Problem Set | Physics - Wyatt's Notes

Problem 1. Starting from Maxwell’s equations in differential form, derive the continuity Equation $\nabla \cdot \mathbf{J} + \partial\rho/\partial t = 0$ . Explain why this result Requires the displacement current term.

Solution

Take the divergence of the Ampere-Maxwell law:

$\nabla \cdot (\nabla \times \mathbf{B}) = \mu_0\nabla \cdot \mathbf{J} + \mu_0\varepsilon_0\frac{\partial}{\partial t}(\nabla \cdot \mathbf{E})$

Since $\nabla \cdot (\nabla \times \mathbf{B}) = 0$ and $\nabla \cdot \mathbf{E} = \rho/\varepsilon_0$ :

$0 = \mu_0\nabla \cdot \mathbf{J} + \mu_0\varepsilon_0\frac{\partial}{\partial t}\!\left(\frac{\rho}{\varepsilon_0}\right) = \mu_0\!\left(\nabla \cdot \mathbf{J} + \frac{\partial\rho}{\partial t}\right)$

$\nabla \cdot \mathbf{J} + \frac{\partial\rho}{\partial t} = 0$

Without the displacement current term, we would obtain $\nabla \cdot \mathbf{J} = 0$ Which Violates charge conservation whenever $\partial\rho/\partial t \neq 0$ (e.g., inside a Charging capacitor).

Cross-reference: Section 1.3, Section 4.5.

Problem 2. A point charge $q$ is placed at the centre of a dielectric sphere of radius $R$ And permittivity $\varepsilon$ . Find $\mathbf{D}$ , $\mathbf{E}$ And $\mathbf{P}$ everywhere. Determine the bound surface charge density.

Solution

By spherical symmetry, $\mathbf{D}$ is radial. Use Gauss’s law for $\mathbf{D}$ with a Spherical Gaussian surface of radius $r$ :

For $r \lt R$ : $D \cdot 4\pi r^2 = q \implies \mathbf{D} = \frac{q}{4\pi r^2}\,\hat{\mathbf{r}}$ .

$\mathbf{E}_{\mathrm{in} = \frac{\mathbf{D}}{\varepsilon} = \frac{q}{4\pi\varepsilon r^2}\,\hat{\mathbf{r}}}$ .

$\mathbf{P} = \varepsilon_0\chi_e\mathbf{E}_{\mathrm{in} = \left(\varepsilon - \varepsilon_0\right)\frac{q}{4\pi\varepsilon r^2}\,\hat{\mathbf{r}}}$ .

For $r \gt R$ : $D \cdot 4\pi r^2 = q \implies \mathbf{D} = \frac{q}{4\pi r^2}\,\hat{\mathbf{r}}$ .

$\mathbf{E}_{\mathrm{out} = \frac{\mathbf{D}}{\varepsilon_0} = \frac{q}{4\pi\varepsilon_0 r^2}\,\hat{\mathbf{r}}}$ .

$\mathbf{P}_{\mathrm{out} = \mathbf{0}}$ (vacuum).

Bound surface charge: $\sigma_b = \mathbf{P}\cdot\hat{\mathbf{r}}\big|_{r=R} = \left(\varepsilon - \varepsilon_0\right)\frac{q}{4\pi\varepsilon R^2}$ .

Bound volume charge: $\rho_b = -\nabla \cdot \mathbf{P} = -\frac{1}{r^2}\frac{\partial}{\partial r}(r^2 P_r) = 0$ for $r \lt R$ .

Cross-reference: Section 2.10.

Problem 3. An infinitely long cylindrical shell of radius $R$ carries a uniform surface Charge density $\sigma$ . Find the electric field everywhere.

Solution

By cylindrical symmetry, $\mathbf{E}$ is radial and depends only on $r$ . Use a Gaussian Cylinder of radius $r$ and length $L$ .

For $r \lt R$ : no charge enclosed, so $\mathbf{E} = \mathbf{0}$ .

For $r \gt R$ : the enclosed charge is $Q_{\mathrm{enc} = \sigma \cdot 2\pi R L}$ .

