An alternating sinusoidal current of frequency  $\omega =1000s^{-1}$ flows in the winding of a straight solenoid whose cross-sectional radius is equal to R = 6.0 cm. Find the ratio of peak values of electric and magnetic energies within the solenoid. If your answer is $n\times {10}^{-15},$ find n=?

$\text{ Here }\mathrm{I}={\mathrm{I}}_{\mathrm{m}}\cos \omega \mathrm{t}\text{, then the peak magnetic energy is }$

${\mathrm{W}}_{\mathrm{m}}=\frac{1}{2}{\mathrm{LI}}_{\mathrm{m}}^2=\frac{1}{2}{\mu }_0{\mathrm{n}}^2{\mathrm{I}}_{\mathrm{m}}^2\pi {\mathrm{R}}^2\mathrm{d}$

Changing magnetic field induces an electric field which by Faraday's law is given by 

$\mathrm{E}\cdot 2\mathrm{\pi r}=-\frac{\mathrm{d}}{\mathrm{dt}}\int \overrightarrow{\mathrm{B}}\cdot \mathrm{d}\overrightarrow{\mathrm{S}}={\mathrm{\pi r}}^2{\mathrm{\mu }}_0{\mathrm{nIm}}_{\mathrm{m}}\mathrm{\omega sin}\mathrm{\omega t}$

$\mathrm{E}=\frac{1}{2}{\mathrm{r}}_0{\mathrm{nI}}_{\mathrm{m}}\mathrm{\omega sin}\mathrm{\omega t}$

The associated peak electric energy is 

${\mathrm{W}}_{\mathrm{e}}=\int \frac{1}{2}{\mathrm{\epsilon }}_0{\mathrm{E}}^2{\mathrm{d}}^3\mathrm{r}=\frac{1}{8}{\mathrm{\epsilon }}_0{\mathrm{\mu }}_0^2{\mathrm{n}}^2{\mathrm{I}}_{\mathrm{m}}^2{\mathrm{\omega }}^2{\sin }^2\mathrm{\omega t}\times \frac{{\mathrm{\pi R}}^4}{2}\mathrm{d}$

$\text{ Hence }\frac{{\mathrm{W}}_{\mathrm{e}}}{{\mathrm{W}}_{\mathrm{m}}}=\frac{1}{8}{\mathrm{\epsilon }}_0{\mathrm{\mu }}_0(\mathrm{\omega R}{)}^2=\frac{1}{8}{\left(\frac{\mathrm{\omega R}}{\mathrm{c}}\right)}^2=5\times {10}^{-15}$