Chapter 6

Magnetic Vector Potential

6.1 The Vector Potential A

Since $\nabla \cdot \mathbf{B} = 0$, we can always write $\mathbf{B} = \nabla \times \mathbf{A}$for a vector potential $\mathbf{A}$. Substituting into Ampere's law:

$$\nabla \times (\nabla \times \mathbf{A}) = \mu_0 \mathbf{J}$$$$\nabla(\nabla \cdot \mathbf{A}) - \nabla^2 \mathbf{A} = \mu_0 \mathbf{J}$$

Choosing the Coulomb gauge $\nabla \cdot \mathbf{A} = 0$:

$$\boxed{\nabla^2 \mathbf{A} = -\mu_0 \mathbf{J}}$$

This is a vector Poisson equation — identical in form to the scalar Poisson equation for $V$. The solution is:

$$\mathbf{A}(\mathbf{r}) = \frac{\mu_0}{4\pi}\int \frac{\mathbf{J}(\mathbf{r}')}{|\mathbf{r} - \mathbf{r}'|}\,d\tau'$$

6.1.1 Gauge Freedom

The vector potential is not unique: we can add the gradient of any scalar function$\lambda$ without changing $\mathbf{B}$:

$$\mathbf{A} \to \mathbf{A} + \nabla\lambda, \qquad V \to V - \frac{\partial\lambda}{\partial t}$$

Physical quantities ($\mathbf{E}$, $\mathbf{B}$) are gauge-invariant. Common gauge choices:

Coulomb gauge

$$\nabla \cdot \mathbf{A} = 0$$

Natural for magnetostatics and radiation

Lorenz gauge

$$\nabla \cdot \mathbf{A} + \mu_0\epsilon_0\frac{\partial V}{\partial t} = 0$$

Manifestly relativistic; best for wave equations

6.2 Magnetic Multipole Expansion

For a localized current loop, the vector potential at large distances expands as:

$$\mathbf{A}(\mathbf{r}) = \frac{\mu_0}{4\pi}\left[\frac{\mathbf{m} \times \hat{r}}{r^2} + \cdots\right]$$

The leading term is the magnetic dipole with moment:

$$\mathbf{m} = I\int d\mathbf{a} = I\mathbf{a}$$

The magnetic field of a pure magnetic dipole $\mathbf{m} = m\hat{z}$ at large distances is:

$$\mathbf{B}_{\rm dip} = \frac{\mu_0 m}{4\pi r^3}(2\cos\theta\,\hat{r} + \sin\theta\,\hat{\theta})$$

Note: there is no magnetic monopole term (since $\nabla \cdot \mathbf{B} = 0$ always). The dipole is the lowest non-vanishing term.

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