$E \cdot 2\pi r L = \frac{\sigma \cdot 2\pi R L}{\varepsilon_0}$

$\mathbf{E} = \frac{\sigma R}{\varepsilon_0 r}\,\hat{\mathbf{r}}$

At the surface ( $r = R^+$ ): $E = \sigma/\varepsilon_0$ Which is the discontinuity expected From the surface charge.

Cross-reference: Section 2.2, Section 2.6.

Problem 4. A conducting sphere of radius $a$ carries charge $Q$ and is surrounded by a Concentric conducting spherical shell of inner radius $b$ and outer radius $c$ carrying charge $-Q$ . Find $V(r)$ everywhere.

Solution

By spherical symmetry, $\mathbf{E}$ is radial. Use Gauss’s law with spherical Gaussian Surfaces.

$r \lt a$ : $\mathbf{E} = \mathbf{0}$ (conductor interior), so $V = V_a$ (constant).

$a \lt r \lt b$ : $E \cdot 4\pi r^2 = Q/\varepsilon_0 \implies \mathbf{E} = \frac{Q}{4\pi\varepsilon_0 r^2}\,\hat{\mathbf{r}}$ .

$V(r) = -\int_a^r E\,dr' + V_a = \frac{Q}{4\pi\varepsilon_0}\!\left(\frac{1}{r} - \frac{1}{a}\right) + V_a$

$b \lt r \lt c$ : $\mathbf{E} = \mathbf{0}$ (conductor), so $V = V_b$ (constant).

$V_b = \frac{Q}{4\pi\varepsilon_0}\!\left(\frac{1}{b} - \frac{1}{a}\right) + V_a$ .

$r \gt c$ : $E \cdot 4\pi r^2 = (Q - Q)/\varepsilon_0 = 0 \implies \mathbf{E} = \mathbf{0}$ So $V = 0$ (choosing $V(\infty) = 0$ ).

Since $V_c = 0$ and $V_c = V_b$ (same conductor), $V_b = 0$ :

$V_a = \frac{Q}{4\pi\varepsilon_0}\!\left(\frac{1}{a} - \frac{1}{b}\right)$

This is the capacitance of the spherical capacitor: $C = Q/V_a = 4\pi\varepsilon_0 ab/(b-a)$ .

Cross-reference: Section 2.3, Section 2.4.

Problem 5. The potential on the surface of a sphere of radius $R$ is $V(\theta) = V_0\cos\theta$ . Find the potential inside and outside the sphere.

Solution

Inside ( $r \lt R$ ), solve Laplace’s equation by separation of variables in spherical Coordinates. The general azimuthally symmetric solution is:

$V(r,\theta) = \sum_{l=0}^{\infty}\left(A_l r^l + \frac{B_l}{r^{l+1}}\right)P_l(\cos\theta)$

For $r \lt R$ : finiteness at $r = 0$ requires $B_l = 0$ .

$V_{\mathrm{in} = \sum_{l=0}^{\infty} A_l\,r^l\,P_l(\cos\theta)}$

Boundary condition at $r = R$ : $V_{\mathrm{in}(R,\theta) = V_0\cos\theta = V_0 P_1(\cos\theta)}$ .

By orthogonality of Legendre polynomials, only $l = 1$ contributes: $A_1 = V_0/R$ .

$V_{\mathrm{in} = \frac{V_0}{R}\,r\cos\theta = \frac{V_0}{R}\,z}$

For $r \gt R$ : $V \to 0$ as $r \to \infty$ requires $A_l = 0$ .

$V_{\mathrm{out} = \sum_{l=0}^{\infty}\frac{B_l}{r^{l+1}}P_l(\cos\theta)}$

Matching at $r = R$ : $B_1/R^2 = V_0 \implies B_1 = V_0 R^2$ .

$V_{\mathrm{out} = \frac{V_0 R^2}{r^2}\cos\theta}$

The interior field is uniform: $\mathbf{E}_{\mathrm{in} = -\nabla V_{\mathrm{in} = -(V_0/R)\,\hat{\mathbf{z}}}}$ .

Cross-reference: Section 2.4, Section 2.7.

