Complete Electromagnetism

Vector Analysis

Core ideas

Electromagnetism is written in the language of fields. Gradients describe how scalar potentials change, divergence measures sources and sinks, curl measures circulation, and integral theorems connect local differential laws to global flux and circulation laws. Delta functions represent point charges and currents cleanly.

For review, be able to move between differential and integral forms, use divergence and Stokes’ theorems, interpret boundary normals, and work in Cartesian, cylindrical, and spherical symmetry. Keep the physical question visible: identify the degrees of freedom, the conserved quantities, the approximation being made, and the observable that would be measured.

Mathematical spine

$$\int_V\nabla\cdot\bm A\,d^3x=\oint_{\partial V}\bm A\cdot d\bm a,\qquad \int_S(\nabla\times\bm A)\cdot d\bm a=\oint_{\partial S}\bm A\cdot d\bm \ell$$

Worked example

Verify the divergence theorem for $\bm A = x\,\hat{\bm x} + y\,\hat{\bm y} + z\,\hat{\bm z}$ over a sphere of radius $R = 2\,\text{m}$.

First, $\nabla\cdot\bm A = \partial_x x + \partial_y y + \partial_z z = 3$. The volume integral is

$$\int_V (\nabla\cdot\bm A)\,d^3x = 3\cdot\frac{4}{3}\pi R^3 = 4\pi(2)^3 = 32\pi\;\text{m}^3 .$$

On the surface, $d\bm a = \hat{\bm r}\,R^2\sin\theta\,d\theta\,d\phi$ and $\bm A\cdot d\bm a = R\,\hat{\bm r}\cdot\hat{\bm r}\,R^2\,d\Omega = R^3\,d\Omega = 8\,d\Omega$. Integrating over the full solid angle $4\pi$ gives $32\pi\;\text{m}^3$, confirming the theorem.

Problems with Solutions

Problem 1. Compute the divergence of $\bm A = x^2y\,\hat{\bm x} + yz\,\hat{\bm y} + z^3\,\hat{\bm z}$ at the point $(1,2,3)$.

Solution.

$$\nabla\cdot\bm A = \partial_x(x^2y) + \partial_y(yz) + \partial_z(z^3) = 2xy + z + 3z^2 .$$

At $(1,2,3)$: $2(1)(2) + 3 + 3(9) = 4 + 3 + 27 = 34$.

Problem 2. Use Stokes’ theorem to evaluate $\oint_C \bm A\cdot d\bm\ell$ for $\bm A = -y\,\hat{\bm x} + x\,\hat{\bm y}$ around a circle of radius $R = 3\,\text{m}$ in the $xy$-plane.

Solution. $(\nabla\times\bm A)_z = \partial_x x - \partial_y(-y) = 2$. The area of the disk is $\pi R^2 = 9\pi\;\text{m}^2$. By Stokes’ theorem,

$$\oint_C \bm A\cdot d\bm\ell = \int_S (\nabla\times\bm A)\cdot d\bm a = 2\cdot 9\pi = 18\pi\;\text{m}^2\!\!\cdot\!\text{m}^{-1} \approx 56.5\;\text{m} .$$

Section summary. Vector analysis is the grammar of field laws.

Electrostatics

Core ideas

Electrostatics studies fields produced by stationary charge. Coulomb’s law gives the field of point charges, superposition builds general charge distributions, and Gauss’s law solves high-symmetry problems. Electrostatic fields are conservative, so work is path independent and energy can be stored in the field.

For review, be able to compute fields from charge distributions, choose Gaussian surfaces, relate force, field, and potential energy, and use boundary conditions at charged surfaces. Keep the physical question visible: identify the degrees of freedom, the conserved quantities, the approximation being made, and the observable that would be measured.

Mathematical spine

$$\bm E(\bm r)=\frac{1}{4\pi\epsilon_0}\int \rho(\bm r')\frac{\bm r-\bm r'}{|\bm r-\bm r'|^3}\,d^3r',\qquad \nabla\cdot\bm E=\rho/\epsilon_0$$

Worked example

A solid insulating sphere of radius $R = 5.0\,\text{cm}$ carries a uniformly distributed total charge $Q = 10\,\text{nC}$. Find the electric field at $r = 10\,\text{cm}$ from the center.

By Gauss’s law, outside a spherically symmetric charge distribution the field is that of a point charge:

$$E = \frac{1}{4\pi\epsilon_0}\frac{Q}{r^2} = \frac{(8.99\times 10^9\,\text{N}\cdot\text{m}^2/\text{C}^2)(10\times 10^{-9}\,\text{C})}{(0.10\,\text{m})^2} \approx 8.99\times 10^3\;\text{N/C} .$$

The field points radially outward. Inside the sphere ($r<R$) the enclosed charge grows as $r^3$ and $E\propto r$.

Problems with Solutions

Problem 1. Two point charges $q_1 = +3.0\,\mu\text{C}$ and $q_2 = -5.0\,\mu\text{C}$ are placed $20\,\text{cm}$ apart in vacuum. What is the magnitude of the electrostatic force on $q_1$?

