$3\, m$ की दूरी पर स्थित किसी $100\, W$ बल्ब से आ रहे विकिरण द्वारा उत्पन्न विध्यूत एवं चुंबकीय क्षेत्रों की गणना कीजिए। आप यह जानते हैं कि बल्ब की दक्षता $2.5 \, \%$ है और यह एक बिंदु स्रोत है।

The bulb, as a point source, radiates light in all directions uniformly. At a distance of $3\; m ,$ the surface area of the surrounding sphere is

$A=4 \pi r^{2}=4 \pi(3)^{2}=113 \,m ^{2}$

The intensity I at this distance is

$I=\frac{\text { Power }}{\text { Area }}=\frac{100 \,W \times 2.5 \%}{113\, m ^{2}}$

$=0.022 \,W / m ^{2}$

Half of this intensity is provided by the electric field and half by the magnetic field.

$\frac{1}{2} I=\frac{1}{2}\left(\varepsilon_{0} E_{r m s}^{2} c\right)$

$=\frac{1}{2}\left(0.022 \,W / m ^{2}\right)$

$E_{m s} =\sqrt{\frac{0.022}{\left(8.85 \times 10^{-12}\right)\left(3 \times 10^{8}\right)}} \,V / m$

The value of $E$ found above is the root mean square value of the electric field. since the electric field in a light beam is sinusoidal, the peak electric field, $E_{0}$ is

$E_{0}=\sqrt{2} E_{ rms }=\sqrt{2} \times 2.9 \,V / m$

$=4.07 \,V / m$

Thus, you see that the electric field strength of the light that you use for reading is fairly large. Compare it with electric field strength of $TV$ or $FM$ waves, which is of the order of a few microvolts per metre.

Now, let us calculate the strength of the magnetic field. It is

$B_{r m s}=\frac{E_{m s}}{c}=\frac{2.9 \,Vm ^{-1}}{3 \times 10^{8}\, m s ^{-1}}=9.6 \times 10^{-9} \,T$

Again, since the field in the light beam is sinusoidal, the peak magnetic field is $B_{0}=\sqrt{2} B_{m s}=1.4 \times 10^{-8}\,T$.

Note that although the energy in the magnetic field is equal to the energy in the electric field, the magnetic field strength is evidently very weak.