Solution:
Formula: \(e = L \frac{dI}{dt}\). Substituting \(L = 0.02\text{ H}\), \(dI = 4\text{ A}\) and \(dt = 2 \times 10^{-2}\text{ s}\), we get \(e = 0.02 \times 200 = 4\text{ V}\).
A coil has a self-inductance of 0.02 H. The current through it, is allowed to change at the rate of 4 A in \(2 \times 10^{-2}\text{ s}\). The e.m.f induced in the coil will be
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