Magnetic vector potential

The magnetic vector potential A\mathbf{A} is a vector potential function whose Curl is the magnetic field B\mathbf{B}:

∇×A=B\nabla\times\mathbf{A}=\mathbf{B}

Divergence and curl

The curl of the magnetic vector potential is obvious: it's just the definition. The Divergence is more interesting. A vector potential remains such even after another function with no curl is added onto it

A~=A+Fwhere∇×F=0\tilde{\mathbf{A}}=\mathbf{A}+\mathbf{F}\quad\text{where}\quad \nabla\times\mathbf{F}=0

but a function whose curl is zero is always the Gradient of a scalar function

F=∇V\mathbf{F}=\nabla V

This allows us to state that the divergence of the vector potential is always zero:

∇⋅A=0\nabla\cdot\mathbf{A}=0

(same as the magnetic field). To rigorously prove this is true, suppose our original potential, A0\mathbf{A}_{0}, is not divergenceless. If we add to it the gradient of some function GG we get the new divergence

∇⋅A=∇⋅A0+∇2V\nabla\cdot\mathbf{A}=\nabla\cdot\mathbf{A}_{0}+\nabla ^{2}V

For this to be zero, we must have ∇2V=−∇⋅A0\nabla ^{2}V=-\nabla\cdot\mathbf{A}_{0}. This is just Poisson's equation (if a bit foreign-looking) with f=−∇⋅A0f=-\nabla\cdot\mathbf{A}_{0}. If ∇⋅A0\nabla\cdot\mathbf{A}_{0} goes to zero at infinity, solution then is

V=14π∫V∇⋅A0r dτ′V=\frac{1}{4\pi}\int_{\mathcal{V}} \frac{\nabla\cdot\mathbf{A}_{0}}{\mathfrak{r}}\ d\tau'

which is the exact function that makes the ∇2V=−∇⋅A0\nabla ^{2}V=-\nabla\cdot\mathbf{A}_{0} or, in other words, makes the divergence of A\mathbf{A} always zero. Since VV obeys Poisson's equation, and we can always solve Poisson's equation in some way or another, we can always find a VV for which the divergence of A\mathbf{A} is zero. In fact, we could pick a VV for which ∇⋅A\nabla\cdot\mathbf{A} is nonzero, but it's typically just plain easier to set ∇⋅A=0\nabla\cdot\mathbf{A}=0 and pretend it doesn't exist, so we might as well.

This gives us permission to express Ampere's law like this:

∇×B=∇×(∇×A)=∇(∇⋅A)−∇2A=μ0J\nabla\times\mathbf{B}=\nabla\times(\mathbf{\nabla\times\mathbf{A}})=\nabla(\cancel{ \nabla\cdot\mathbf{A} })-\nabla ^{2}\mathbf{A}=\mu_{0}\mathbf{J}

and so

∇2A=−μ0J\boxed{\nabla ^{2}\mathbf{A}=-\mu_{0}\mathbf{J}}

These are just three Poisson's equation, one for each coordinate, which we know how to solve of J\mathbf{J} goes to zero at infinity. We get

A(r)=μ04π∫I(r′)r dl′,A(r)=μ04π∫K(r)′r da′,A(r)=μ04π∫J(r′)r dτ′\mathbf{A}(\mathbf{r})=\frac{\mu_{0}}{4\pi}\int \frac{\mathbf{I}(\mathbf{r}')}{\mathfrak{r}}\ dl',\quad \mathbf{A}(\mathbf{r})=\frac{\mu_{0}}{4\pi}\int \frac{\mathbf{K}(\mathbf{r})'}{\mathfrak{r}}\ da',\quad \mathbf{A}(\mathbf{r})=\frac{\mu_{0}}{4\pi}\int \frac{\mathbf{J}(\mathbf{r}')}{\mathfrak{r}}\ d\tau'

for line, surface and volume current densities.

Boundary conditions

Unlike the magnetic field, the vector potential does not exhibit a discontinuity across a surface current density K\mathbf{K}. In fact, the lack of divergence implies that the normal component of the vector is continuous, whereas the curl

∮A⋅dr=∫B⋅da=Φ\oint \mathbf{A}\cdot d\mathbf{r}=\int \mathbf{B}\cdot d\mathbf{a}=\Phi

tells us that the tangential components are also continuous because the Amperian loop that we're using can be shrunk to be arbitrarily small, therefore continuously reducing the flux through the surface until it is zero. Therefore we can state

Aabove=Abelow\mathbf{A}_\text{above}=\mathbf{A}_\text{below}

However, the (normal) derivative of A\mathbf{A} does inherit B\mathbf{B}'s discontinuity:

∂Aabove∂n−∂Abelow∂n=−μ0K\frac{ \partial \mathbf{A}_\text{above} }{ \partial n } -\frac{ \partial \mathbf{A}_\text{below} }{ \partial n } =-\mu_{0}\mathbf{K}

Multipole expansion

The vector potential can be expanded in a multipole series in the same way the electric potential can. Given the position vector r\mathbf{r}, the source vector r′\mathbf{r}' and the angle α\alpha between the two, the vector potential of a current loop can be written as