Solution. Using Coulomb’s law,

$$F = \frac{1}{4\pi\epsilon_0}\frac{|q_1q_2|}{r^2} = (8.99\times 10^9)\frac{(3.0\times 10^{-6})(5.0\times 10^{-6})}{(0.20)^2} \approx 3.37\;\text{N} .$$

Problem 2. An infinite plane sheet has uniform surface charge density $\sigma = -2.0\,\mu\text{C/m}^2$. Find the electric field magnitude and direction.

Solution. By symmetry and Gauss’s law with a pillbox,

$$E = \frac{|\sigma|}{2\epsilon_0} = \frac{2.0\times 10^{-6}}{2(8.85\times 10^{-12})} \approx 1.13\times 10^5\;\text{N/C} .$$

The field points toward the sheet because the charge is negative.

Section summary. Static charge creates conservative electric fields.

Potentials

Core ideas

Because electrostatic fields have zero curl, they can be written as the gradient of a scalar potential. Poisson and Laplace equations turn field problems into boundary value problems. Green functions, image charges, and separation of variables are the main tools for conductors and prescribed boundaries.

For review, be able to solve simple Poisson/Laplace problems, use uniqueness, apply image charges, and interpret equipotentials and boundary conditions. Keep the physical question visible: identify the degrees of freedom, the conserved quantities, the approximation being made, and the observable that would be measured.

Mathematical spine

$$\bm E=-\nabla V,\qquad \nabla^2V=-\rho/\epsilon_0,\qquad V(\bm r)=\frac{1}{4\pi\epsilon_0}\int\frac{\rho(\bm r')}{|\bm r-\bm r'|}\,d^3r'$$

Worked example

A point charge $q = 5.0\,\text{nC}$ is placed a distance $d = 4.0\,\text{cm}$ above an infinite grounded conducting plane ($V=0$). Find the potential at a point on the plane directly below the charge.

By the method of images, replace the plane with an image charge $-q$ at $z=-d$. The potential on the plane ($z=0$) from the real and image charges is

$$V = \frac{1}{4\pi\epsilon_0}\left(\frac{q}{d} + \frac{-q}{d}\right) = 0 ,$$

which satisfies the boundary condition. At any point $(x,y,0)$ the cancellation is exact because the distances to $+q$ and $-q$ are equal.

Problems with Solutions

Problem 1. A parallel-plate capacitor has plate separation $d = 2.0\,\text{mm}$ and voltage $V_0 = 100\,\text{V}$. Find the potential $V(z)$ between the plates (taking $V=0$ at the grounded plate) and the electric field.

Solution. With no free charge between the plates, $\nabla^2V=0$ gives $V(z) = V_0 z/d$. Then

$$\bm E = -\nabla V = -\frac{V_0}{d}\,\hat{\bm z} = -\frac{100}{0.002}\,\hat{\bm z} = -5.0\times 10^4\;\text{V/m}\,\hat{\bm z} .$$

Problem 2. A spherical shell of radius $R = 10\,\text{cm}$ is held at potential $V_0 = 200\,\text{V}$. Find the potential outside the shell.

Solution. Laplace’s equation with spherical symmetry gives $V(r) = A/r + B$. As $r\to\infty$, $V\to 0$, so $B=0$. At $r=R$, $V=V_0$, giving $A=V_0R$. Hence

$$V(r) = \frac{V_0R}{r} = \frac{20.0\,\text{V}\cdot\text{m}}{r} .$$

Section summary. Potentials make electrostatics a boundary value problem.

Electric Fields in Matter

Core ideas

Matter polarizes in electric fields. Bound charge is described by polarization $\bm P$, while the displacement field $\bm D$ separates free charge from bound response. Linear dielectrics are summarized by permittivity, but the physical picture is microscopic dipoles plus boundary conditions.

For review, be able to compute bound charge, use $\bm D$ for free charge, apply dielectric boundary conditions, and compare vacuum and material energy densities. Keep the physical question visible: identify the degrees of freedom, the conserved quantities, the approximation being made, and the observable that would be measured.

Mathematical spine

$$\rho_b=-\nabla\cdot\bm P,\qquad \bm D=\epsilon_0\bm E+\bm P,\qquad \nabla\cdot\bm D=\rho_{\rm free}$$

Worked example

A parallel-plate capacitor has plate area $A = 100\,\text{cm}^2$, separation $d = 2.0\,\text{mm}$, and is connected to a $V = 100\,\text{V}$ battery. The gap is filled with a linear dielectric of susceptibility $\chi_e = 3$ ($\kappa = 4$). Find $\bm E$, $\bm D$, and the bound surface charge density $\sigma_b$.