Problem 6. Prove the uniqueness theorem for Neumann boundary conditions: the solution to $\nabla^2 V = -\rho/\varepsilon_0$ in a volume $\mathcal{V}$ is unique up to an additive Constant when $\partial V/\partial n$ is specified on $\mathcal{S}$ .

Solution

Suppose $V_1$ and $V_2$ both satisfy Poisson’s equation with the same Neumann boundary Condition $\partial V_1/\partial n = \partial V_2/\partial n$ on $\mathcal{S}$ . Define $U = V_1 - V_2$ . Then $\nabla^2 U = 0$ in $\mathcal{V}$ and $\partial U/\partial n = 0$ on $\mathcal{S}$ .

Apply Green’s first identity with $\phi = \psi = U$ :

$\int_{\mathcal{V}}\lvert\nabla U\rvert^2\,dV = \oint_{\mathcal{S}} U\,\frac{\partial U}{\partial n}\,dA = 0$

Since the integrand $\lvert\nabla U\rvert^2 \geq 0$ We conclude $\nabla U = \mathbf{0}$ In $\mathcal{V}$ So $U$ is constant throughout $\mathcal{V}$ .

$V_1 = V_2 + C$ for some constant $C$ . The solution is unique up to an additive constant. (The constant is physically irrelevant since only potential differences matter.) $\blacksquare$

Cross-reference: Section 2.7.

Problem 7. A point charge $q$ is placed at distance $a$ from the centre of a grounded Conducting sphere of radius $R$ ( $a \gt R$ ). Find the image charge location and magnitude, And determine the force on $q$ .

Solution

Place $q$ at distance $a$ along the $z$ -axis. The image charge $q'$ is at distance $b$ Along the $z$ -axis (inside the sphere).

For $V = 0$ on the sphere ( $r = R$ ), we need:

$\frac{q}{d_1} + \frac{q'}{d_2} = 0 \quad \mathrm{for\ all\ }\theta$

Where $d_1^2 = R^2 + a^2 - 2Ra\cos\theta$ and $d_2^2 = R^2 + b^2 - 2Rb\cos\theta$ .

The ratio $d_2/d_1$ must be constant. Setting $b = R^2/a$ :

$\frac{d_2^2}{d_1^2} = \frac{R^2 + R^4/a^2 - 2R^3\cos\theta/a}{R^2 + a^2 - 2Ra\cos\theta} = \frac{R^2}{a^2}$

This is constant (independent of $\theta$ ). With $q'/q = -R/a$ :

$q' = -\frac{qR}{a}, \quad b = \frac{R^2}{a}$

The force on $q$ is the Coulomb force due to $q'$ :

$\mathbf{F} = \frac{qq'}{4\pi\varepsilon_0(a-b)^2}\,\hat{\mathbf{z}} = \frac{q(-qR/a)}{4\pi\varepsilon_0(a-R^2/a)^2}\,\hat{\mathbf{z}} = -\frac{q^2R}{4\pi\varepsilon_0 a(a^2-R^2)}\,\hat{\mathbf{z}}$

The negative sign indicates attraction toward the sphere. $\blacksquare$

Cross-reference: Section 2.8.

Problem 8. A charge $+q$ is at $z = +d/2$ and $-q$ is at $z = -d/2$ . Compute the Electric dipole moment and find the potential to dipole order at a point in the $xy$ -plane At distance $r$ from the origin.

Solution

The dipole moment:

$\mathbf{p} = \sum_i q_i\mathbf{r}_i = q\!\left(\frac{d}{2}\right)\hat{\mathbf{z}} + (-q)\!\left(-\frac{d}{2}\right)\hat{\mathbf{z}} = qd\,\hat{\mathbf{z}}$

The dipole potential:

$V_1(\mathbf{r}) = \frac{1}{4\pi\varepsilon_0}\frac{\mathbf{p}\cdot\hat{\mathbf{r}}}{r^2}$

In the $xy$ -plane, $\hat{\mathbf{r}} = \cos\phi\,\hat{\mathbf{x}} + \sin\phi\,\hat{\mathbf{y}}$ So $\mathbf{p}\cdot\hat{\mathbf{r}} = qd\,\hat{\mathbf{z}}\cdot\hat{\mathbf{r}} = 0$ .

