Magnetic Vector Potential

Step 1: The mathematical theorem (Helmholtz decomposition)

Any vector field can be decomposed into a curl-free part and a divergence-free part. The Helmholtz theorem states that a divergence-free field ($\nabla \cdot \mathbf{B} = 0$) can always be written as the curl of some other vector field.

Step 2: Direct construction from Biot-Savart

From the Biot-Savart law, write $\hat{\mathscr{r}}/\mathscr{r}^2 = -\nabla(1/\mathscr{r})$:

$$\mathbf{B} = \frac{\mu_0}{4\pi}\int \mathbf{J}(\mathbf{r}') \times \frac{\hat{\mathscr{r}}}{\mathscr{r}^2}\,d\tau' = -\frac{\mu_0}{4\pi}\int \mathbf{J}(\mathbf{r}') \times \nabla\!\left(\frac{1}{\mathscr{r}}\right)d\tau'$$

Step 3: Use the product rule for curl

Since $\mathbf{J}(\mathbf{r}')$ does not depend on $\mathbf{r}$, apply $\mathbf{F} \times \nabla g = \nabla \times (g\mathbf{F}) - g(\nabla \times \mathbf{F})$. With $\nabla \times \mathbf{J}(\mathbf{r}') = 0$ (since $\mathbf{J}$ depends on $\mathbf{r}'$, not $\mathbf{r}$):

$$\mathbf{J} \times \nabla\!\left(\frac{1}{\mathscr{r}}\right) = -\nabla \times \left(\frac{\mathbf{J}}{\mathscr{r}}\right)$$

Step 4: Factor the curl out of the integral

Since the curl acts on $\mathbf{r}$ but the integration is over $\mathbf{r}'$:

$$\mathbf{B} = \frac{\mu_0}{4\pi}\nabla \times \int \frac{\mathbf{J}(\mathbf{r}')}{\mathscr{r}}\,d\tau'$$

Step 5: Identify A

This is precisely $\mathbf{B} = \nabla \times \mathbf{A}$ with:

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

The existence of $\mathbf{A}$ is guaranteed whenever $\nabla \cdot \mathbf{B} = 0$, which is always true in electrodynamics.

Step 1: Substitute B = curl A into Ampere's law

$$\nabla \times (\nabla \times \mathbf{A}) = \mu_0 \mathbf{J}$$

Step 2: Expand using the BAC-CAB identity

The vector identity $\nabla \times (\nabla \times \mathbf{A}) = \nabla(\nabla \cdot \mathbf{A}) - \nabla^2\mathbf{A}$ gives:

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

Step 3: Exploit gauge freedom

The potential $\mathbf{A}$ is not unique: if $\mathbf{A}$ works, so does $\mathbf{A}' = \mathbf{A} + \nabla\lambda$ for any scalar $\lambda$, since $\nabla \times (\nabla\lambda) = 0$. Under this transformation:

$$\nabla \cdot \mathbf{A}' = \nabla \cdot \mathbf{A} + \nabla^2\lambda$$

Step 4: Choose the Coulomb gauge

We can always find a $\lambda$ such that $\nabla^2\lambda = -\nabla \cdot \mathbf{A}$ (a Poisson equation for $\lambda$, which always has a solution). This makes:

$$\boxed{\nabla \cdot \mathbf{A} = 0} \qquad \text{(Coulomb gauge)}$$

Step 5: Simplify Ampere's law in the Coulomb gauge

With $\nabla \cdot \mathbf{A} = 0$, the first term vanishes and we obtain:

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

Step 6: Solution by analogy with Poisson's equation for V

This is three uncoupled scalar Poisson equations ($\nabla^2 A_x = -\mu_0 J_x$, etc.). By direct analogy with $\nabla^2 V = -\rho/\epsilon_0$ whose solution is $V = \frac{1}{4\pi\epsilon_0}\int\frac{\rho}{\mathscr{r}}\,d\tau'$, the solution is:

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

Step 7: Verify the Coulomb gauge is satisfied

Taking the divergence of $\mathbf{A}$: $\nabla \cdot \mathbf{A} = \frac{\mu_0}{4\pi}\int \mathbf{J}(\mathbf{r}') \cdot \nabla\!\left(\frac{1}{\mathscr{r}}\right)d\tau'$. Integrating by parts and using $\nabla' \cdot \mathbf{J} = 0$ (steady currents), this vanishes, confirming self-consistency.

