Along solenoid of radius $R$ and $n$ turns per unit length carries a time-varying current $I=I_0\cos (\omega t)$. What is the magnitude of the induced electric field at a distance $r$ from the axis, where $r<R$?

Solution

1. The magnetic field inside along solenoid is given by $$\begin{array}{rl}B& ={\mu }_{0}nI\\ & ={\mu }_{0}n{I}_0\cos (\omega t)\end{array}$$.
2. Apply Faraday's Law in integral form, $\oint \vec{E}\cdot d\vec{l}=-\frac{d{\Phi }_B}{dt}$, over a circular path of radius $r<R$.
3. The left side is $E(2\pi r)$ and the magnetic flux is $$\begin{array}{rl}{\Phi }_B& =B(\pi r^2)\\ & ={\mu }_{0}n{I}_0\cos (\omega t)\pi r^2\end{array}$$.
4. Differentiating the flux with respect to time gives $\frac{d{\Phi }_B}{dt}=-{\mu }_{0}n{I}_0\omega \sin (\omega t)\pi r^2$.
5. Equating the terms, $E(2\pi r)={\mu }_{0}n{I}_0\omega \sin (\omega t)\pi r^2$, which simplifies to $E=\frac{1}{2}{\mu }_{0}n{I}_0\omega r\sin (\omega t)$.
6. Hence the answer is (A).