Module 3 : MAGNETIC FIELD
Lecture 19 : Time Varying Field
 

\includegraphics{fig3.47.eps}

Equating these, we get for $r\le R$  ,\begin{displaymath}E = -\frac{r}{2}\frac{dB}{dt} = -\frac{B_0\omega}{2}r\cos\omega t\end{displaymath}

For $r>R$, the flux is $\Phi = \pi R^2B(t)$, so that \begin{displaymath}{\cal E} = -\frac{d\Phi}{dt} = -\pi R^2\frac{dB}{dt}\end{displaymath}

and the electric field foir $r>R$ is  \begin{displaymath}E = -\frac{R^2}{2r}\frac{dB}{dt} = -\frac{B\omega R^2}{2r}\cos\omega t\end{displaymath}

Exercise 1

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