Question 101: EM waves

Question

An electric bulb is rated as 200 W. What will be the peak magnetic field at 4 m distance produced by the radiations coming from this bulb? Consider this bulb as a point source with 3.5% efficiency.

Accepted Answer

The total power (P
T
) of the light bulb is given as 200 W, but only 3.5% of this power is actually emitted as radiation, which we will call P.
So, Effective power output of the bulb
$
\mathrm{P}=\frac{3 \cdot 5}{100} 	imes 200=7 \mathrm{~W}
$
$
\begin{aligned}
	ext { Intensity } \mathrm{I} & =\frac{\mathrm{P}}{4 \pi r^2}=\frac{7}{4 \pi(4)^2} \mathrm{~W} / \mathrm{m}^2 \\
	ext { Intensity } \mathrm{I} & =	ext { Average energy density } 	imes c \\
& =\frac{1}{2} \in_0 \mathrm{E}_0^2 c \\
& =\frac{1}{2} \frac{\mathrm{B}_0^2}{\mu_0} c
\end{aligned}
$
$
\begin{aligned}
	herefore \quad \mathrm{B}_0 & =\sqrt{\frac{2 \mu_0 \mathrm{I}}{c}} \\
& =\sqrt{\frac{2 	imes 4 \pi 	imes 10^{-7} 	imes 7}{3 	imes 10^8 	imes 4 \pi 	imes 16}} \\
& =1.71 	imes 10^{-8} \mathrm{~T}
\end{aligned}
$

Answer

1.19 $	imes$ 10
$-$8
T

Answer

1.71 $	imes$ 10
$-$8
T

Answer

0.84 $	imes$ 10
$-$8
T

Answer

3.36 $	imes$ 10
$-$8
T

EM waves

An electric bulb is rated as 200 W. What will be the peak magnetic field at 4 m distance produced by the radiations coming from this bulb? Consider this bulb as a point source with 3.5% efficiency.

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