A solenoid has $n=2000\text{ turns/m}$ and across-sectional area $A=0.01{\text{ m}}^2$. A small coil of $N=50$ turns is wound tightly around the center of the solenoid. If the current $I$ in the solenoid changes from $2\text{ A}$ to $4\text{ A}$ in $\Delta t=0.1\text{ s}$, what is the induced EMF $\epsilon$ in the small coil?

Solution

1. The magnetic field inside the solenoid is given by $B={\mu }_{0}nI$, where ${\mu }_0=4\pi \times 10^{-7}\text{ T}\cdot \text{m/A}$, $n=2000\text{ turns/m}$, and $I$ is the current.
2. According to Faraday's Law, the induced EMF is $$\begin{array}{rl}\epsilon & =N\frac{\Delta \Phi }{\Delta t}\\ & =NA{\mu }_{0}n\frac{\Delta I}{\Delta t}\end{array}$$.
3. Substitute the given values: $\epsilon =50\times 0.01\times (4\pi \times 10^{-7})\times 2000\times \frac{4-2}{0.1}$.
4. Calculating the expression yields $\epsilon =0.2512\text{ V}$.
5. Hence the answer is (B).