Answer
a) $450V/m$
b) $7.20W/m^2$, $540W/m^2$
Work Step by Step
(a) We know that
$E_{max}=cB_{max}$
$\implies E_{max}=(3\times 10^8m/s)(1.5\times 10^{-6}T)=450V/m$
Since the intensity is directly proportional to the square of the maximum electric field, thus the intensity of wave 2 is greater than that of wave 1.
(b) We can calculate the intensity of each wave as follows:
$I_1=c\epsilon_{\circ}E^2$
We plug in the known values to obtain:
$I_1=(3\times 10^8m/s)(8.85\times 10^{-12}N.m^2/C^2)(52V/m)^2=7.20W/m^2$
and $I_2=c\epsilon_{\circ}E_{max}^2$
We plug in the known values to obtain:
$I_2=(3\times 10^8m/s)(8.85\times 10^{-12}Nm^2/C^2)(450V/m^2)$
$I_2=540W/m^2$
