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Chapter 27: Problem 69

An electric field and a magnetic field have the same energy density. Find an expression for the ratio \(E / B\) and evaluate this ratio numerically. What are its units? Is your answer close to any of the fundamental constants listed inside the front cover?

Short Answer

Expert verified
The ratio of the electric field to the magnetic field, \(E / B\), is \(2.998 x 10^8 m/s\), with units of speed. This number is indeed quite close to the value of the speed of light in vacuum, which is a fundamental constant.

Step by step solution

01

Write Down the Energy Density Formulas for Electric and Magnetic Fields

The energy density (u) of an electric field (E) is given by \(u_E = \frac{1}{2} \epsilon_0 E^2\) and the energy density of a magnetic field (B) is given by \(u_B = \frac{1}{2}\frac{B^2}{\mu_0}\). Here, \(\epsilon_0\) is the permittivity of free space and \(\mu_0\) is the permeability of free space.
02

Equate the Energy Densities and Solve for the Ratio \(E / B\)

Since the energy densities are equal, we can set the two equations equal to each other and solve for the ratio \(E / B\): \(\frac{1}{2} \epsilon_0 E^2 = \frac{1}{2}\frac{B^2}{\mu_0}\)Solving for \(E / B\) gives \(E / B = \sqrt{\frac{\mu_0}{\epsilon_0}}\).
03

Evaluate the Ratio Numerically

The values for \(\mu_0\) and \(\epsilon_0\) are known, \(\mu_0 = 4\pi x 10^{-7} T m/A\) and \(\epsilon_0 = 8.85 x 10^{-12} C^2/N m^2\). Plugging these into the equation for \(E / B\) gives \(E / B = \sqrt{\frac{4\pi x 10^{-7} T m/A}{8.85 x 10^{-12} C^2/N m^2}} = 2.998 x 10^8 m/s\).
04

Identify the Units

The units of the ratio \(E / B\) are the units of E divided by the units of B. In the SI system, the unit of E is Newton per Coulomb (N/C), and the unit of B is Tesla (T). Therefore, the units of speed, meters per second (m/s), are obtained as a result.

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