Vector magnetic potential

The vector magnetic potential $\mathbf{A}$ is a vector field whose curl is the magnetic flux density:

$\mathbf{B}=\nabla \times \mathbf{A}\text{(Wb/m or Tcdotpm)}.$

It exists for any $\mathbf{B}$ field because $\nabla \cdot \mathbf{B}=0$ (no magnetic monopoles), and any divergence-free vector field can be written as the curl of another vector field. The $\mathbf{A}$ field plays the role for magnetic problems that the scalar potential $V$ plays for electric problems.

Why use it

A magnetic field $\mathbf{B}$ has three scalar components but is constrained by $\nabla \cdot \mathbf{B}=0$, leaving two independent components per point. Working with $\mathbf{A}$ instead has its own complication — $\mathbf{A}$ has three components, but it’s only defined up to a gauge (you can add $\nabla \psi$ to $\mathbf{A}$ without changing $\mathbf{B}$). Net independent information matches.

Practical reasons for $\mathbf{A}$:

1. Linear in current. Each current element contributes additively to $\mathbf{A}$, just as each charge element contributes additively to $V$.
2. The result $\mathbf{B}=\nabla \times \mathbf{A}$ automatically satisfies $\nabla \cdot \mathbf{B}=0$ — one Maxwell equation comes for free.
3. Natural object for time-varying problems. In dynamics, $\mathbf{E}$ is not the gradient of a scalar alone; it splits as $\mathbf{E}=-\nabla V-{\partial }_t\mathbf{A}$.

Integral expression

For a current distribution in an unbounded space, the vector potential is

$\mathbf{A}(\mathbf{R})=\frac{{\mu }_0}{4\pi }\int \frac{\mathbf{J}({\mathbf{R}}^{\prime })}{|\mathbf{R}-{\mathbf{R}}^{\prime }|}d{v}^{\prime }.$

For line, surface, volume currents respectively:

$\mathbf{A}=\frac{{\mu }_0}{4\pi }{\int }_C\frac{Id{\mathbf{l}}^{\prime }}{R},\mathbf{A}=\frac{{\mu }_0}{4\pi }{\int }_S\frac{{\mathbf{J}}_{s}d{s}^{\prime }}{R},\mathbf{A}=\frac{{\mu }_0}{4\pi }{\int }_V\frac{\mathbf{J}d{v}^{\prime }}{R}.$

These are direct analogs of the scalar potential integrals — every current bit contributes like a “source,” weighted by $1/R$.

Once $\mathbf{A}$ is computed, take its curl to get $\mathbf{B}$. The computational advantage: integrate a scalar at each point in $\mathbf{A}$, then differentiate. Often easier than integrating the more complex $1/R^2$ cross product in Biot-Savart law.

Gauge freedom

If $\mathbf{A}$ gives $\mathbf{B}$, so does $\mathbf{A}+\nabla \psi$ for any scalar $\psi$, since $\nabla \times \nabla \psi =0$. This gauge freedom is the price you pay for the unification: $\mathbf{A}$ isn’t uniquely determined by the physics, only $\mathbf{B}$ is.

A common gauge choice is the Coulomb gauge: $\nabla \cdot \mathbf{A}=0$. In this gauge, the integral expression above is the unique $\mathbf{A}$ for the given current distribution. The Coulomb gauge is convenient in magnetostatics.

For time-varying problems, the Lorenz gauge is more useful: $\nabla \cdot \mathbf{A}+\mu ϵ{\partial }_{t}V=0$. This gauge makes the potentials’ wave equations symmetric: ${\nabla }^2\mathbf{A}-\mu ϵ{\partial }_t^2\mathbf{A}=-\mu \mathbf{J}$.

In time-varying problems

In dynamic problems, the Electric field is no longer a pure gradient. By Faraday’s law $\nabla \times \mathbf{E}=-{\partial }_t\mathbf{B}=-\nabla \times {\partial }_t\mathbf{A}$, so the combination $\mathbf{E}+{\partial }_t\mathbf{A}$ is curl-free and can be written as $-\nabla V$:

$\mathbf{E}=-\nabla V-\frac{\partial \mathbf{A}}{\partial t}.$

The pair $(V,\mathbf{A})$ — four scalar fields — is the natural object in dynamical EM. See Electromagnetic potentials for the full treatment, gauge freedom, and the wave equations they obey.

Worked example: long straight wire

A wire along $\hat{\mathbf{z}}$ carrying current $I$. By symmetry $\mathbf{A}=A_z(r)\hat{\mathbf{z}}$ — direction along the current, magnitude depending only on $r$.

The Biot-Savart integral for $A_z$ diverges if you integrate over an infinite wire — formal artifact of the unbounded geometry. The standard workaround: differentiate first to get $\mathbf{B}$, or use the known $\mathbf{B}$ field and find an $\mathbf{A}$ that produces it.

We know $\mathbf{B}=\hat{\mathbf{\varphi }}{\mu }_{0}I/(2\pi r)$. Try $\mathbf{A}=A_z(r)\hat{\mathbf{z}}$. Curl in cylindrical: $\nabla \times \mathbf{A}=-(d{A}_z/dr)\hat{\mathbf{\varphi }}$. Match: $-d{A}_z/dr={\mu }_{0}I/(2\pi r)$, so

$A_z(r)=-\frac{{\mu }_{0}I}{2\pi }\ln r+\text{const}.$

The constant is arbitrary (gauge). Logarithmic growth at large $r$ — the integral diverges as expected, but $\mathbf{B}=-d{A}_z/dr$ is finite and well-defined.