The electric field is fixed by the battery:

$$E = \frac{V}{d} = \frac{100}{0.002} = 5.0\times 10^4\;\text{V/m} .$$

The polarization is $P = \epsilon_0\chi_e E = (8.85\times 10^{-12})(3)(5.0\times 10^4) = 1.33\times 10^{-6}\;\text{C/m}^2$. The displacement field is

$$D = \epsilon_0 E + P = \epsilon_0\kappa E = (8.85\times 10^{-12})(4)(5.0\times 10^4) \approx 1.77\times 10^{-6}\;\text{C/m}^2 .$$

The bound surface charge on the dielectric facing the positive plate is $\sigma_b = +P = 1.33\,\mu\text{C/m}^2$; on the opposite face it is $-1.33\,\mu\text{C/m}^2$.

Problems with Solutions

Problem 1. A coaxial cable has inner radius $a = 1.0\,\text{mm}$ and outer radius $b = 3.0\,\text{mm}$. The space between conductors is filled with a dielectric of $\kappa = 2.5$. If the inner conductor carries free linear charge density $\lambda = 4.0\,\text{nC/m}$, find $D$ and $E$ at $r = 2.0\,\text{mm}$.

Solution. Cylindrical symmetry and Gauss’s law for $\bm D$ give $D(2\pi r L) = \lambda L$, so

$$D = \frac{\lambda}{2\pi r} = \frac{4.0\times 10^{-9}}{2\pi(2.0\times 10^{-3})} \approx 3.18\times 10^{-7}\;\text{C/m}^2 .$$

Then $E = D/(\kappa\epsilon_0) = 3.18\times 10^{-7}/(2.5\cdot 8.85\times 10^{-12}) \approx 1.44\times 10^4\;\text{V/m}$.

Problem 2. At a flat boundary between two dielectrics ($\kappa_1=2$, $\kappa_2=5$) the electric field in region 1 makes an angle $\theta_1 = 30^\circ$ with the normal. Find the angle $\theta_2$ in region 2.

Solution. Boundary conditions give $E_{1\parallel}=E_{2\parallel}$ and $D_{1\perp}=D_{2\perp}$, so $\kappa_1 E_{1\perp} = \kappa_2 E_{2\perp}$. Using $\tan\theta = E_\parallel/E_\perp$,

$$\frac{\tan\theta_2}{\tan\theta_1} = \frac{E_{2\perp}}{E_{1\perp}} = \frac{\kappa_1}{\kappa_2} = \frac{2}{5} .$$

Thus $\tan\theta_2 = (2/5)\tan 30^\circ = 0.231$, giving $\theta_2 \approx 13.0^\circ$.

Section summary. Dielectrics modify fields through polarization.

Magnetostatics

Core ideas

Magnetostatics describes steady currents and time-independent magnetic fields. The Biot—Savart law gives fields from currents, Ampere’s law solves symmetric systems, and $\nabla\cdot\bm B=0$ states that magnetic monopoles are absent in classical electromagnetism.

For review, be able to compute fields of wires, loops, and solenoids, choose Amperian loops, use vector potential, and understand magnetic force on currents. Keep the physical question visible: identify the degrees of freedom, the conserved quantities, the approximation being made, and the observable that would be measured.

Mathematical spine

$$\bm B(\bm r)=\frac{\mu_0}{4\pi}\int\frac{\bm J(\bm r')\times(\bm r-\bm r')}{|\bm r-\bm r'|^3}\,d^3r',\qquad \nabla\times\bm B=\mu_0\bm J$$

Worked example

A long straight wire carries a steady current $I = 5.0\,\text{A}$. Find the magnetic field at a perpendicular distance $r = 10\,\text{cm}$ from the wire.

Choose a circular Amperian loop of radius $r$ centered on the wire. By symmetry $\bm B$ is tangential and constant in magnitude on the loop. Ampère’s law gives

$$\oint \bm B\cdot d\bm\ell = B(2\pi r) = \mu_0 I_{\rm enc} = \mu_0 I .$$

Solving for $B$,

$$B = \frac{\mu_0 I}{2\pi r} = \frac{(4\pi\times 10^{-7}\,\text{T}\cdot\text{m/A})(5.0\,\text{A})}{2\pi(0.10\,\text{m})} = 1.0\times 10^{-5}\;\text{T} = 10\;\mu\text{T} .$$

Problems with Solutions

Problem 1. A circular loop of radius $R = 5.0\,\text{cm}$ carries current $I = 2.0\,\text{A}$. What is the magnetic field magnitude at the center of the loop?

Solution. Using the Biot—Savart law, every element contributes parallel to the axis at the center. Integrating gives

$$B = \frac{\mu_0 I}{2R} = \frac{(4\pi\times 10^{-7})(2.0)}{2(0.050)} = 2.51\times 10^{-5}\;\text{T} \approx 25\;\mu\text{T} .$$

Problem 2. Two long parallel wires each carry $I = 3.0\,\text{A}$ in the same direction and are separated by $d = 20\,\text{cm}$. Find the force per unit length on one wire.

Solution. The field at one wire due to the other is $B = \mu_0 I/(2\pi d)$. The force per unit length is

$$\frac{F}{L} = \frac{\mu_0 I^2}{2\pi d} = \frac{(4\pi\times 10^{-7})(3.0)^2}{2\pi(0.20)} = 9.0\times 10^{-6}\;\text{N/m} .$$

The force is attractive because the currents are parallel.

