32. A long solenoid has 1000 turns. When a current of $4 A$ flows through it, the magnetic flux linked with each turn of the solenoid is $4 \times 10^{-3} Wb$. The self inductance of the solenoid is: (2016 - I)
Solution:
Total magnetic flux linked with the solenoid is $N \Phi = L I$.
Given $N = 1000$, $\Phi = 4 \times 10^{-3} Wb$, and $I = 4 A$.
$L = \frac{N \Phi}{I} = \frac{1000 \times 4 \times 10^{-3}}{4} = 1 H$.
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