Maxwell's Equations

🌐 Maxwell's Equations

Part 1 of 7 — The Four Laws of Electromagnetism

The Four Equations

The AP course numbers Maxwell's equations as you meet them during the year:

Physical Meaning

You will use the first three, and the current term of the fourth, quantitatively. Maxwell's addition (the changing-$\vec{E}$ term) is needed only as an idea: a changing electric field produces a magnetic field.

🔑 Maxwell's equations are the complete set of laws of classical electromagnetism. (They also predict electromagnetic waves, which are not assessed on the AP Physics C: E&M exam.)

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📝 Worked Example — Gauss's Law with Calculus

A point charge $Q = +4.0 \text{ nC}$ sits at the center of a sphere of radius $r = 0.20 \text{ m}$. Find the electric flux through the sphere, then the field magnitude on its surface.

Step 1 — Apply Gauss's Law. The closed-surface integral gives the enclosed charge over $\varepsilon_0$: $\Phi_E = \oint \vec{E} \cdot d\vec{A} = \frac{Q_{enc}}{\varepsilon_0} = \frac{4.0 \times 10^{-9}}{8.85 \times 10^{-12}} \approx 452 \text{ N}\cdot\text{m}^2/\text{C}$

Step 2 — Exploit spherical symmetry. By symmetry $\vec{E}$ is radial and constant in magnitude over the surface, so $\oint \vec{E} \cdot d\vec{A} = E \oint dA = E(4\pi r^2)$.

Step 3 — Solve for the field. Setting $E(4\pi r^2) = \frac{Q}{\varepsilon_0}$ gives the familiar point-charge result: $E = \frac{1}{4\pi\varepsilon_0}\frac{Q}{r^2} = \frac{(8.99 \times 10^9)(4.0 \times 10^{-9})}{(0.20)^2} \approx 899 \text{ N/C}$

🔑 Gauss's Law is most powerful when symmetry lets you pull $E$ outside the surface integral.

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🧲 Gauss's Law for Magnetism

Part 2 of 7 — Gauss's Law for Magnetism and Closed Field Lines

Maxwell's Second Equation

$\oint \vec{B} \cdot d\vec{A} = 0$

The net magnetic flux through any closed surface is zero: every magnetic field line that enters a closed surface also leaves it. The reason is that magnetic fields come from dipoles — current loops, moving charges, and aligned atomic dipoles — never from isolated poles (monopoles).

Bar Magnets and Earth

• Outside a bar magnet the field points from the north pole to the south pole; inside it runs from south back to north, closing each loop.
• Cut a magnet in half and you get two complete dipoles — no lone north or south pole ever appears.
• Earth's field is approximately that of a dipole. A compass's north end points toward geographic north, so the pole near geographic north is a magnetic south pole.

Sign Convention for Closed Surfaces

Area vectors on a closed surface point outward: flux leaving the surface counts as positive, flux entering counts as negative. Gauss's law for magnetism says these always cancel.

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📝 Worked Example — Flux Through Closed Surfaces

(a) Uniform field. A closed cylinder of radius $0.050 \text{ m}$ has its axis along a uniform field $B = 0.40 \text{ T}$. Find the flux through each part.

• Cap where the field leaves: $\Phi = +BA = (0.40)\pi(0.050)^2 = +3.14 \times 10^{-3} \text{ Wb}$
• Cap where the field enters: $\Phi = -3.14 \times 10^{-3} \text{ Wb}$
• Curved side: $\vec{B}$ runs along the surface, so $\vec{B} \cdot d\vec{A} = 0$

Total: $0$, as Gauss's law for magnetism demands.

(b) Nonuniform field. A closed surface is made of a flat disk and a dome. $2.5 \times 10^{-4} \text{ Wb}$ of magnetic flux enters through the disk. Without knowing anything else about the field, the dome must carry exactly $2.5 \times 10^{-4} \text{ Wb}$ outward, because the total must be zero.

🔑 For any open surface, the magnetic flux depends only on its boundary: any two surfaces sharing the same edge carry the same flux. That is why "the flux through a loop" is well defined in Faraday's law.

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⚡ Faraday's Law and Induced Electric Fields

Part 3 of 7 — Faraday's Law as an Induced Electric Field

Maxwell's Third Equation

$\oint \vec{E} \cdot d\vec{l} = -\frac{d\Phi_B}{dt}$

A changing magnetic flux creates an electric field that circulates around the region where the flux is changing. The field exists whether or not a wire is there; a conducting loop simply lets it push charges around, and the EMF in the loop is $\mathcal{E} = \oint \vec{E} \cdot d\vec{l}$. For a coil of $N$ turns, $\mathcal{E} = -N\frac{d\Phi_B}{dt}$, and Lenz's law gives the direction.

Two Kinds of Electric Field

Both kinds exert the same force $\vec{F} = q\vec{E}$ on a charge.

