Discussion
A coil has 2000 turns. A current of 2A causes a flux 4 mWb to link the coil. What is the self-inductance of the coil?
Answer: B
The self-inductance is 4 H.
Self-inductance is the flux linkage per ampere: \(L = N\Phi / I\).
\[ \begin{aligned} L &= \frac{2000 \times 4 \times 10^{-3}}{2} \\ &= \frac{8}{2} = 4\ \text{H} \end{aligned} \]
Forgetting to divide by the current gives 8 H, which is not offered; 2 H and 1 H come from dividing by the wrong quantity.
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