Question 69: EM waves

Question

For the plane electromagnetic wave given by $$E=E_{0} \sin (\omega t-k x)$$ and $$B=B_{0} \sin (\omega t-k x)$$, the ratio of average electric energy density to average magnetic energy density is

Accepted Answer

The average energy density of an electromagnetic wave is equally shared between the electric field and the magnetic field. This means that the average electric energy density is equal to the average magnetic energy density.
The average electric energy density ($u_E$) is given by:
$ u_E = \frac{1}{2} \varepsilon_0 E^2 $
and the average magnetic energy density ($u_B$) is given by:
$ u_B = \frac{1}{2\mu_0} B^2 $
where $\varepsilon_0$ is the permittivity of free space, $\mu_0$ is the permeability of free space, $E$ is the electric field strength, and $B$ is the magnetic field strength.
In an electromagnetic wave, $E$ and $B$ are related by the equation:
$ E = cB $
where $c$ is the speed of light in vacuum.
Substituting this into the energy density equations gives:
$ u_E = \frac{1}{2} \varepsilon_0 (cB)^2 $
and
$ u_B = \frac{1}{2\mu_0} B^2 $
Since $c = \frac{1}{\sqrt{\varepsilon_0 \mu_0}}$, we can rewrite $u_E$ as:
$ u_E = \frac{1}{2} \frac{1}{\mu_0} B^2 = u_B $
Therefore, the ratio of the average electric energy density to the average magnetic energy density is 1.

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EM waves

For the plane electromagnetic wave given by $E=E_{0} \sin (\omega t-k x) and B=B_{0} \sin (\omega t-k x)$, the ratio of average electric energy density to average magnetic energy density is

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