Section summary. Steady currents produce divergence-free magnetic fields.

Magnetic Fields in Matter

Core ideas

Magnetized matter contains microscopic current loops. Magnetization $\bm M$ produces bound volume and surface currents, while $\bm H$ separates free current from material response. Paramagnets, diamagnets, and ferromagnets differ in how $\bm M$ responds to applied field.

For review, be able to compute bound currents, use $\bm H$ boundary conditions, distinguish susceptibility regimes, and recognize hysteresis as nonlinear material response. Keep the physical question visible: identify the degrees of freedom, the conserved quantities, the approximation being made, and the observable that would be measured.

Mathematical spine

$$\bm J_b=\nabla\times\bm M,\qquad \bm K_b=\bm M\times\hat{\bm n},\qquad \bm H=\bm B/\mu_0-\bm M$$

Worked example

A toroidal solenoid with $N = 500$ turns carries current $I = 2.0\,\text{A}$. The mean radius is $r = 10\,\text{cm}$ and the core is a linear magnetic material of susceptibility $\chi_m = 500$. Find $\bm H$, $\bm M$, and $\bm B$ inside the core.

By Ampère’s law for $\bm H$, $\oint\bm H\cdot d\bm\ell = N I$, giving

$$H = \frac{NI}{2\pi r} = \frac{500(2.0)}{2\pi(0.10)} \approx 1.59\times 10^3\;\text{A/m} .$$

The magnetization is $M = \chi_m H = 500(1.59\times 10^3) = 7.96\times 10^5\;\text{A/m}$. The magnetic field is

$$B = \mu_0(H+M) = \mu_0(1+\chi_m)H = (4\pi\times 10^{-7})(501)(1.59\times 10^3) \approx 1.00\;\text{T} .$$

Problems with Solutions

Problem 1. An infinitely long cylinder of radius $R = 2.0\,\text{cm}$ has uniform magnetization $\bm M = M_0\,\hat{\bm z}$ with $M_0 = 8.0\times 10^5\,\text{A/m}$. Find the bound surface current density $\bm K_b$.

Solution. The bound surface current is $\bm K_b = \bm M\times\hat{\bm n}$. At the curved surface $\hat{\bm n}=\hat{\bm s}$, so

$$\bm K_b = M_0\,\hat{\bm z}\times\hat{\bm s} = -M_0\,\hat{\bm\phi} = -8.0\times 10^5\;\text{A/m}\,\hat{\bm\phi} .$$

There is no bound volume current because $\bm M$ is uniform ($\nabla\times\bm M=0$). The cylinder acts like a solenoid with $K_b$.

Problem 2. A long solenoid with $n = 500\,\text{turns/m}$ carries $I = 1.5\,\text{A}$ and is filled with a material of relative permeability $\mu_r = 800$. Find $\bm B$ inside.

Solution. For a long solenoid, $H = nI = 500(1.5) = 750\;\text{A/m}$. Then

$$B = \mu_0\mu_r H = (4\pi\times 10^{-7})(800)(750) \approx 0.754\;\text{T} .$$

Section summary. Magnetic media are handled by separating free and bound currents.

Electrodynamics

Core ideas

Time-dependent electric and magnetic fields generate one another. Faraday induction and Maxwell’s displacement current complete the static laws and enforce charge conservation. Maxwell’s equations are the compact local statement of classical electromagnetism.

For review, be able to write all four Maxwell equations, convert between integral and differential forms, derive continuity, and identify quasistatic limits. Keep the physical question visible: identify the degrees of freedom, the conserved quantities, the approximation being made, and the observable that would be measured.

Mathematical spine

$$\nabla\cdot\bm E=\rho/\epsilon_0,\quad \nabla\cdot\bm B=0,\quad \nabla\times\bm E=-\partial_t\bm B,\quad \nabla\times\bm B=\mu_0\bm J+\mu_0\epsilon_0\partial_t\bm E$$

Worked example

A parallel-plate circular capacitor of radius $R = 5.0\,\text{cm}$ and plate separation $d = 2.0\,\text{mm}$ is being charged by a current $I = 2.0\,\text{A}$. Find the magnetic field at a point $r = 3.0\,\text{cm}$ from the center between the plates.

The displacement current density is uniform:

$$J_d = \epsilon_0\frac{\partial E}{\partial t} = \frac{I}{\pi R^2} = \frac{2.0}{\pi(0.050)^2} \approx 255\;\text{A/m}^2 .$$

The displacement current enclosed by a circular loop of radius $r$ is $I_{d,{\rm enc}} = J_d(\pi r^2) = I(r^2/R^2) = 2.0(0.03/0.05)^2 = 0.72\,\text{A}$. Using the Ampère—Maxwell law,

$$B(2\pi r) = \mu_0 I_{d,{\rm enc}} \quad\Rightarrow\quad B = \frac{\mu_0(0.72)}{2\pi(0.03)} \approx 4.8\times 10^{-6}\;\text{T} .$$

Problems with Solutions

Problem 1. A circular conducting loop of radius $a = 4.0\,\text{cm}$ lies in the $xy$-plane. A uniform magnetic field $\bm B = B_0\sin(\omega t)\,\hat{\bm z}$ with $B_0 = 0.10\,\text{T}$ and $\omega = 100\,\text{rad/s}$ passes through it. Find the induced EMF at $t = 0$.

