Answer
$E = 2200~V/m$
$B = 7.3\times 10^{-6}~T$
Work Step by Step
We can find the intensity at the surface of the bulb:
$I = \frac{P}{4\pi~d^2}$
$I = \frac{100~W}{(4\pi)~(0.035~m)^2}$
$I = 6496~W/m^2$
We can find the electric field strength:
$I = \frac{1}{2}~E^2~\epsilon_0~c$
$E^2 = \frac{2~I}{\epsilon_0~c}$
$E = \sqrt{\frac{2~I}{\epsilon_0~c}}$
$E = \sqrt{\frac{(2)(6496~W/m^2)}{(8.854\times 10^{-12}~F/m)~(3.0\times 10^8~m/s)}}$
$E = 2200~V/m$
We can find the magnetic field strength:
$B = \frac{E}{c}$
$B = \frac{2200~V/m}{3.0\times 10^8~m/s}$
$B = 7.3\times 10^{-6}~T$
