Question

Assume a bulb of efficiency $2.5\%$ as a point source. The peak values of electric and magnetic fields produced by the radiation coming from a $100\, W$ bulb at a distance of $3\, m$ is respectively

• A. $2.5 \,V \,m^{-1}$, $3.6 \times 10^{-8}\, T$
• B. $4.2 \,V \,m^{-1}$, $2.8 \times 10^{-8}\, T$
• C. $4.08 \,V \,m^{-1}$, $1.36 \times 10^{-8}\, T$
• D. $3.6 \,V \,m^{-1}$, $4.2 \times 10^{-8}\, T$

Answer 1

Answer: (C)

$4.08 \,V \,m^{-1}$, $1.36 \times 10^{-8}\, T$

Answer 2

Here intensity, $I = \frac{power}{area}$ $= \frac{100 \times 2.5}{4\pi \left(3\right)^{2} \times 100}$ $= \frac{2.5}{36\pi} W\,m^{-2}$ Half of this intensity belongs to electric field and half of that to magnetic field. $\therefore\quad \frac{I}{2} = \frac{I}{4} \varepsilon_{0}\,E^{2}_{0}\,c$ or $\quad E_{0}= \sqrt{\frac{2I}{\varepsilon_{0}c}}$ $= \sqrt{\frac{2 \times \frac{2.5}{36\pi}}{\frac{1}{4\pi \times9\times 10^{9}} \times 3 \times 10^{8}}}$ $= 4.08\,V\,m^{-1}$ $\therefore\quad B_{0} = \frac{E_{0}}{c} = \frac{4.08}{3 \times 10^{8}}$ $= 1.36 \times 10^{-8}\,T$