91Ó°ÊÓ

The power transported per unit area is known as the intensity .

The formula used to calculate the intensity ( l ) is:

$l=\frac{P}{A}$

Where, A is area measured in the direction perpendicular to the energy and P is the power in watts.

The formula used to determine the amplitude of electric and magnetic fields of the wave are:

$\begin{array}{r}{E}_{max}=\sqrt{\frac{2l}{{\mathrm{Î\mu }}_{0}C}}\\ B_{max}=\frac{E_{max}}{c}\end{array}$

Where,${\mathrm{Î\mu }}_0=8.85×{10}^{-12}C/N.m^2$ and is the speed of light that is equal to $3.0×{10}^{8}m/s$.

Given that,

$$\begin{gathered} P=75W \\ r=3×{10}^{-2}m \end{gathered}$$

Given that, only 5% of the energy goes to visible-light. So, we use the 5% of power ( P ) .

The formula used to calculate the intensity ( l ) is:

$\begin{array}{r}I=\frac{P}{A}\\ =\frac{5\mathrm{\%}P}{4\mathrm{π}\left(3×{10}^{−2}\right)}\\ =\frac{\left(0.05\right)×75}{113×{10}^{−4}}\\ =331\mathrm{W}/{\mathrm{m}}^2\end{array}$

Hence, the intensity of visible-light at the surface of the bulb$330W/m^2$ .

The amplitude of electric field is:

$E_{max}=\sqrt{\frac{2I}{{\mathrm{Î\mu }}_{0}c}}$

Substitute the values

$\begin{array}{r}{E}_{max}=\sqrt{\frac{2×\left(331\right)}{\left(8.85×{10}^{−12}\right)\left(3×{10}^8\right)}}\\ =500\mathrm{V}/\mathrm{m}\end{array}$

The amplitude of magnetic field is:

$B_{max}=\frac{E_{max}}{c}$

Substitute the values

$\begin{array}{r}{B}_{max}=\frac{500}{3×{10}^8}\\ =1.7×{10}^{−6}\mathrm{T}\end{array}$

Hence, the amplitudes of the electric and magnetic fields at the surface, for a sinusoidal wave with intensity $330W/m^2$are 500 V/m and$1.7×{10}^{-6}T$ respectively.