Answer
(a) $9.95\times 10^8\frac{J}{m^3}$
(b) $1.50\times 10^{10}\frac{V}{m}$
Work Step by Step
We know that
$\mu_B=\frac{B^2}{2\mu_{\circ}}$
We plug in the known values to obtain:
$\mu_B=\frac{(50.0)^2}{2(4\pi\times 10^{-7})}$
$\mu_B=9.95\times 10^8\frac{J}{m^3}$
(b) We know that
$E=\sqrt{\frac{2\mu_B}{\epsilon_{\circ}}}$
We plug in the known values to obtain:
$E=\sqrt{\frac{2(9.95\times 10^8)}{8.85\times 10^{-12}}}$
$E=1.50\times 10^{10}\frac{V}{m}$
