The self-inductance of a solenoid is $L=5\text{ mH}$. If the current changes from $I_1=2\text{ A}$ to $I_2=4\text{ A}$ in $\Delta t=0.01\text{ s}$, what is the induced EMF $\mathcal{E}$?

Solution

1. The formula for induced EMF is $\mathcal{E}=-L\frac{\Delta I}{\Delta t}$.
2. Calculate the change in current: $$\begin{array}{rl}\Delta I& =4\text{ A}-2\text{ A}\\ & =2\text{ A}\end{array}$$.
3. Substitute values: $\mathcal{E}=5\times 10^{-3}\text{ H}\times \frac{2\text{ A}}{0.01\text{ s}}$.
4. Calculate the result: $$\begin{array}{rl}\mathcal{E}& =0.005\times 200\\ & =1\text{ V}\end{array}$$.
5. Hence the answer is (A).