Answer
(a) $1.1V/m$
(b) $3.3KW/m^2$
(c) $1.6KW/m^2$
Work Step by Step
(a) We know that
$E=cB$
We plug in the known values to obtain:
$E=(3.00\times 10^8)(3.7\times 10^{-6})$
$E=1.1V/m$
(b) As $I=\frac{c}{\mu_{\circ}}B^2$
We plug in the known values to obtain:
$I=\frac{3.00\times 10^8}{4\pi\times 10^{-7}}(3.7\times 10^{-6})^2$
$I=3.3KW/m^2$
(c) We can find the average intensity as
$I_{avg}=\frac{c}{\mu{\circ}}B_{rms}^2$
We plug in the known values to obtain:
$I_{avg}=\frac{3.00\times 10^8}{4\pi\times 10^{-7}}(\frac{3.7\times 10^{-6}}{\sqrt2})=1.6KW/m^2$