Solution. By Faraday’s law,

$$\mathcal{E} = -\frac{d\Phi_B}{dt} = -\pi a^2 B_0\omega\cos(\omega t) .$$

At $t=0$, $\cos(0)=1$, so

$$|\mathcal{E}| = \pi(0.040)^2(0.10)(100) \approx 5.0\times 10^{-2}\;\text{V} = 50\;\text{mV} .$$

Problem 2. Starting from the two Maxwell inhomogeneous equations, derive the continuity equation $\nabla\cdot\bm J + \partial_t\rho = 0$.

Solution. Take the divergence of $\nabla\times\bm B = \mu_0\bm J + \mu_0\epsilon_0\partial_t\bm E$. Since $\nabla\cdot(\nabla\times\bm B)=0$,

$$0 = \mu_0\nabla\cdot\bm J + \mu_0\epsilon_0\partial_t(\nabla\cdot\bm E) .$$

Using $\nabla\cdot\bm E = \rho/\epsilon_0$ gives $0 = \mu_0\nabla\cdot\bm J + \mu_0\partial_t\rho$, which yields the continuity equation upon dividing by $\mu_0$.

Section summary. Maxwell’s equations unify static and time-dependent fields.

Conservation Laws

Core ideas

Fields carry energy, momentum, and angular momentum. Poynting’s theorem describes local energy exchange between matter and fields. The Maxwell stress tensor expresses electromagnetic forces as momentum flux through surfaces.

For review, be able to derive Poynting’s theorem, compute energy density and intensity, use stress tensor for pressure or force, and interpret field momentum. Keep the physical question visible: identify the degrees of freedom, the conserved quantities, the approximation being made, and the observable that would be measured.

Mathematical spine

$$u=\frac{\epsilon_0E^2}{2}+\frac{B^2}{2\mu_0},\qquad \bm S=\frac{1}{\mu_0}\bm E\times\bm B,\qquad \partial_tu+\nabla\cdot\bm S=-\bm J\cdot\bm E$$

Worked example

A plane electromagnetic wave in vacuum has electric field amplitude $E_0 = 100\,\text{V/m}$. Find the time-averaged energy density and the magnitude of the Poynting vector (intensity).

The magnetic field amplitude is $B_0 = E_0/c = 100/(3.0\times 10^8) = 3.33\times 10^{-7}\,\text{T}$. The instantaneous energy density is

$$u = \frac{\epsilon_0E_0^2}{2}\cos^2(kx-\omega t) + \frac{B_0^2}{2\mu_0}\cos^2(kx-\omega t) = \epsilon_0 E_0^2\cos^2(kx-\omega t) .$$

Averaging over a cycle ($\langle\cos^2\rangle = 1/2$),

$$\langle u\rangle = \frac{\epsilon_0 E_0^2}{2} = \frac{(8.85\times 10^{-12})(100)^2}{2} \approx 4.43\times 10^{-8}\;\text{J/m}^3 .$$

The intensity is $I = \langle S\rangle = c\langle u\rangle = (3.0\times 10^8)(4.43\times 10^{-8}) \approx 13.3\;\text{W/m}^2$.

Problems with Solutions

Problem 1. A parallel-plate capacitor with circular plates of radius $R = 5.0\,\text{cm}$ is being charged so that $E(t) = E_0 t/\tau$ with $E_0 = 1.0\times 10^5\,\text{V/m}$ and $\tau = 1.0\,\text{ms}$. Find the Poynting vector magnitude at the curved edge and the rate of energy inflow.

Solution. The displacement current gives a magnetic field at the edge $B = \mu_0\epsilon_0 (r/2)\partial_t E = (R/2c^2)(E_0/\tau)$. The Poynting vector is radially inward:

$$S = \frac{1}{\mu_0}EB = \frac{\epsilon_0 R E_0^2 t}{2\tau^2} .$$

At $t=\tau$, $S = \epsilon_0 R E_0^2/(2\tau) \approx 2.21\times 10^{-2}\;\text{W/m}^2$. The power entering through the cylindrical edge ($2\pi R d$, with $d$ the plate spacing) equals the rate of change of stored field energy $u\cdot\pi R^2 d$.

Problem 2. Sunlight at Earth’s orbit has intensity $I = 1.35\,\text{kW/m}^2$. What is the radiation pressure on a perfectly absorbing surface?