Step 1: Expand 1/|r - r'| for r >> r'

Using the standard multipole expansion of the separation distance:

$$\frac{1}{|\mathbf{r}-\mathbf{r}'|} = \frac{1}{r}\sum_{n=0}^{\infty}\left(\frac{r'}{r}\right)^n P_n(\cos\alpha)$$

where $\alpha$ is the angle between $\mathbf{r}$ and $\mathbf{r}'$, and $P_n$ are Legendre polynomials.

Step 2: Write A as a multipole series (for a line current)

For a current loop, $\mathbf{A} = \frac{\mu_0 I}{4\pi}\oint \frac{d\boldsymbol{\ell}'}{|\mathbf{r}-\mathbf{r}'|}$. Substituting the expansion:

$$\mathbf{A} = \frac{\mu_0 I}{4\pi}\sum_{n=0}^{\infty}\frac{1}{r^{n+1}}\oint (r')^n P_n(\cos\alpha)\,d\boldsymbol{\ell}'$$

Step 3: The monopole term (n = 0) vanishes

For $n = 0$: $P_0 = 1$, so the integral is $\oint d\boldsymbol{\ell}' = 0$ for any closed loop. There is no magnetic monopole contribution to $\mathbf{A}$.

Step 4: The dipole term (n = 1)

For $n = 1$: $P_1(\cos\alpha) = \cos\alpha = \hat{r}\cdot\hat{r}' = \hat{r}\cdot\mathbf{r}'/r'$. The integral becomes:

$$\mathbf{A}_{\text{dip}} = \frac{\mu_0 I}{4\pi r^2}\oint (\hat{r}\cdot\mathbf{r}')\,d\boldsymbol{\ell}'$$

Step 5: Apply the vector identity for closed loop integrals

There is a useful identity: for any closed loop, $\oint(\mathbf{c}\cdot\mathbf{r}')d\boldsymbol{\ell}' = -\mathbf{c} \times \int d\mathbf{a}' = \mathbf{a} \times \mathbf{c}$, where $\mathbf{a} = \int d\mathbf{a}'$ is the vector area of the loop. With $\mathbf{c} = \hat{r}$:

$$\oint(\hat{r}\cdot\mathbf{r}')d\boldsymbol{\ell}' = -\hat{r} \times \int d\mathbf{a}'$$

Step 6: Identify the magnetic dipole moment

Defining $\mathbf{m} = I\int d\mathbf{a}' = I\mathbf{a}$ (current times the vector area of the loop):

$$\mathbf{A}_{\text{dip}} = \frac{\mu_0 I}{4\pi r^2}(-\hat{r} \times \mathbf{a}) = \frac{\mu_0}{4\pi}\frac{\mathbf{m} \times \hat{r}}{r^2}$$

Step 7: Final result

$$\boxed{\mathbf{A}_{\text{dip}}(\mathbf{r}) = \frac{\mu_0}{4\pi}\frac{\mathbf{m} \times \hat{r}}{r^2}}$$

In spherical coordinates with $\mathbf{m} = m\hat{z}$: $\mathbf{A}_{\text{dip}} = \frac{\mu_0 m \sin\theta}{4\pi r^2}\hat{\phi}$.

