The self-inductance of a solenoid of length L | vedclass

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(A) The magnetic field inside a long solenoid is given by $B = \mu_0 n I$,where $n = \frac{N}{L}$ is the number of turns per unit length.Thus,$B = \fr

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$\frac{\mu_0 N^2 A}{L}$

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(A) The magnetic field inside a long solenoid is given by $B = \mu_0 n I$,where $n = \frac{N}{L}$ is the number of turns per unit length.Thus,$B = \frac{\mu_0 N I}{L}$.The magnetic flux $\phi$ through each turn is $\phi = B \cdot A = \frac{\mu_0 N I A}{L}$.The total magnetic flux linkage is $\Phi = N \phi = \frac{\mu_0 N^2 I A}{L}$.By definition,the self-inductance $L_{ind}$ is given by $\Phi = L_{ind} I$.Comparing the two expressions,we get $L_{ind} = \frac{\mu_0 N^2 A}{L}$.

The self-inductance of a solenoid of length $L$,area of cross-section $A$,and having $N$ turns is:

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