Solution. For absorption, the radiation pressure equals the momentum flux $p = I/c$:

$$p = \frac{1.35\times 10^3}{3.0\times 10^8} = 4.5\times 10^{-6}\;\text{Pa} .$$

Section summary. Conservation laws track energy and momentum flow in fields.

Electromagnetic Waves

Core ideas

In empty space Maxwell’s equations imply self-propagating transverse waves moving at $c$. In media, waves refract, reflect, attenuate, and disperse depending on material response. Boundary conditions determine Fresnel coefficients and waveguide modes.

For review, be able to derive the wave equation, identify polarization and intensity, apply boundary conditions at interfaces, and relate index to phase velocity. Keep the physical question visible: identify the degrees of freedom, the conserved quantities, the approximation being made, and the observable that would be measured.

Mathematical spine

$$\nabla^2\bm E-\frac{1}{c^2}\partial_t^2\bm E=0,\qquad c=\frac{1}{\sqrt{\mu_0\epsilon_0}},\qquad \bm B=\frac{1}{\omega}\bm k\times\bm E$$

Worked example

A monochromatic electromagnetic wave in vacuum has wavelength $\lambda = 500\,\text{nm}$. Find its frequency, wave number, and the magnetic field amplitude if the electric field amplitude is $E_0 = 80\,\text{V/m}$.

The frequency is

$$f = \frac{c}{\lambda} = \frac{3.00\times 10^8}{500\times 10^{-9}} = 6.00\times 10^{14}\;\text{Hz} .$$

The wave number is $k = 2\pi/\lambda = 1.26\times 10^7\,\text{m}^{-1}$. The angular frequency is $\omega = 2\pi f = 3.77\times 10^{15}\,\text{rad/s}$. The magnetic field amplitude follows from $\bm B_0 = (\bm k\times\bm E_0)/\omega$:

$$B_0 = \frac{E_0}{c} = \frac{80}{3.00\times 10^8} = 2.67\times 10^{-7}\;\text{T} = 0.267\;\mu\text{T} .$$

Problems with Solutions

Problem 1. Light of vacuum wavelength $\lambda_0 = 600\,\text{nm}$ enters glass with refractive index $n = 1.5$. Find the wavelength, frequency, and speed in the glass.

Solution. The frequency does not change at the interface: $f = c/\lambda_0 = 5.00\times 10^{14}\,\text{Hz}$. The speed in glass is $v = c/n = 2.00\times 10^8\,\text{m/s}$. The wavelength becomes $\lambda = \lambda_0/n = 400\,\text{nm}$.

Problem 2. A plane wave in vacuum is normally incident on a perfect conductor at $x=0$. The incident electric field is $\bm E_i = E_0\cos(kx-\omega t)\,\hat{\bm y}$. Write the total electric field and explain the standing wave pattern.

Solution. The reflected field must cancel the incident field at the surface, so $\bm E_r = -E_0\cos(kx+\omega t)\,\hat{\bm y}$. The total field is

$$\bm E = E_0[\cos(kx-\omega t) - \cos(kx+\omega t)]\,\hat{\bm y} = 2E_0\sin(kx)\sin(\omega t)\,\hat{\bm y} .$$

Nodes occur where $\sin(kx)=0$ ($x=0, \lambda/2, \lambda, \dots$) and antinodes where $|\sin(kx)|=1$ ($x=\lambda/4, 3\lambda/4, \dots$).

Section summary. Light is an electromagnetic wave predicted by Maxwell’s equations.

Potentials and Fields

Core ideas

Scalar and vector potentials encode the fields with gauge freedom. Gauge transformations change potentials without changing $\bm E$ or $\bm B$. Lorenz gauge makes relativistic covariance and retarded solutions transparent; Coulomb gauge separates instantaneous constraint from radiation degrees of freedom.

For review, be able to use gauge transformations, derive retarded potentials, distinguish Lorenz and Coulomb gauges, and explain why potentials are not unique. Keep the physical question visible: identify the degrees of freedom, the conserved quantities, the approximation being made, and the observable that would be measured.

Mathematical spine

$$\bm B=\nabla\times\bm A,\qquad \bm E=-\nabla V-\partial_t\bm A,\qquad V' = V-\partial_t\chi,\quad \bm A'=\bm A+\nabla\chi$$

Worked example

In a region with uniform $\bm B = B_0\,\hat{\bm z}$ ($B_0 = 0.5\,\text{T}$), one possible vector potential is $\bm A = B_0 x\,\hat{\bm y}$ (Landau gauge). Show that a gauge transformation with $\chi = -\frac{1}{2}B_0 xy$ produces the symmetric gauge $\bm A' = \frac{1}{2}B_0(-y\,\hat{\bm x} + x\,\hat{\bm y})$, and verify that both give the same $\bm B$.

