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Ch. 2 Solutions

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Table of Contents

Basic

Medium

Basic

B-1. Numerical Calculation of the Speed of Light

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Problem: Calculate \(c\) using \(\mu_0 = 4\pi \times 10^{-7}\) T·m/A and \(\varepsilon_0 = 8.854 \times 10^{-12}\) F/m.

Solution:

\(\mu_0 \varepsilon_0 = (4\pi \times 10^{-7})(8.854 \times 10^{-12})\)

\(= 4 \times 3.1416 \times 8.854 \times 10^{-19}\)

\(= 12.566 \times 8.854 \times 10^{-19}\)

\(= 1.113 \times 10^{-17} \;\text{s}^2/\text{m}^2\)

\(c = \frac{1}{\sqrt{1.113 \times 10^{-17}}} = \frac{1}{1.055 \times 10^{-8.5}}\)

\(\boxed{c \approx 2.998 \times 10^8 \;\text{m/s}}\)

This agrees with the measured value of the speed of light, \(c = 2.998 \times 10^8\) m/s.

Key Point: \(\mu_0\) was determined independently from magnetic experiments, and \(\varepsilon_0\) from electrical experiments. Combining them yields the speed of light—this was the basis for Maxwell's conviction that "light is an electromagnetic wave."

B-2. From Potentials to Maxwell's Equations

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Problem: Show that \(\nabla \cdot \mathbf{B} = 0\) follows automatically from \(\mathbf{B} = \nabla \times \mathbf{A}\).

Solution:

From the vector identity (Appendix A.6):

\(\nabla \cdot (\nabla \times \mathbf{A}) = 0\)

This is an identity that holds for any vector field \(\mathbf{A}\).

Substituting \(\mathbf{B} = \nabla \times \mathbf{A}\):

\(\boxed{\nabla \cdot \mathbf{B} = \nabla \cdot (\nabla \times \mathbf{A}) = 0}\)

Key point: Maxwell's second equation \(\nabla \cdot \mathbf{B} = 0\) (no magnetic monopoles exist) is automatically satisfied simply by expressing the magnetic field in terms of the vector potential \(\mathbf{A}\). In other words, the potential formulation "builds in" part of Maxwell's equations. This is not merely a rewriting but reflects the structure of the theory.

Verification of the identity (optional): To verify in components, let \(\mathbf{A} = (A_x, A_y, A_z)\):

\(\nabla \times \mathbf{A} = \left(\frac{\partial A_z}{\partial y} - \frac{\partial A_y}{\partial z},\; \frac{\partial A_x}{\partial z} - \frac{\partial A_z}{\partial x},\; \frac{\partial A_y}{\partial x} - \frac{\partial A_x}{\partial y}\right)\)

Taking the divergence of this:

\(\nabla \cdot (\nabla \times \mathbf{A}) = \frac{\partial}{\partial x}\left(\frac{\partial A_z}{\partial y} - \frac{\partial A_y}{\partial z}\right) + \frac{\partial}{\partial y}\left(\frac{\partial A_x}{\partial z} - \frac{\partial A_z}{\partial x}\right) + \frac{\partial}{\partial z}\left(\frac{\partial A_y}{\partial x} - \frac{\partial A_x}{\partial y}\right)\)

Expanding yields 6 terms, which all cancel in pairs by the symmetry of mixed partial derivatives (\(\frac{\partial^2}{\partial x \partial y} = \frac{\partial^2}{\partial y \partial x}\)), giving zero.

Medium

M-1. Derivation of Electromagnetic Wave Speed

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Problem: From the 3rd and 4th Maxwell's equations (in vacuum), derive the wave equation for the electric field \(\mathbf{E}\) and show that the wave speed is \(c = 1/\sqrt{\mu_0 \varepsilon_0}\).

Solution:

Maxwell's equations in vacuum (\(\rho = 0\), \(\mathbf{j} = 0\)):

\(\nabla \times \mathbf{E} = -\frac{\partial \mathbf{B}}{\partial t} \qquad \text{(3rd equation)}\)

\(\nabla \times \mathbf{B} = \mu_0 \varepsilon_0 \frac{\partial \mathbf{E}}{\partial t} \qquad \text{(4th equation)}\)

Apply \(\nabla \times\) to both sides of the 3rd equation:

\(\nabla \times (\nabla \times \mathbf{E}) = -\frac{\partial}{\partial t}(\nabla \times \mathbf{B})\)

Using the vector identity (Appendix A.6) on the left-hand side:

\(\nabla \times (\nabla \times \mathbf{E}) = \nabla(\nabla \cdot \mathbf{E}) - \nabla^2 \mathbf{E}\)

In vacuum, \(\nabla \cdot \mathbf{E} = 0\) (Gauss's law with \(\rho = 0\)), so:

\(\nabla \times (\nabla \times \mathbf{E}) = -\nabla^2 \mathbf{E}\)

Substituting the 4th equation into the right-hand side:

\(-\frac{\partial}{\partial t}(\nabla \times \mathbf{B}) = -\mu_0 \varepsilon_0 \frac{\partial^2 \mathbf{E}}{\partial t^2}\)

Combining:

\(-\nabla^2 \mathbf{E} = -\mu_0 \varepsilon_0 \frac{\partial^2 \mathbf{E}}{\partial t^2}\)

\(\boxed{\nabla^2 \mathbf{E} = \mu_0 \varepsilon_0 \frac{\partial^2 \mathbf{E}}{\partial t^2}}\)

This has the form of the wave equation \(\nabla^2 \mathbf{E} = \frac{1}{c^2}\frac{\partial^2 \mathbf{E}}{\partial t^2}\). By comparison:

\(\frac{1}{c^2} = \mu_0 \varepsilon_0 \qquad \therefore \quad \boxed{c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}}}\)

Key point: Maxwell derived the speed of light solely from the experimental constants of electricity and magnetism, \(\mu_0\) and \(\varepsilon_0\). He was not studying light—the true nature of light was revealed as a byproduct of unifying electromagnetism.