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Chapter 22: Problem 41

Prove that, in an EM wave traveling in vacuum, the electric and magnetic energy densities are equal; that is, prove that,$$\frac{1}{2} \epsilon_{0} E^{2}=\frac{1}{2 \mu_{0}} B^{2}$$ at any point and at any instant of time.,

Short Answer

Expert verified
Answer: Yes, in an electromagnetic wave traveling in a vacuum, the electric and magnetic energy densities are equal at any given point and time.

Step by step solution

01

Identify the formulas for electric and magnetic energy densities

In an EM wave, the electric energy density (u_E) and magnetic energy density (u_B) are given by the following formulas: $$ u_E = \frac{1}{2}\epsilon_0 E^2 $$ and $$ u_B = \frac{1}{2\mu_0}B^2 $$ Where: - \(E\) is the electric field - \(B\) is the magnetic field - \(\epsilon_0\) is the vacuum permittivity (\(8.854\times10^{-12} F/m\)) - \(\mu_0\) is the vacuum permeability (\(4\pi\times10^{-7} Tm/A\))
02

Establish the relationship between \(E\) and \(B\) in an EM wave

In an EM wave propagating through a vacuum, the electric field (E) and magnetic field (\(B\)) are perpendicular to each other, and their magnitudes are related by the following equation: $$ E = cB $$ Where \(c\) is the speed of light in vacuum (\(3\times10^8 m/s\)).
03

Substitute the relationship between \(E\) and \(B\) into the formulas for energy densities

From the relationship above, we can express \(B\) in terms of \(E\). So we get: $$ B = \frac{E}{c} $$ Now, substitute this value of \(B\) into the magnetic energy density equation: $$ u_B = \frac{1}{2\mu_0}\left(\frac{E}{c}\right)^2 $$
04

Simplify the equation for magnetic energy density

Now, simplify the equation for magnetic energy density: $$ u_B = \frac{1}{2\mu_0}\frac{E^2}{c^2} $$
05

Prove that the energy densities are equal

Now, we just need to prove that \(u_E\) is equal to \(u_B\). From step 4, we have: $$ u_B = \frac{1}{2\mu_0}\frac{E^2}{c^2} $$ Note that \(c^2 = \frac{1}{\epsilon_0\mu_0}\), so we can write the equation for \(u_B\) as follows: $$ u_B = \frac{1}{2}\frac{E^2}{\epsilon_0} $$ Comparing the equations for \(u_E\) and \(u_B\), we can see that they are equal: $$ \frac{1}{2}\epsilon_0 E^2 = \frac{1}{2}\epsilon_0 E^2 $$ Thus, we can conclude that in an EM wave traveling in a vacuum, the electric and magnetic energy densities are equal at any given point and time.

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Most popular questions from this chapter

Using Faraday's law, show that if a magnetic dipole antenna's axis makes an angle \(\theta\) with the magnetic field of an EM wave, the induced emf in the antenna is reduced from its maximum possible value by a factor of cos $\theta .$ [Hint: Assume that, at any instant, the magnetic field everywhere inside the loop is uniform. \(]\)
A sinusoidal EM wave has an electric field amplitude $E_{\mathrm{m}}=32.0 \mathrm{mV} / \mathrm{m} .$ What are the intensity and average energy density? [Hint: Recall the relationship between amplitude and rms value for a quantity that varies sinusoidally.]
In order to study the structure of a crystalline solid, you want to illuminate it with EM radiation whose wavelength is the same as the spacing of the atoms in the crystal \((0.20 \mathrm{nm}) .\) (a) What is the frequency of the EM radiation? (b) In what part of the EM spectrum (radio, visible, etc. \()\) does it lie?
By expressing \(\epsilon_{0}\) and \(\mu_{0}\) in base SI units $(\mathrm{kg}, \mathrm{m}, \mathrm{s}, \mathrm{A})$ prove that the only combination of the two with dimensions of speed is $\left(\boldsymbol{\epsilon}_{0} \mu_{0}\right)^{-1 / 2}$.
An unpolarized beam of light (intensity \(I_{0}\) ) is moving in the \(x\) -direction. The light passes through three ideal polarizers whose transmission axes are (in order) at angles \(0.0^{\circ}, 45.0^{\circ},\) and \(30.0^{\circ}\) counterclockwise from the \(y\) -axis in the \(y z\) -plane. (a) What is the intensity and polarization of the light that is transmitted by the last polarizer? (b) If the polarizer in the middle is removed, what is the intensity and polarization of the light transmitted by the last polarizer?
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