The transformed potential is

$$\bm A' = \bm A + \nabla\chi = B_0 x\,\hat{\bm y} - \frac{1}{2}B_0(y\,\hat{\bm x} + x\,\hat{\bm y}) = \frac{B_0}{2}(-y\,\hat{\bm x} + x\,\hat{\bm y}) .$$

Taking the curl:

$$(\nabla\times\bm A')_z = \partial_x\left(\frac{B_0 x}{2}\right) - \partial_y\left(-\frac{B_0 y}{2}\right) = \frac{B_0}{2} + \frac{B_0}{2} = B_0 .$$

Similarly $(\nabla\times\bm A)_z = \partial_x(B_0 x) = B_0$. Both yield the same physical field.

Problems with Solutions

Problem 1. Given potentials $V = 0$ and $\bm A = A_0\sin(kx - \omega t)\,\hat{\bm z}$ with $A_0 = 10^{-3}\,\text{V}\cdot\text{s/m}$, $k = 1.0\times 10^7\,\text{m}^{-1}$, and $\omega = 3.0\times 10^{15}\,\text{rad/s}$, find the electric and magnetic fields.

Solution.

$$\bm B = \nabla\times\bm A = -\partial_x A_z\,\hat{\bm y} = -A_0 k\cos(kx-\omega t)\,\hat{\bm y} .$$ $$\bm E = -\nabla V - \partial_t\bm A = -(-\omega A_0\cos(kx-\omega t))\,\hat{\bm z} = \omega A_0\cos(kx-\omega t)\,\hat{\bm z} .$$

Numerically, $E_0 = \omega A_0 = 3.0\times 10^{12}\,\text{V/m}$ and $B_0 = A_0 k = 1.0\times 10^4\,\text{T}$. (These unphysical magnitudes simply illustrate the algebra.)

Problem 2. Show that the Lorenz gauge condition $\nabla\cdot\bm A + \mu_0\epsilon_0\partial_t V = 0$ is preserved under a gauge transformation if the gauge function $\chi$ satisfies the wave equation $\nabla^2\chi - \mu_0\epsilon_0\partial_t^2\chi = 0$.

Solution. Under $V' = V - \partial_t\chi$ and $\bm A' = \bm A + \nabla\chi$,

$$\nabla\cdot\bm A' + \mu_0\epsilon_0\partial_t V' = (\nabla\cdot\bm A + \nabla^2\chi) + \mu_0\epsilon_0(\partial_t V - \partial_t^2\chi) .$$

If the original potentials satisfy the Lorenz condition, the right-hand side becomes $\nabla^2\chi - \mu_0\epsilon_0\partial_t^2\chi$, which vanishes when $\chi$ obeys the wave equation.

Section summary. Potentials reveal gauge freedom and causal propagation.

Radiation

Core ideas

Accelerated charges radiate. Far from the source, radiation fields fall as $1/r$, are transverse, and carry power through the Poynting flux. Dipole radiation is the leading approximation for many antennas and atoms, with angular patterns fixed by acceleration direction.

For review, be able to identify near and radiation zones, use Larmor power, estimate dipole radiation, and connect angular distribution to polarization. Keep the physical question visible: identify the degrees of freedom, the conserved quantities, the approximation being made, and the observable that would be measured.

Mathematical spine

$$P=\frac{\mu_0q^2a^2}{6\pi c},\qquad \left\langle P\right\rangle_{\rm dipole}=\frac{\mu_0\omega^4p_0^2}{12\pi c}$$

Worked example

An oscillating electric dipole has moment $\bm p(t) = p_0\cos(\omega t)\,\hat{\bm z}$ with amplitude $p_0 = 1.0\times 10^{-9}\,\text{C}\cdot\text{m}$ and angular frequency $\omega = 1.0\times 10^9\,\text{rad/s}$. Calculate the time-averaged total radiated power.

Using the dipole radiation formula,

$$\langle P\rangle = \frac{\mu_0\omega^4 p_0^2}{12\pi c} = \frac{(4\pi\times 10^{-7})(10^9)^4(10^{-9})^2}{12\pi(3.0\times 10^8)} .$$

Simplifying the numerator: $4\pi\times 10^{-7}\times 10^{36}\times 10^{-18} = 4\pi\times 10^{11}$. The denominator is $36\pi\times 10^8$. Thus

$$\langle P\rangle = \frac{4\pi\times 10^{11}}{36\pi\times 10^8} = \frac{10^3}{9} \approx 111\;\text{W} .$$

Problems with Solutions

Problem 1. An electron ($q = 1.6\times 10^{-19}\,\text{C}$) in a classical hydrogen atom orbits with radius $r = 5.3\times 10^{-11}\,\text{m}$ and speed $v = 2.2\times 10^6\,\text{m/s}$. Estimate the radiated power using the Larmor formula.

Solution. The centripetal acceleration is $a = v^2/r = (2.2\times 10^6)^2/(5.3\times 10^{-11}) \approx 9.1\times 10^{22}\,\text{m/s}^2$. Then

$$P = \frac{\mu_0 q^2 a^2}{6\pi c} = \frac{(4\pi\times 10^{-7})(1.6\times 10^{-19})^2(9.1\times 10^{22})^2}{6\pi(3.0\times 10^8)} \approx 4.7\times 10^{-8}\;\text{W} .$$

(This classical instability motivated the quantum theory of atoms.)