Therefore $V_1 = 0$ in the $xy$ -plane. The first non-zero contribution comes from the Quadrupole term ( $\sim 1/r^3$ ). $\blacksquare$

Cross-reference: Section 2.9.

Problem 9. A dielectric slab of permittivity $\varepsilon$ and thickness $d$ is inserted Between the plates of a parallel-plate capacitor with plate separation $D \gt d$ and plate Area $A$ Carrying free charge $\pm Q$ . Find the capacitance.

Solution

Let the plates be at $x = 0$ and $x = D$ With the slab occupying $0 \lt x \lt d$ . Since $Q$ is fixed, $D_n = \sigma_f = Q/A$ is the same in both regions.

In the dielectric ( $0 \lt x \lt d$ ): $E_1 = D/\varepsilon = Q/(\varepsilon A)$ .

In vacuum ( $d \lt x \lt D$ ): $E_2 = D/\varepsilon_0 = Q/(\varepsilon_0 A)$ .

The potential difference:

$V = E_1 d + E_2(D - d) = \frac{Q}{A}\!\left(\frac{d}{\varepsilon} + \frac{D-d}{\varepsilon_0}\right)$

The capacitance:

$C = \frac{Q}{V} = \frac{\varepsilon_0 A}{D - d + d/\varepsilon_r}$

Where $\varepsilon_r = \varepsilon/\varepsilon_0$ . For $d = D$ (fully filled): $C = \varepsilon_r\varepsilon_0 A/D$ Which is $\varepsilon_r$ times the vacuum capacitance.

Cross-reference: Section 2.10.

Problem 10. Find the magnetic field at the centre of a square loop of side $a$ carrying Current $I$ using the Biot-Savart law.

Solution

By symmetry, each side contributes equally. Consider one side from $(a/2, -a/2, 0)$ to $(a/2, a/2, 0)$ . For this side, $d\mathbf{l} = dy\,\hat{\mathbf{y}}$ and $\mathbf{r} = (a/2)\hat{\mathbf{x}} - y\hat{\mathbf{y}}$ So $r = \sqrt{(a/2)^2 + y^2}$ .

$d\mathbf{B} = \frac{\mu_0 I}{4\pi}\frac{d\mathbf{l} \times \mathbf{r}}{r^3} = \frac{\mu_0 I}{4\pi}\frac{dy\,\hat{\mathbf{y}} \times [(a/2)\hat{\mathbf{x}} - y\hat{\mathbf{y}}]}{r^3}$

$= \frac{\mu_0 I}{4\pi}\frac{(-a/2)\,dy}{[(a/2)^2+y^2]^{3/2}}\,\hat{\mathbf{z}}$

$B_{\mathrm{one\ side} = \frac{\mu_0 I}{4\pi}\!\left(-\frac{a}{2}\right)\!\int_{-a/2}^{a/2}\frac{dy}{[(a/2)^2+y^2]^{3/2}}}$

Using $\int dy/(s^2+y^2)^{3/2} = y/[s^2\sqrt{s^2+y^2}]$ with $s = a/2$ :

$B_{\mathrm{one\ side} = \frac{\mu_0 I}{4\pi}\!\left(-\frac{a}{2}\right)\frac{2}{(a/2)\sqrt{(a/2)^2+(a/2)^2}} \cdot 2 = -\frac{\mu_0 I}{\pi a}\cdot\frac{1}{\sqrt{2}} \cdot 2 = -\frac{\mu_0 I}{\pi a}\sqrt{2}}$

The magnitude from all four sides: $B = 4 \times \frac{\sqrt{2}\,\mu_0 I}{\pi a} = \frac{2\sqrt{2}\,\mu_0 I}{\pi a}$ .

$\mathbf{B} = -\frac{2\sqrt{2}\,\mu_0 I}{\pi a}\,\hat{\mathbf{z}}$

(by the right-hand rule, into the page for counterclockwise current). $\blacksquare$

Cross-reference: Section 3.1.

