Electrodynamics

A changing magnetic field induces an electric field:

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

Lenz’s Law: The induced EMF opposes the change in flux that produced it.

Example. A circular loop of radius $R$ in a uniform magnetic field $\mathbf{B}(t) = B_0 \cos(\omega t)\,\hat{\mathbf{z}}$.

The flux: $\Phi_B = \pi R^2 B_0 \cos(\omega t)$.

The induced EMF: $\mathcal{E} = -\frac{d\Phi_B}{dt} = \pi R^2 B_0 \omega \sin(\omega t)$.

Maxwell’s key insight: Ampere’s law $\nabla \times \mathbf{B} = \mu_0 \mathbf{J}$ is inconsistent with The continuity equation. Adding the displacement current term $\mu_0 \varepsilon_0 \partial \mathbf{E}/\partial t$ Resolves this:

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

Problem. A parallel-plate capacitor with circular plates of radius $R$ is being charged by a Current $I$. Find the magnetic field between the plates at distance $r$ from the axis.

Solution. Between the plates, $\mathbf{J} = 0$But there is a changing electric field. The Displacement current density is $J_d = \varepsilon_0 \frac{\partial E}{\partial t}$.

$E = \frac{\sigma}{\varepsilon_0} = \frac{Q}{\pi R^2 \varepsilon_0}$So $\frac{\partial E}{\partial t} = \frac{I}{\pi R^2 \varepsilon_0}$.

By symmetry, use an Amperian loop of radius $r \lt R$:

$\oint \mathbf{B} \cdot d\mathbf{l} = \mu_0 \varepsilon_0 \frac{\partial}{\partial t}\int \mathbf{E} \cdot d\mathbf{A}$

$B \cdot 2\pi r = \mu_0 \varepsilon_0 \cdot \frac{I}{\pi R^2 \varepsilon_0} \cdot \pi r^2 = \frac{\mu_0 I r^2}{R^2}$

$B = \frac{\mu_0 I r}{2\pi R^2}$

$\blacksquare$

When a conductor moves through a magnetic field, the Lorentz force on the charges produces an EMF:

$\mathcal{E} = \oint (\mathbf{v} \times \mathbf{B}) \cdot d\mathbf{l}$

This is consistent with the flux rule $\mathcal{E} = -d\Phi_B/dt$ since changing the Circuit’s geometry or position changes the flux.

Problem with Ampere’s original law. The original Ampere’s law was $\nabla \times \mathbf{B} = \mu_0 \mathbf{J}$. Taking the divergence:

$\nabla \cdot (\nabla \times \mathbf{B}) = 0 = \mu_0 \nabla \cdot \mathbf{J}$

This requires $\nabla \cdot \mathbf{J} = 0$ at all times, which contradicts the continuity Equation $\nabla \cdot \mathbf{J} = -\partial\rho/\partial t$ whenever charge density changes.

Resolution. Use Gauss’s law to rewrite the continuity equation:

$\nabla \cdot \mathbf{J} = -\frac{\partial\rho}{\partial t} = -\frac{\partial}{\partial t}(\varepsilon_0 \nabla \cdot \mathbf{E}) = -\nabla \cdot \left(\varepsilon_0\frac{\partial \mathbf{E}}{\partial t}\right)$

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

This suggests modifying Ampere’s law to:

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

Now taking the divergence gives zero identically, consistent with charge conservation. The Term $\mu_0 \varepsilon_0\,\partial\mathbf{E}/\partial t$ is the displacement current.

Physical interpretation. The displacement current represents the time-varying electric field That produces a magnetic field just as a real current does. It is essential inside capacitors, Where $\mathbf{J} = 0$ but $\partial\mathbf{E}/\partial t \neq 0$.

Mutual inductance. When circuit 1 produces flux $\Phi_{21}$ through circuit 2:

$M = \frac{\Phi_{21}}{I_1}$

The EMF induced in circuit 2 by a changing current in circuit 1:

$\mathcal{E}_2 = -M\frac{dI_1}{dt}$

Self-inductance. A circuit carrying current $I$ produces flux $\Phi$ through itself:

$L = \frac{N\Phi}{I}$

The back-EMF:

$\mathcal{E} = -L\frac{dI}{dt}$

Energy stored in an inductor:

$U = \frac{1}{2}LI^2$

Example: Solenoid. A long solenoid of length $\ell$ with $N$ turns, cross-sectional area $A$:

$L = \frac{\mu_0 N^2 A}{\ell}$