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📝 Worked Example — Induced Field In and Around a Solenoid

A long solenoid of radius $R = 0.020 \text{ m}$ has $n = 1000$ turns per meter. Its current increases at $\frac{dI}{dt} = 50 \text{ A/s}$. Find the induced electric field at (a) $r = 0.010 \text{ m}$ (inside) and (b) $r = 0.040 \text{ m}$ (outside).

Step 1 — Rate of change of $B$ inside. From Ampère's law, $B = \mu_0 n I$, so $\frac{dB}{dt} = \mu_0 n \frac{dI}{dt} = (4\pi \times 10^{-7})(1000)(50) = 6.28 \times 10^{-2} \text{ T/s}$

Step 2 — Use symmetry. On a circle of radius $r$ centered on the axis, $\vec{E}$ is tangent to the circle with constant magnitude, so $\oint \vec{E} \cdot d\vec{l} = E(2\pi r)$.

Step 3 — Inside ($r < R$). The circle encloses flux $B\pi r^2$: $E(2\pi r) = \pi r^2 \frac{dB}{dt} ;\Rightarrow; E = \frac{r}{2}\frac{dB}{dt} = \frac{0.010}{2}(6.28 \times 10^{-2}) = 3.1 \times 10^{-4} \text{ V/m}$

Step 4 — Outside ($r > R$). $B \approx 0$ outside, but the circle still encloses the solenoid's flux $B\pi R^2$: $E(2\pi r) = \pi R^2 \frac{dB}{dt} ;\Rightarrow; E = \frac{R^2}{2r}\frac{dB}{dt} = \frac{(0.020)^2}{2(0.040)}(6.28 \times 10^{-2}) = 3.1 \times 10^{-4} \text{ V/m}$

🔑 Inside, $E$ grows linearly with $r$; outside, it falls as $1/r$. There is an induced electric field outside the solenoid even though the magnetic field there is essentially zero.

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🔁 Ampère's Law and Maxwell's Addition

Part 4 of 7 — Ampère's Law for Symmetric Currents and Maxwell's Addition

Maxwell's Fourth Equation, Current Term

$\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{enc}$

Ampère's law always holds, but it solves for $B$ only when symmetry makes $B$ constant along the Amperian loop (or perpendicular to it on some segments). The AP exam uses it for exactly these shapes:

Several conductors? Use superposition: add the fields of each, or let the Amperian loop count the net enclosed current.

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📝 Worked Example — A Hollow Cylindrical Conductor

A long tube has inner radius $a = 1.0 \text{ mm}$ and outer radius $b = 3.0 \text{ mm}$ and carries $I = 12 \text{ A}$ spread uniformly over its cross-section. Find $B$ at $r = 2.0 \text{ mm}$ (inside the metal).

Step 1 — Enclosed current. The current is uniform over the metal's area $\pi(b^2 - a^2)$, so a circle of radius $r$ encloses the fraction of that area between $a$ and $r$: $I_{enc} = I\,\frac{r^2 - a^2}{b^2 - a^2} = 12\cdot\frac{4.0 - 1.0}{9.0 - 1.0} = 12 \cdot \frac{3}{8} = 4.5 \text{ A}$

Step 2 — Ampère's law. $B(2\pi r) = \mu_0 I_{enc}$: $B = \frac{\mu_0 I_{enc}}{2\pi r} = \frac{(2 \times 10^{-7})(4.5)}{2.0 \times 10^{-3}} = 4.5 \times 10^{-4} \text{ T}$

Check the regions: $B = 0$ in the hollow ($r < a$); $B$ grows through the metal; $B = \mu_0 I/(2\pi r)$ outside ($r > b$), just like a thin wire.

Maxwell's Addition — Qualitatively

While a capacitor charges, no charge crosses the gap between its plates, yet a magnetic field circles the gap just as it circles the wires. The source is the changing electric field between the plates: Maxwell's addition says a changing $\vec{E}$ creates $\vec{B}$, much as a moving charge does. The AP exam does not ask you to calculate with this term; you need only the idea.

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🧭 Gaussian Surfaces vs. Amperian Loops

Part 5 of 7 — Choosing Gaussian Surfaces and Amperian Loops

The Same Strategy, Two Laws

Matching the Symmetry

When Symmetry Fails

Both laws are always true, but they give a field only with spherical, cylindrical, or planar symmetry (for $\vec{E}$) or the long-wire, solenoid, slab, and cylinder shapes (for $\vec{B}$). Otherwise:

• Few point charges → superpose Coulomb fields.
• Ring, arc, or finite line of charge → integrate $dE = k\,dq/r^2$.
• Current segment, arc, or single loop → use the Biot–Savart law.

Concept Check 🎯

📝 Worked Example — Two Coaxial Problems Side by Side

(a) Concentric charged cylinders (Gauss). A long inner cylinder carries charge $+\lambda$ per length; a thin outer coaxial shell carries $-\lambda$ per length.