Problem 2. A short dipole antenna oriented along $z$ radiates at $\omega = 2\pi\times 100\,\text{MHz}$. At a distance $r = 1.0\,\text{km}$ in the $xy$-plane (perpendicular to the dipole), the measured electric field amplitude is $E_0 = 10\,\text{mV/m}$. Find the magnetic field amplitude and the time-averaged Poynting vector magnitude.

Solution. In the radiation zone $B_0 = E_0/c = (10\times 10^{-3})/(3.0\times 10^8) = 3.33\times 10^{-11}\,\text{T} = 33.3\,\text{fT}$. The intensity is

$$\langle S\rangle = \frac{E_0 B_0}{2\mu_0} = \frac{E_0^2}{2\mu_0 c} = \frac{(10^{-2})^2}{2(4\pi\times 10^{-7})(3.0\times 10^8)} \approx 1.33\times 10^{-7}\;\text{W/m}^2 .$$

Section summary. Radiation is energy carried away by fields from accelerated charges.

Electrodynamics and Relativity

Core ideas

Electric and magnetic fields are frame-dependent parts of one electromagnetic field tensor. Lorentz transformations mix $\bm E$ and $\bm B$, while charge conservation becomes four-current conservation. Covariant notation makes Maxwell’s equations compact and shows why magnetism is tied to relativity.

For review, be able to use four-vectors, transform fields between inertial frames, write Maxwell equations covariantly, and identify Lorentz invariants. Keep the physical question visible: identify the degrees of freedom, the conserved quantities, the approximation being made, and the observable that would be measured.

Mathematical spine

$$F^{\mu\nu}=\partial^\mu A^\nu-\partial^\nu A^\mu,\qquad \partial_\mu F^{\mu\nu}=\mu_0J^\nu,\qquad \partial_{[\alpha}F_{\beta\gamma]}=0$$

Worked example

In inertial frame $S$, a region has $\bm E = E_0\,\hat{\bm y}$ with $E_0 = 100\,\text{kV/m}$ and $\bm B = 0$. Frame $S'$ moves at velocity $\bm v = 0.60c\,\hat{\bm x}$ relative to $S$. Find the fields measured in $S'$.

With $\gamma = 1/\sqrt{1-0.60^2} = 1.25$, the field transformations are

$$E'_y = \gamma(E_y - vB_z) = 1.25(100\,\text{kV/m}) = 125\,\text{kV/m} ,$$ $$B'_z = \gamma\left(B_z - \frac{v}{c^2}E_y\right) = -1.25\frac{(0.60c)}{c^2}(10^5) = -\frac{0.75\times 10^5}{3.0\times 10^8} = -2.5\times 10^{-4}\;\text{T} .$$

A purely electric field in $S$ appears as a mixture of $\bm E'$ and $\bm B'$ in $S'$, illustrating that $\bm E$ and $\bm B$ are frame-dependent components of $F^{\mu\nu}$.

Problems with Solutions

Problem 1. In frame $S$, $\bm E = 3.0\times 10^5\,\text{V/m}\,\hat{\bm x}$ and $\bm B = 0.40\,\text{T}\,\hat{\bm y}$. Show that $|\bm E| < c|\bm B|$ and find the velocity of a frame $S'$ in which $\bm E' = 0$.

Solution. Since $3.0\times 10^5 < (3.0\times 10^8)(0.40) = 1.2\times 10^8$, a frame exists where $\bm E'=0$. The condition $E'_\parallel=0$ is automatic because $\bm E$ and $\bm B$ are perpendicular. We need $\bm E'_\perp = \gamma(\bm E + \bm v\times\bm B)=0$, so $\bm v\times\bm B = -\bm E$. Choosing $\bm v$ along $\bm E\times\bm B \propto \hat{\bm z}$:

$$v = \frac{E}{B} = \frac{3.0\times 10^5}{0.40} = 7.5\times 10^5\;\text{m/s} = 750\;\text{km/s} .$$

Problem 2. A line charge with linear charge density $\lambda = 2.0\,\mu\text{C/m}$ moves at speed $v = 0.80c$ along the $x$-axis. Write the four-current $J^\mu$ in the lab frame.

Solution. The proper charge density is $\lambda_0 = \gamma\lambda$ where $\gamma = 1/\sqrt{1-0.8^2} = 5/3$. In the lab frame the charge density is $\rho = \lambda\delta(y)\delta(z)$ and the current density is $J_x = \rho v = \lambda v\delta(y)\delta(z)$. The four-current is

$$J^\mu = (c\rho, J_x, 0, 0) = \lambda\delta(y)\delta(z)(c, v, 0, 0) .$$

Numerically, $J^0/c = \rho = 2.0\,\mu\text{C/m}$ and $J^1 = (2.0\,\mu\text{C/m})(0.80c) = 480\,\text{A}$ (per unit cross-sectional area factor from the delta functions).

Section summary. Electromagnetism is naturally a relativistic field theory.