Step 1: Flat current loop

For a planar loop carrying current $I$ enclosing area $\mathbf{a}$ (vector area with direction given by the right-hand rule):

$$\mathbf{m} = I\mathbf{a} = IA\,\hat{n}$$

Step 2: Generalize to volume currents

Replace $I\,d\boldsymbol{\ell}' \to \mathbf{J}\,d\tau'$. The dipole term in the multipole expansion of $\mathbf{A}$ for a volume current is:

$$\mathbf{A}_{\text{dip}} = \frac{\mu_0}{4\pi r^2}\int (\hat{r}\cdot\mathbf{r}')\,\mathbf{J}(\mathbf{r}')\,d\tau'$$

Step 3: Use the vector identity for volume integrals

For a steady current distribution ($\nabla \cdot \mathbf{J} = 0$), one can show that:

$$\int (\mathbf{c}\cdot\mathbf{r}')\mathbf{J}\,d\tau' = -\frac{1}{2}\mathbf{c}\times\int(\mathbf{r}'\times\mathbf{J})\,d\tau'$$

for any constant vector $\mathbf{c}$. This identity requires integration by parts and the continuity equation.

Step 4: Substitute with c = r-hat

Setting $\mathbf{c} = \hat{r}$:

$$\mathbf{A}_{\text{dip}} = \frac{\mu_0}{4\pi r^2}\left(-\frac{1}{2}\hat{r}\times\int \mathbf{r}'\times\mathbf{J}\,d\tau'\right) = \frac{\mu_0}{4\pi}\frac{\mathbf{m}\times\hat{r}}{r^2}$$

Step 5: Identify the general magnetic dipole moment

$$\boxed{\mathbf{m} = \frac{1}{2}\int \mathbf{r}' \times \mathbf{J}(\mathbf{r}')\,d\tau'}$$

Step 6: Verify consistency with the loop formula

For a thin wire carrying current $I$, replace $\mathbf{J}\,d\tau' \to I\,d\boldsymbol{\ell}'$:

$$\mathbf{m} = \frac{I}{2}\oint \mathbf{r}' \times d\boldsymbol{\ell}' = I\mathbf{a}$$

since $\frac{1}{2}\oint \mathbf{r}' \times d\boldsymbol{\ell}'$ is precisely the vector area enclosed by the loop.

Step 1: Torque on a rectangular current loop in a uniform field

Consider a rectangular loop (sides $a$ and $b$) carrying current $I$ in a uniform field $\mathbf{B}$. The forces on opposite sides are equal and opposite, but they create a couple. For sides of length $a$ perpendicular to $\mathbf{B}$, each experiences force $F = IaB$. The lever arm is $(b/2)\sin\theta$ on each side, so:

$$N = 2 \cdot IaB \cdot \frac{b}{2}\sin\theta = Iab\,B\sin\theta = mB\sin\theta$$

Step 2: Write the torque in vector form

The torque tends to align $\mathbf{m}$ with $\mathbf{B}$. In vector notation:

$$\boxed{\mathbf{N} = \mathbf{m} \times \mathbf{B}}$$

This result generalizes to any shape of loop and any orientation.

Step 3: Energy of a dipole in an external field

The work done rotating the dipole from angle $\theta_0$ to $\theta$ against the torque gives the potential energy:

$$U = -\int N\,d\theta = -mB\cos\theta = -\mathbf{m}\cdot\mathbf{B}$$

Step 4: Force from a non-uniform field

In a non-uniform field, the force is the negative gradient of the potential energy. For a rigid dipole (fixed $\mathbf{m}$):

$$\mathbf{F} = -\nabla U = \nabla(\mathbf{m}\cdot\mathbf{B})$$

Step 5: Expand using the vector identity

Using $\nabla(\mathbf{m}\cdot\mathbf{B}) = (\mathbf{m}\cdot\nabla)\mathbf{B} + \mathbf{m}\times(\nabla\times\mathbf{B}) + \cdots$. Since $\mathbf{m}$ is constant and in free space $\nabla\times\mathbf{B} = 0$ (away from free currents):

$$\mathbf{F} = (\mathbf{m}\cdot\nabla)\mathbf{B}$$

Step 6: Component form

For a dipole $\mathbf{m} = m\hat{z}$ in an axially symmetric field:

$$F_z = m\frac{\partial B_z}{\partial z}$$

$$\boxed{\mathbf{F} = \nabla(\mathbf{m}\cdot\mathbf{B})}$$

The dipole is attracted toward regions of stronger field if $\mathbf{m}$ is parallel to $\mathbf{B}$ (paramagnets), and repelled if antiparallel (diamagnets).