Problem 11. A toroid with $N$ turns, inner radius $a$ And outer radius $b$ carries current $I$ . Find the magnetic field everywhere.

Solution

By symmetry, $\mathbf{B}$ is tangential and depends only on $r$ (distance from the axis of Symmetry). Apply Ampere’s law to a circular loop of radius $r$ .

For $r \lt a$ : no current is enclosed, so $\mathbf{B} = \mathbf{0}$ .

For $a \lt r \lt b$ : the Amperian loop encloses all $N$ turns.

$B \cdot 2\pi r = \mu_0 N I \implies \mathbf{B} = \frac{\mu_0 N I}{2\pi r}\,\hat{\boldsymbol{\phi}}$

For $r \gt b$ : the net enclosed current is $NI - NI = 0$ So $\mathbf{B} = \mathbf{0}$ .

The field is confined entirely within the toroid, unlike a solenoid where the field extends Beyond the ends. $\blacksquare$

Cross-reference: Section 3.2, Section 3.4.

Problem 12. A circular loop of radius $R$ carries current $I$ . Find the magnetic dipole Moment and the field on the axis at distance $z$ from the centre. Show that the result Reduces to the dipole field for $z \gg R$ .

Solution

The magnetic dipole moment: $\mathbf{m} = I\pi R^2\,\hat{\mathbf{z}}$ .

From the Biot-Savart law, every element $d\mathbf{l}$ is perpendicular to $\hat{\mathbf{r}}$ So $d\mathbf{B} = \frac{\mu_0 I}{4\pi}\frac{dl}{R^2+z^2}$ . By symmetry, only the axial component Survives:

$B_z = \frac{\mu_0 I}{4\pi(R^2+z^2)}\frac{R}{\sqrt{R^2+z^2}} \cdot 2\pi R = \frac{\mu_0 I R^2}{2(R^2+z^2)^{3/2}}$

For $z \gg R$ : $(R^2+z^2)^{3/2} \approx z^3(1 + 3R^2/2z^2) \approx z^3$ .

$B_z \approx \frac{\mu_0 I R^2}{2z^3} = \frac{\mu_0}{4\pi}\frac{2m}{z^3}$

The dipole field formula gives, on the axis ( $\theta = 0$ ):

$\mathbf{B}_{\mathrm{dip} = \frac{\mu_0}{4\pi}\frac{2\mathbf{m}}{z^3}}$

This matches. $\blacksquare$

Cross-reference: Section 3.5.

Problem 13. A long straight wire along the $z$ -axis carries current $I$ . Find the vector Potential $\mathbf{A}$ and verify that $\nabla \times \mathbf{A}$ gives the correct $\mathbf{B}$ .

Solution

By cylindrical symmetry, $\mathbf{A}$ can only depend on $s$ (the radial distance) and must Point along $\hat{\mathbf{z}}$ (parallel to the current).

$\mathbf{A}(s) = -\frac{\mu_0 I}{2\pi}\ln\!\left(\frac{s}{s_0}\right)\hat{\mathbf{z}}$

Where $s_0$ is an arbitrary reference distance (gauge-dependent).

Verify: $\mathbf{B} = \nabla \times \mathbf{A}$ .

In cylindrical coordinates, $\nabla \times (A_z\,\hat{\mathbf{z}}) = -\frac{\partial A_z}{\partial s}\,\hat{\boldsymbol{\phi}}$ .

$B_\phi = -\frac{\partial}{\partial s}\!\left(-\frac{\mu_0 I}{2\pi}\ln\frac{s}{s_0}\right) = \frac{\mu_0 I}{2\pi s}$

$\mathbf{B} = \frac{\mu_0 I}{2\pi s}\,\hat{\boldsymbol{\phi}}$

This matches the Ampere’s law result. $\blacksquare$

Cross-reference: Section 3.3, Section 3.6.

Problem 14. An iron ring of mean radius $R = 10$ cm, cross-sectional area $A = 4\ \mathrm{cm}^2$ And relative permeability $\mu_r = 500$ has $N = 200$ turns carrying current $I = 2$ A. Find $B$ , $H$ , $M$ And the total flux through the ring.

