Problem 24.22

24.22.
The total power that comes from the bulb is
P = V I
So, the amount that goes into EM radiation is given by
P = .01 V I
$P = .01 \frac{V^{2}}{R}$
P = 1.31 W
If the power is radiated isotropically, then the intensity is given by
$I = \frac{P}{A}$
One meter away, we have
$I = \frac{1.31 W}{4\pi (1 m)^{2}}$
I = 0.104 W/m²
This is the average of the Poynting vector and also,
$I = S_{av} = \frac{c}{2\mu_{0}} B_{max}^{2}$
Solving for B_max, we have
$B_{max} = \sqrt{\frac{2\mu_{0}I}{c}}$
Putting in the numbers, we have
$B_{max} = 2.95 \times 10^{-3} T$