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The self inductance of a solenoid of length $L$, area of cross-section $A$ and having $N$ turns is
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Verified Answer
The correct answer is:
$\frac{\mu_0 N^2 A}{L}$
Given,
length of Solenoid $=2$
Ara of cross - section $=A$
No. of tums $\quad=\mathrm{N}$
We know that total number of tums in solenoid, $N=n \&$ Magnetic field inside tong solenoid, $B=\mu_0 n_i$
So, flux, through $N$ thurs, $\phi_t=N \phi_1 \quad\left\{\theta_1=B A=\mu_0 n i A\right\}$
$\begin{aligned} & =n l \times \mu_0 \cdot n i A \\ & =\mu_0 n^2 l A i\end{aligned}$
$u \sin g_1 \quad \psi_1=2 i$
Inductanes of coil, $\because \mu \cdot n^2 A l i=l i$
$\Rightarrow \quad L \quad=\mu_0 n^2 A l=\mu_0 \frac{N^2}{l^2} A l=\frac{\mu_0 N^2 A}{l}$
Hence, option (1) is correct
length of Solenoid $=2$
Ara of cross - section $=A$
No. of tums $\quad=\mathrm{N}$
We know that total number of tums in solenoid, $N=n \&$ Magnetic field inside tong solenoid, $B=\mu_0 n_i$
So, flux, through $N$ thurs, $\phi_t=N \phi_1 \quad\left\{\theta_1=B A=\mu_0 n i A\right\}$
$\begin{aligned} & =n l \times \mu_0 \cdot n i A \\ & =\mu_0 n^2 l A i\end{aligned}$
$u \sin g_1 \quad \psi_1=2 i$
Inductanes of coil, $\because \mu \cdot n^2 A l i=l i$
$\Rightarrow \quad L \quad=\mu_0 n^2 A l=\mu_0 \frac{N^2}{l^2} A l=\frac{\mu_0 N^2 A}{l}$
Hence, option (1) is correct
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