Question

Treat the bulb as a point source of electromagnetic radiation, estimate the maximum magnitudes of the electric and magnetic fields of the light that is incident on this page because of the visible light coming from your desk lamp. Assume only $5\%$ light is efficient in converting electrical energy to visible light.
(Hint: assume $60\,W$ bulb and a distance of $0.3\,m$ )

• A. $45\,Vm^{- 1},1.5\times 10^{- 7}T$
• B. $4.6\,Vm^{- 1},1.5\times 10^{- 7}T$
• C. $0.35\,Vm^{- 1},1.5\times 10^{- 7}T$
• D. $0.045\,Vm^{- 1},1.5\times 10^{- 2}T$

Answer 1

Answer: (A)

Intensity, $I=\frac{\left(0 . 05 P\right)}{4 \pi r^{2}} \, \, \, , \, \text{ where} \, P=5\% \, \, \text{ of } \, 60W$
$\Rightarrow I=\frac{\left(0 . 05 \times 60\right)}{4 \pi \times 0 . 3^{2}}=2.65Wm^{- 2}$
Also,
$I=\frac{1}{2}ε_{0}E_{0}^{2}\times c$ and $I=\frac{B_{0}^{2}}{2 \mu _{_{0}}}c$
Electric field Amplitude, $\Rightarrow E_{0}=\sqrt{\frac{2 I}{\epsilon _{0} c}}=\sqrt{\frac{2 \times 2 . 65}{8 . 85 \times 10^{- 12} \times 3 \times 10^{8}}}=45Vm^{- 1}$
Magnetic field Amplitude, $B_{0}=\sqrt{\frac{2 \mu _{0} I}{c}}=\sqrt{\frac{2 \times 4 \pi \times 10^{- 7} \times 2 . 65}{3 \times 10^{8}}}=1.5\times 10^{- 7}T$