All of classical electricity and magnetism fits in four equations, named for Maxwell, who completed them in the 1860s. Each one can be written two ways: as a statement about fluxes and circulations (the integral form, close to experiments) or about divergence and curl at each point (the differential form). Stokes' theorem and the divergence theorem translate between them; Gauss's law was the first example.
| Law | Integral form | Differential form |
|---|---|---|
| Gauss's law | $\iint_S \mathbf{E} \cdot d\mathbf{S} = Q_{\text{enc}}/\varepsilon_0$ | $\nabla \cdot \mathbf{E} = \rho/\varepsilon_0$ |
| No magnetic charges | $\iint_S \mathbf{B} \cdot d\mathbf{S} = 0$ | $\nabla \cdot \mathbf{B} = 0$ |
| Faraday's law | $\oint_C \mathbf{E} \cdot d\mathbf{r} = -\frac{d}{dt}\iint_S \mathbf{B} \cdot d\mathbf{S}$ | $\nabla \times \mathbf{E} = -\frac{\partial \mathbf{B}}{\partial t}$ |
| Ampère–Maxwell law | $\oint_C \mathbf{B} \cdot d\mathbf{r} = \mu_0 I_{\text{enc}} + \mu_0\varepsilon_0\frac{d}{dt}\iint_S \mathbf{E} \cdot d\mathbf{S}$ | $\nabla \times \mathbf{B} = \mu_0\mathbf{J} + \mu_0\varepsilon_0\frac{\partial \mathbf{E}}{\partial t}$ |
Key idea
Divergence equations correspond to flux through closed surfaces (divergence theorem). Curl equations correspond to circulation around loops (Stokes' theorem).
Why it works
Faraday, from integral to differential form: by Stokes, $\oint_C \mathbf{E} \cdot d\mathbf{r} = \iint_S (\nabla \times \mathbf{E}) \cdot d\mathbf{S}$, so $\iint_S \left(\nabla \times \mathbf{E} + \frac{\partial \mathbf{B}}{\partial t}\right) \cdot d\mathbf{S} = 0$ for every small surface $S$, in every orientation. A continuous field whose flux through every tiny disk is ${}0$ must be $\mathbf{0}$.
Why Maxwell added a term: take the divergence of the Ampère–Maxwell law. The left side is $\nabla \cdot (\nabla \times \mathbf{B}) = 0$. The right side becomes $$\mu_0\left(\nabla \cdot \mathbf{J} + \varepsilon_0\frac{\partial}{\partial t}\nabla \cdot \mathbf{E}\right) = \mu_0\left(\nabla \cdot \mathbf{J} + \frac{\partial \rho}{\partial t}\right),$$ which is ${}0$ by conservation of charge. Without the $\frac{\partial \mathbf{E}}{\partial t}$ term, the equation would force $\nabla \cdot \mathbf{J} = 0$, which fails while a capacitor charges.
Why Maxwell added a term: take the divergence of the Ampère–Maxwell law. The left side is $\nabla \cdot (\nabla \times \mathbf{B}) = 0$. The right side becomes $$\mu_0\left(\nabla \cdot \mathbf{J} + \varepsilon_0\frac{\partial}{\partial t}\nabla \cdot \mathbf{E}\right) = \mu_0\left(\nabla \cdot \mathbf{J} + \frac{\partial \rho}{\partial t}\right),$$ which is ${}0$ by conservation of charge. Without the $\frac{\partial \mathbf{E}}{\partial t}$ term, the equation would force $\nabla \cdot \mathbf{J} = 0$, which fails while a capacitor charges.
Worked examples
Worked example: The field of a wire
Use Ampère's law to find the magnetic field at distance $r$ from a long straight wire carrying a steady current $I$.
- By symmetry $\mathbf{B}$ circles the wire with constant size $B(r)$ on each circle around it.
- $\oint \mathbf{B} \cdot d\mathbf{r} = B \cdot 2\pi r = \mu_0 I$, so $B = \frac{\mu_0 I}{2\pi r}$.
Answer: $\frac{\mu_0 I}{2\pi r}$
Worked example: An induced voltage
A circular loop of radius ${}0.1$ m lies in the $xy$-plane in the field $\mathbf{B} = \langle 0, 0, 0.5t \rangle$ teslas. Find $\oint_C \mathbf{E} \cdot d\mathbf{r}$, with $C$ counterclockwise seen from above.
- Magnetic flux through the disk (upward normal): ${}0.5t \cdot \pi(0.1)^2 = 0.005\pi t$.
- Faraday: $$\oint_C \mathbf{E} \cdot d\mathbf{r} = -\frac{d}{dt}(0.005\pi t) = -0.005\pi \approx -0.0157$$ volts.
Answer: $-0.005\pi$ V
Worked example: Could it be a magnetic field?
Which of $\langle x, y, -2z \rangle$ and $\langle x, y, z \rangle$ could be a magnetic field?
- $\nabla \cdot \langle x, y, -2z \rangle = 1 + 1 - 2 = 0$: allowed.
- $\nabla \cdot \langle x, y, z \rangle = 3 \ne 0$: it would need magnetic charges, so no.
Answer: only $\langle x, y, -2z \rangle$