Calculate the average value of the Poynting vector, \(S_{\text {ave }}\) for an electromagnetic wave having an electric field of amplitude \(100 . \mathrm{V} / \mathrm{m}\) a) What is the average energy density of this wave? b) How large is the amplitude of the magnetic field?

The Poynting vector, \(\boldsymbol{S}\), represents the power, or the rate of energy flow, per unit area in an electromagnetic wave. It can be found using the formula: $$\boldsymbol{S} = \frac{1}{\mu_0}\boldsymbol{E} \times \boldsymbol{B}$$ Where \(\boldsymbol{E}\) and \(\boldsymbol{B}\) are the electric and magnetic fields, respectively, and \(\mu_0\) is the permeability of free space (\(4 \pi \times 10^{-7} \,\mathrm{Tm/A}\)). However, we want to find the average value of the Poynting vector, which can be obtained as follows: $$S_{\text{ave}} = \frac{1}{2} \epsilon_0 c |E_0|^2$$ Where \(\epsilon_0\) is the permittivity of free space (\(8.85 \times 10^{-12} \,\mathrm{F/m}\)), \(c\) is the speed of light (\(3 \times 10^8 \,\mathrm{m/s}\)), and \(E_0\) is the amplitude of the electric field. Given the electric field amplitude of \(100 \,\mathrm{V/m}\), we can find \(S_{\text{ave}}\).

The total energy density, \(u\), of an electromagnetic wave is given by the sum of the electric and magnetic energy densities: $$u = u_\text{E} + u_\text{B}$$ Since the electric and magnetic fields are in phase, their average energy densities are equal and can be calculated as follows: $$u_\text{E} = \frac{1}{2}\epsilon_0 |E_0|^2$$ We can now find the average energy density of the wave.

To find the amplitude of the magnetic field, \(B_0\), we can use the relationship between the electric and magnetic fields in an electromagnetic wave: $$|E_0| = c |B_0|$$ Using the given value of \(E_0\) and the speed of light \(c\), we can find the amplitude of the magnetic field. Calculations:

Using the given values, we can calculate \(S_{\text{ave}}\): $$S_{\text{ave}} = \frac{1}{2} \times 8.85 \times 10^{-12} \times 3 \times 10^8 \times (100)^2$$ $$S_{\text{ave}} \approx 1.327 \times 10^{-3} \,\mathrm{W/m^2}$$

To calculate the average energy density of the wave, we can calculate \(u_\text{E}\) and then find the total energy density: $$u_\text{E} = \frac{1}{2} \times 8.85 \times 10^{-12} \times (100)^2$$ $$u_\text{E} \approx 4.425 \times 10^{-9} \,\mathrm{J/m^3}$$ Since \(u_\text{E} = u_\text{B}\), the total energy density \(u\) is: $$u = 2u_\text{E} = 2 \times 4.425 \times 10^{-9} \,\mathrm{J/m^3}$$ $$u \approx 8.85 \times 10^{-9} \,\mathrm{J/m^3}$$

Using the given values, we can find the amplitude of the magnetic field: $$|B_0| = \frac{|E_0|}{c} = \frac{100}{3 \times 10^8}$$ $$|B_0| \approx 3.33 \times 10^{-7} \,\mathrm{T}$$ To summarize: a) The average energy density of the wave is approximately \(8.85 \times 10^{-9} \,\mathrm{J/m^3}.\) b) The amplitude of the magnetic field is approximately \(3.33 \times 10^{-7} \,\mathrm{T}.\)