A solenoid of length $l$, radius $r$, and $n$ turns per unit length carries a current $I=I_0\cos (\omega t)$. The induced electric field at a distance $r/2$ from the axis inside the solenoid is?

Solution

1. The magnetic field inside the solenoid is given by $$\begin{array}{rl}B& ={\mu }_{0}nI\\ & ={\mu }_{0}n{I}_0\cos (\omega t)\end{array}$$.
2. According to Faraday's Law, $\oint \vec{E}\cdot d\vec{l}=-\frac{d{\Phi }_B}{dt}$, where the flux through a circle of radius $r/2$ is ${\Phi }_B=B\pi (r/2{)}^2$.
3. Substituting the expressions, we get $E(2\pi (r/2))=-\frac{d}{dt}[{\mu }_{0}n{I}_0\cos (\omega t)\pi (r^2/4)]$.
4. Evaluating the derivative yields $E(\pi r)={\mu }_{0}n{I}_0\omega \sin (\omega t)\frac{\pi r^2}{4}$.
5. Solving for $E$ gives $E=\frac{{\mu }_{0}n{I}_0\omega r\sin (\omega t)}{4}$.
6. Hence the answer is (A).