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

Intensity at a distance $r$ from a point source of power $P$ is given by 

$I=\frac{P}{4\pi r^2}$

So intensity at a distance of $3 m$ from the bulb with $2.5\%$ efficiency will be 

$$\begin{gathered} I=\frac{100}{4\pi {\left(3\right)}^2}\times \frac{2.5}{100} \\ =0.022 W/m^2 \end{gathered}$$

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

$I_E=\frac{I}{2}=0.011 W/m^2$

$I_E$ is given by $\frac{1}{2}{\epsilon }_0{E}^{2}c$

$I_E=\frac{1}{2}{\epsilon }_0{E}^{2}c$ or $E=\sqrt{\frac{2I_E}{{\epsilon }_{0}c}}$

Substituting the values, we have 

$$\begin{gathered} E=\sqrt{\frac{2\times 0.011}{\left(8.85\times {10}^{-12}\right)\left(3\times {10}^8\right)}} \\ =2.9 V/m \end{gathered}$$

From the equation , $c=\frac{E}{B}$

$$\begin{gathered} B=\frac{E}{c}=\frac{2.9}{3.0\times {10}^8} \\ =9.6\times {10}^{-9}T \end{gathered}$$