• Between them: a coaxial Gaussian cylinder encloses $+\lambda L$, so $E(2\pi rL) = \lambda L/\varepsilon_0$ and $E = \dfrac{\lambda}{2\pi\varepsilon_0 r}$.
• Outside both: the enclosed charge is $\lambda L - \lambda L = 0$, so $E = 0$.

(b) Coaxial cable (Ampère). An inner wire carries current $I$ one way; the outer conducting shell carries $I$ back.

• Between them: a circular Amperian loop encloses $I$, so $B(2\pi r) = \mu_0 I$ and $B = \dfrac{\mu_0 I}{2\pi r}$.
• Outside both: the net enclosed current is $I - I = 0$, so $B = 0$.

Numbers for (b): with $I = 5.0 \text{ A}$ at $r = 2.0 \text{ mm}$: $B = \dfrac{(2 \times 10^{-7})(5.0)}{2.0 \times 10^{-3}} = 5.0 \times 10^{-4} \text{ T}$.

🔑 Same symmetry, same steps: pick the surface or loop, find what it encloses, pull the field out of the integral.

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🛠️ Maxwell Workshop

Part 6 of 7 — Practice

Which Equation Does the Job?

Many problems chain two or three of these: one law gives a field, another uses how that field changes.

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📝 Worked Example — One Solenoid, Three Equations

A long solenoid of radius $R = 0.030 \text{ m}$ has $n = 2000$ turns per meter. At one instant its current is $I = 4.0 \text{ A}$ and increasing at $10 \text{ A/s}$.

(a) Field inside — Ampère's law. $B = \mu_0 n I = (4\pi \times 10^{-7})(2000)(4.0) \approx 1.0 \times 10^{-2} \text{ T}$

(b) Flux through a closed box that straddles the windings — Gauss's law for magnetism. Whatever the box's shape, the net flux is $0$: field lines that enter it inside the solenoid leave it elsewhere.

(c) Induced electric field at $r = 0.060 \text{ m}$, outside — Faraday's law. $\frac{dB}{dt} = \mu_0 n \frac{dI}{dt} = (4\pi \times 10^{-7})(2000)(10) = 2.5 \times 10^{-2} \text{ T/s}$ A circle of radius $r$ encloses flux $B\pi R^2$, so $E = \frac{R^2}{2r}\frac{dB}{dt} = \frac{(0.030)^2}{2(0.060)}(2.5 \times 10^{-2}) \approx 1.9 \times 10^{-4} \text{ V/m}$

🔑 Inside the field region $E \propto r$; outside it the enclosed flux is fixed at $B\pi R^2$, so $E \propto 1/r$.

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📋 Maxwell's Equations Review

Part 7 of 7 — Final Summary

The Big Picture

Maxwell's four equations describe ALL of classical electromagnetism:

1. Gauss for E: $\oint \vec{E} \cdot d\vec{A} = Q_{enc}/\varepsilon_0$ — charges create electric fields
2. Gauss for B: $\oint \vec{B} \cdot d\vec{A} = 0$ — no magnetic monopoles; field lines close
3. Faraday: $\oint \vec{E} \cdot d\vec{l} = -d\Phi_B/dt$ — changing magnetic flux creates a circulating electric field
4. Ampère + Maxwell's addition: currents create magnetic fields (quantitative), and so does a changing electric field (qualitative only)

🔑 "Maxwell's equations are the most beautiful equations in physics." — Richard Feynman

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📝 Worked Example — Synthesis: Wire Field → Flux → EMF

A long straight wire carries a current that increases as $I(t) = (2.0 \text{ A/s})\,t$. A square loop of side $\ell = 0.10 \text{ m}$ lies in the same plane, with its near side $a = 0.050 \text{ m}$ from the wire and its far side at $0.15 \text{ m}$. Find the EMF induced in the loop.

Step 1 — Field of the wire (Ampère's law). $B(r) = \dfrac{\mu_0 I}{2\pi r}$, perpendicular to the loop's plane.

Step 2 — Flux by integration. Strips of width $dr$ and length $\ell$: $\Phi_B = \int_{a}^{3a} \frac{\mu_0 I}{2\pi r}\,\ell\,dr = \frac{\mu_0 I \ell}{2\pi}\ln 3 = (2 \times 10^{-7})(0.10)(1.10)\,I = (2.2 \times 10^{-8})\,I$

Step 3 — EMF (Faraday's law). $|\mathcal{E}| = \frac{d\Phi_B}{dt} = (2.2 \times 10^{-8})\frac{dI}{dt} = (2.2 \times 10^{-8})(2.0) = 4.4 \times 10^{-8} \text{ V}$

Step 4 — Direction (Lenz's law). The flux into the loop is increasing, so the induced current circulates to make a field opposite to the wire's field inside the loop.

🔑 One problem, three laws: Ampère gives $B$, integration gives $\Phi_B$, Faraday gives $\mathcal{E}$.

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