Question

An electromagnetic wave propagating in vacuum has electric and magnetic fields given by \(\vec{E}(\vec{x}, t)=\vec{E}_{0} \cos (\vec{k} \cdot \vec{x}-\omega t)\) and \(\vec{B}(\vec{x}, t)=\vec{B}_{0} \cos (k \cdot \vec{x}-\omega t),\) where \(\vec{B}_{0}\) is given by \(\vec{B}_{0}=\vec{k} \times \vec{E}_{0} / \omega\) and the wave vector \(\vec{k}\) is perpendicular to both \(\vec{E}_{0}\) and \(\vec{B}_{0}\). The magnitude of \(\vec{k}\) and the angular frequency \(\omega\) satisfy the dispersion relation, \(\omega / \vec{k} \mid=\left(\mu_{0} \epsilon_{0}\right)^{-1 / 2}\) where \(\mu_{0}\) and \(\epsilon_{0}\) are the permeability and permittivity of free space, respectively. Such a wave transports energy in both its electric and magnetic fields. Calculate the ratio of the energy densities of the magnetic and electric fields, \(u_{B} / u_{E}\), in this wave. Simplify your final answer as much as possible.

Short answer

Question: Determine the ratio of energy densities of the magnetic and electric fields, \(u_B/u_E\), given the expressions for the electric and magnetic fields and the relationship between their amplitudes. Answer: The ratio of energy densities of the magnetic and electric fields is given by \(\frac{u_B}{u_E} = \frac{1}{\mu_0 \epsilon_0}\).

Step-by-step solution

Write down the expressions for energy densities

The energy densities for the magnetic and electric fields can be expressed as: $$ u_B = \frac{1}{2 \mu_0} B^2(\vec{x}, t) \quad \text{and} \quad u_E = \frac{1}{2 \epsilon_0} E^2(\vec{x}, t) $$

Substitute the expressions for magnetic and electric fields

Insert the given expressions for \(\vec{E}(\vec{x}, t)\) and \(\vec{B}(\vec{x}, t)\) and find the squared magnitudes: $$ u_B = \frac{1}{2 \mu_0} \left( B_0 \cos(k\cdot \vec{x} - \omega t)\right)^2 \quad \text{and} \quad u_E = \frac{1}{2 \epsilon_0} \left( E_0 \cos(\vec{k} \cdot \vec{x} - \omega t)\right)^2 $$

Calculate the ratio of energy densities

Divide \(u_B\) by \(u_E\): $$ \frac{u_B}{u_E} = \frac{\frac{1}{2\mu_0} (B_0\cos(k\cdot \vec{x} - \omega t))^2}{\frac{1}{2\epsilon_0} (E_0\cos(\vec{k}\cdot \vec{x} - \omega t))^2} $$ Simplify the expression by cancelling out the common terms: $$ \frac{u_B}{u_E} = \frac{\epsilon_0}{\mu_0} \frac{B_0^2}{E_0^2} $$

Substitute the expression for \(B_0\) in terms of \(E_0\)

From the given information, we have \(\vec{B}_0 = \frac{\vec{k} \times \vec{E}_0}{\omega}\). We can now substitute this into the previous equation and simplify the ratio: $$ \frac{u_B}{u_E} = \frac{\epsilon_0}{\mu_0} \frac{\left(\frac{k}{\omega} E_0\right)^2}{E_0^2} = \frac{\epsilon_0}{\mu_0} \frac{k^2}{\omega^2} $$

Use the dispersion relation to simplify the result

From the dispersion relation, we know that \(\frac{\omega}{k} = \left(\mu_0 \epsilon_0\right)^{-1/2}\). Therefore, we can write \(\frac{k^2}{\omega^2} = \frac{1}{\mu_0\epsilon_0}\) and substitute this into the ratio: $$ \frac{u_B}{u_E} = \frac{\epsilon_0}{\mu_0} \frac{1}{\mu_0 \epsilon_0} = \frac{1}{\mu_0^2 \epsilon_0^2} \mu_0 \epsilon_0 = \frac{1}{\mu_0 \epsilon_0} $$ So, the final simplified ratio of the energy densities of the magnetic and electric fields is: $$ \frac{u_B}{u_E} = \frac{1}{\mu_0 \epsilon_0} $$