Calculate the electric and magnetic fields produced by the radiation coming from a $100\mathrm{W}$ bulb at a distance of $3\mathrm{m}$. Assume that the efficiency of the bulb is $2.5\%$ and it is a point source.

The bulb, as a point source, radiates light in all directions uniformly. At a distance of $3\mathrm{m}$, the surface area of the surrounding sphere is $A=4\pi r^2=4\pi (3{)}^2=113{\mathrm{m}}^2$ The intensity $I$ at this distance is $I=\frac{\text{ Power }}{\text{ Area }}=\frac{100\mathrm{W}\times 2.5\%}{113{\mathrm{m}}^2}$$=0.022\mathrm{W}/{\mathrm{m}}^2$ Half of this intensity is provided by the electric field and half by the magnetic field. $\frac{1}{2}I=\frac{1}{2}\left({\epsilon }_0{E}_{rms}^{2}c\right)$$=\frac{1}{2}\left(0.022\mathrm{W}/{\mathrm{m}}^2\right)$$E_{rms}=\sqrt{\frac{0.022}{\left(8.85\times 10^{-12}\right)\left(3\times 10^8\right)}}\mathrm{V}/\mathrm{m}$$=2.9\mathrm{V}/\mathrm{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}_{\mathrm{rms}}=\sqrt{2}\times 2.9\mathrm{V}/\mathrm{m}$$=4.07\mathrm{V}/\mathrm{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 $B_{rms}=\frac{E_{rms}}{c}=\frac{2.9{\mathrm{Vm}}^{-1}}{3\times 10^8\mathrm{m}{\mathrm{s}}^{-1}}=9.6\times 10^{-9}\mathrm{T}$ Again, since the field in the light beam is sinusoidal, the peak magnetic field is $B_0=\sqrt{2}{B}_{rms}=1.4\times 10^{-8}$ 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.