The maximum electric field 10 mfrom an isotropic point source of light is 2.0 V/m. What are (a) the maximum value of the magnetic field and (b) the average intensity of the light there? (c) What is the power of the source?

The amplitude of the electric field is, E_m= 2 V/m.

Distance, r = 10 m.

We use the wave speed relation, which is related totheamplitude of electric and magnetic field components. Using this relation, we can find the magnitude ofthemagnetic field component. Using the intensity relation of the wave, we can find its intensity. Using intensity and power relations, we can find the power ofthesource.

The wave speed and amplitudes of electric and magnetic fields are related as

$\mathrm{c}=\frac{{\mathrm{E}}_{\mathrm{m}}}{{\mathrm{B}}_{\mathrm{m}}}$

The amplitude of the magnetic field component inthewave is given by,

${\mathrm{B}}_{\mathrm{m}}=\frac{{\mathrm{E}}_{\mathrm{m}}}{\mathrm{c}}$

Substitute the values in the above expression, and we get,

$$\begin{gathered} {\mathrm{B}}_{\mathrm{m}}=\frac{2\mathrm{V}/\mathrm{m}}{3\times {10}^8\mathrm{m}/\mathrm{s}} \\ =6.7\times {10}^{-9}\mathrm{T} \end{gathered}$$

Thus, the maximum value of the magnetic field is $6.7\times {10}^{-9}\mathrm{T}$ .

The average intensity can be calculated as,

$I_{avg}=\frac{{E_m}^2}{2{\mu }_{0}c}$

Substitute the values in the above expression, and we get,

$$\begin{gathered} I_{avg}=\frac{{\left(2V/m\right)}^2}{2\times \left(4\pi \times {10}^{-7}T.m/A\right)\left(3\times {10}^{8}m/s\right)} \\ =5.3\times {10}^{-3}w/m^2 \end{gathered}$$

Thus, the average intensity of the light is $5.3\times {10}^{-3}{\text{W/m}}^{\text{2}}$ .

The relation between power and average intensity of light is given by,

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

Substitute the values in the above expression, and we get,

$$\begin{gathered} P=4\pi {\left(10m\right)}^2\left(5.3\times {10}^{-3}w/m^2\right) \\ =6.7w \end{gathered}$$

Thus, the power of the source is 6.7 w.