mu_0 \mathrm{~N}^2 \mathrm{IA}}{l}$
Also, self inductance, $\mathrm{L}=\frac{\phi}{\mathrm{I}}=\frac{\frac{\mu_0 \mathrm{~N}^2 \mathrm{IA}}{l}}{\mathrm{I}}$ $\therefore \quad \mathrm{L}=\frac{\mu_0 \mathrm{~N}^2 \mathrm{~A}}{l}$
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