Solution

Apply Ampere’s law for $\mathbf{H}$ around the ring:

$\oint \mathbf{H} \cdot d\mathbf{l} = NI \implies H \cdot 2\pi R = NI$

$H = \frac{NI}{2\pi R} = \frac{200 \times 2}{2\pi \times 0.10} = \frac{400}{0.628} \approx 637\ \mathrm{A}/m$

$B = \mu_0 \mu_r H = 4\pi \times 10^{-7} \times 500 \times 637 \approx 0.40\ \mathrm{T}$

$M = \chi_m H = (\mu_r - 1)H = 499 \times 637 \approx 3.18 \times 10^5\ \mathrm{A}/m$

Total flux: $\Phi = BA = 0.40 \times 4 \times 10^{-4} = 1.6 \times 10^{-4}\ \mathrm{Wb}$ .

Cross-reference: Section 3.7, Section 3.8.

Problem 15. A rectangular conducting loop of width $w = 0.1$ m and length $\ell = 0.2$ m Has resistance $R = 5\ \Omega$ . One end enters a region of uniform magnetic field $B = 0.5$ T (perpendicular to the loop) at velocity $v = 2$ m/s. Find the induced EMF and Current.

Solution

As the loop enters the field with its leading edge at position $x$ inside the field:

$\Phi_B = B \cdot w \cdot x$

$\mathcal{E} = -\frac{d\Phi_B}{dt} = -Bw\frac{dx}{dt} = -Bwv = -0.5 \times 0.1 \times 2 = -0.1\ \mathrm{V}$

The magnitude is $0.1$ V. The current is:

$I = \frac{\lvert\mathcal{E}\rvert}{R} = \frac{0.1}{5} = 0.02\ \mathrm{A}$

By Lenz’s law, the current flows to oppose the increasing flux (counterclockwise when viewed From the direction of $\mathbf{B}$ ).

The magnetic braking force on the leading edge: $F = BIw = 0.5 \times 0.02 \times 0.1 = 0.001$ N (opposing the motion). $\blacksquare$

Cross-reference: Section 4.1, Section 4.4.

Problem 16. A plane electromagnetic wave in vacuum has $\mathbf{E} = 100\cos(kz - \omega t)\,\hat{\mathbf{x}}$ V/m. Find $B_0$ The time-averaged intensity, and the radiation pressure on a perfectly absorbing Surface.

Solution

$B_0 = E_0/c = 100/(3 \times 10^8) = 3.33 \times 10^{-7}$ T.

$\mathbf{B} = 3.33 \times 10^{-7}\cos(kz - \omega t)\,\hat{\mathbf{y}}$ T.

Time-averaged Poynting vector magnitude:

$\langle S \rangle = \frac{E_0^2}{2\mu_0 c} = \frac{100^2}{2 \times 4\pi \times 10^{-7} \times 3 \times 10^8} = \frac{10^4}{754} \approx 13.3\ \mathrm{W}/m^2$

Radiation pressure on a perfect absorber:

$P_{\mathrm{rad} = \frac{\langle S \rangle}{c} = \frac{13.3}{3 \times 10^8} \approx 4.4 \times 10^{-8}\ \mathrm{Pa}}$

Cross-reference: Section 5.2, Section 5.4, Section 5.5.

Problem 17. Find the skin depth for a 1 MHz electromagnetic wave in copper ( $\sigma = 5.96 \times 10^7$ S/m, $\mu_r \approx 1$ ). At what frequency does the skin depth Equal 1 $\mu$ M?

Solution

At $f = 1$ MHz:

$\delta = \sqrt{\frac{2}{\mu_0\sigma\omega}} = \sqrt{\frac{2}{4\pi \times 10^{-7} \times 5.96 \times 10^7 \times 2\pi \times 10^6}}$

$= \sqrt{\frac{2}{4\pi \times 5.96 \times 2\pi^2 \times 10^6}} = \sqrt{\frac{2}{4.70 \times 10^8}} \approx 65.2\ \mu\mathrm{m}$

For $\delta = 1\ \mu$ M:

$1 \times 10^{-6} = \sqrt{\frac{2}{4\pi \times 10^{-7} \times 5.96 \times 10^7 \times 2\pi f}}$

$10^{-12} = \frac{2}{4\pi \times 5.96 \times 2\pi \times f} = \frac{2}{470.4\,f}$

$f = \frac{2}{470.4 \times 10^{-12}} \approx 4.25 \times 10^9\ \mathrm{Hz} = 4.25\ \mathrm{GHz}$

Cross-reference: Section 5.6.

