Question

Self-inductance of a coil is 6 mH and the rate of flow of current in it is $10^3$ A/s. Find the induced emf produced in the coil.

Hint:

Always apply $\mathcal{E} = -L \dfrac{dI}{dt}$. The negative sign indicates that the induced emf opposes the change in current (Lenz’s law). For magnitude only, drop the sign.

Answer 1

The induced emf in a coil is given by \[ \mathcal{E} = L \frac{dI}{dt} \] where \(L\) = self-inductance, and \(\frac{dI}{dt}\) = rate of change of current. Substituting the given values: \[ L = 6 \times 10^{-3}~\text{H}, \frac{dI}{dt} = 10^3~\text{A/s} \] \[ \mathcal{E} = 6 \times 10^{-3} \times 10^3 = 6~\text{V} \] Final Answer: The induced emf produced in the coil is \(\mathbf{6~V}\).