Magnitude of the electric and magnetic fields in an electromagnetic wave radiated by a 200 W bulb at a distance 2 m from it is assuming efficiency of bulb is 5% and it behaves like a point source.
Answer & explanation
Correct answer: option 1
Effective power of the bulb, $P=\frac{5}{100} \times 200=10 W$
Intensity at distance r is given by $I=\frac{P}{4 \pi r^2}=\frac{10}{4 \times 3.14 \times 2^2} W / m^2=\frac{10}{16 \times 3.14} W / m^2$
$=0.199 W / m^2=0.2 W / m^2$
$I=\frac{1}{2} \varepsilon_0 E_0^2 c$
$E_0^2=\frac{2 I}{c \times \varepsilon_0}$
$\Rightarrow E_0=\sqrt{\frac{2 I}{c \times \varepsilon_0}}$
$\Rightarrow E_0=\sqrt{\frac{2 \times 0.2}{3 \times 10^8 \times 8.85 \times 10^{-12}}} \Rightarrow E_0=12.27 N / C$
$\Rightarrow B_0=\frac{E_0}{c}=\frac{12.27}{3 \times 10^8}=\frac{12.27}{3} \times 10^{-8} T=4.09 \times 10^{-8} T$