Problem 18. Verify that Maxwell’s equations in covariant form $\partial_\mu F^{\mu\nu} = \mu_0 J^\nu$ reproduce the Ampere-Maxwell law for $\nu = 1$ .

Solution

For $\nu = 1$ :

$\partial_\mu F^{\mu 1} = \mu_0 J^1 = \mu_0 J_x$

From the field tensor:

$F^{\mu 1} = (E_x/c,\ 0,\ -B_z,\ B_y) \quad \mathrm{for\ }\mu = 0, 1, 2, 3$

So:

$\partial_0 F^{01} + \partial_2 F^{21} + \partial_3 F^{31} = \mu_0 J_x$

$\frac{1}{c}\frac{\partial}{\partial t}\!\left(\frac{E_x}{c}\right) + \frac{\partial}{\partial y}(-B_z) + \frac{\partial}{\partial z}(B_y) = \mu_0 J_x$

$\frac{1}{c^2}\frac{\partial E_x}{\partial t} - \frac{\partial B_z}{\partial y} + \frac{\partial B_y}{\partial z} = \mu_0 J_x$

Using $c^2 = 1/(\mu_0\varepsilon_0)$ and noting that $-(\partial B_z/\partial y) + (\partial B_y/\partial z) = -(\nabla \times \mathbf{B})_x$ :

$\mu_0\varepsilon_0\frac{\partial E_x}{\partial t} - (\nabla \times \mathbf{B})_x = \mu_0 J_x$

$(\nabla \times \mathbf{B})_x = \mu_0\varepsilon_0\frac{\partial E_x}{\partial t} - \mu_0 J_x$

$\nabla \times \mathbf{B} = \mu_0\mathbf{J} + \mu_0\varepsilon_0\frac{\partial\mathbf{E}}{\partial t}$

This is the Ampere-Maxwell law. $\blacksquare$

Cross-reference: Section 7.4, Section 1.1.

flowchart TD
    A[8_Problem Set] --> B[Key Concepts]
    A --> C[Core Principles]
    A --> D[Practical Applications]
    B --> E[Fundamental definitions]
    C --> F[Design patterns]
    D --> G[Real-world usage]

Intuition

Maxwell’s equations are the four laws that govern all classical electromagnetic phenomena. Gauss’s law says electric charges create field lines that spread outward; the magnetostatic version says magnetic field lines always close on themselves. Faraday’s law and the Ampere-Maxwell law describe how changing fields create each other: a changing magnetic field produces an electric field and vice versa. This mutual regeneration is what allows electromagnetic waves to propagate through empty space, carrying energy and momentum without any medium.

Cross-References

Maxwell’s Equations: Every problem in this set originates from or feeds back into the four Maxwell equations in differential form.
Electrodynamics: The continuity equation and displacement current derivations require the time-dependent field equations from electrodynamics.
Electromagnetic Waves: Wave propagation problems in this set build directly on the wave equation derived from Maxwell’s equations.
Calculus
Linear Algebra
Vector Calculus

Common Mistakes

Confusing divergence and curl: Div E = ρ/ε₀ (charges create fields); curl E = -∂B/∂t (changing B creates E). Using the wrong one gives incorrect field solutions.

Forgetting the displacement current term: Maxwell added ε₀∂E/∂t to Ampère’s law. Without it, the equation is inconsistent with charge conservation and cannot predict electromagnetic waves.

Sign errors in Faraday’s law: The minus sign in curl E = -∂B/∂t is Lenz’s law — the induced EMF opposes the change in flux. Dropping the minus sign gives the wrong direction of induced current.

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