A(r)=μ0I4π∮1rdr′=μ0I4π∑n=0∞1rn+1∮(r′)nPn(cos⁡α)dr′\mathbf{A}(\mathbf{r})=\frac{\mu_{0}I}{4\pi}\oint \frac{1}{\mathfrak{r}}d\mathbf{r}'=\frac{\mu_{0}I}{4\pi}\sum_{n=0}^{\infty} \frac{1}{r^{n+1}}\oint(r')^{n}P_{n}(\cos \alpha)d\mathbf{r}'

where P(cos⁡α)P(\cos \alpha) are the Legendre polynomials. Notably, the monopole term is just the displacement around a closed loop, which is always zero:

∮dr=0\oint d\mathbf{r}=0

which is an important confirmation of the fact that magnetic monopoles are not known to exist in nature and, beyond that, Maxwell's equation ∇⋅B=0\nabla\cdot\mathbf{B}=0 implicitly requires this to be so. Since the monopole term vanishes, the smallest terms is the dipole one:

Adip(r)=μ0I4π∮r′cos⁡α dr′=μ0I4π∮(r^⋅r′)dr′\mathbf{A}_\text{dip}(\mathbf{r})=\frac{\mu_{0}I}{4\pi}\oint r'\cos \alpha \ d\mathbf{r}'=\frac{\mu_{0}I}{4\pi}\oint(\hat{\mathbf{r}}\cdot \mathbf{r}')d\mathbf{r}'

This can be written in a more illuminating way by seeing that

∮(r^⋅r′) dr′=−r^×∫da′\oint(\hat{\mathbf{r}}\cdot \mathbf{r}')\ d\mathbf{r}'=-\hat{\mathbf{r}}\times \int d\mathbf{a}'

and thus

Adip(r)=μ04πm×r′r2\boxed{\mathbf{A}_\text{dip}(\mathbf{r})=\frac{\mu_{0}}{4\pi} \frac{\mathbf{m}\times \mathbf{r}'}{r^{2}}}

where m\mathbf{m} is the magnetic dipole moment:

m=I∫da=Ia\mathbf{m}=I\int d\mathbf{a}=I\mathbf{a}

Unlike the electric dipole moment, this one is completely independent on the choice of origin. This is due to the monopole moment being zero all the time. The magnetic field of a pure dipole (with infinitesimal area) is

Bdip(r)=∇×A=μ0m4πr3(2cos⁡θr^+sin⁡θϕ^)\mathbf{B}_\text{dip}(\mathbf{r})=\nabla\times\mathbf{A}=\frac{\mu_{0}m}{4\pi r^{3}}(2\cos \theta \hat{\mathbf{r}}+\sin \theta \hat{\boldsymbol{\phi}})

in Cartesian coordinates. It can be rewritten in coordinate-free form as

Bdip(r)=μ04π1r3[3(m⋅r^)r^−m]\boxed{\mathbf{B}_\text{dip}(\mathbf{r})=\frac{\mu_{0}}{4\pi} \frac{1}{r^{3}}[3(\mathbf{m}\cdot \hat{\mathbf{r}})\hat{\mathbf{r}}-\mathbf{m}]}

Energy of a system

Making a current run through a wire is not free. It requires pushing (i.e. doing work) against the back emf induced by the changing current. This work is, unlike the heat expelled by the Joule effect, recoverable. When the current is removed from the wire, the back emf will do the same amount of work it did originally, just in the opposite direction. In the meantime, you can think of it as being "stored" within the wire.

The work done per unit charge against the back emf of value E\mathcal{E} is simply −E-\mathcal{E}. We also know the current II, which is the amount of charge passing through the wire per unit time. The work done per unit time then is

dWdt=−EI=LIdIdt\frac{dW}{dt}=-\mathcal{E}I=LI \frac{dI}{dt}

where LL is the self inductance of the circuit. Integrating over all time we get

W=12LI2\boxed{W=\frac{1}{2}LI^{2}}

Notably, it is independent on time. It does not matter how fast or slow we change the current. The amount of work we need to do is predetermined by the properties of wire: geometry for LL, desired current intensity for II.

This formula is clean, but can be rewritten to be more informative. The magnetic flux Φ\Phi going through a loop is LILI, but also

LI=Φ=∫B⋅da=∫(∇×A)⋅da=∮A⋅dILI=\Phi=\int \mathbf{B}\cdot d\mathbf{a}=\int(\nabla\times\mathbf{A})\cdot d\mathbf{a}=\oint \mathbf{A}\cdot d\mathbf{I}

Substituting this into the work formula, we get

W=12I∮(A⋅I) dlW=\frac{1}{2}I\oint (\mathbf{A}\cdot I)\ dl

Generalizing to volume currents is pretty easy now:

W=12∫V(A⋅J) dτ\boxed{W=\frac{1}{2}\int_{\mathcal{V}}(\mathbf{A}\cdot \mathbf{J})\ d\tau}

But we know how to express J\mathbf{J} in terms of the magnetic field using Ampere's law:

∇×B=μ0J\nabla\times\mathbf{B}=\mu_{0}\mathbf{J}

so

W=12μ0∮VA⋅(∇×B) dτW=\frac{1}{2\mu_{0}}\oint_{\mathcal{V}}\mathbf{A}\cdot(\nabla\times\mathbf{B})\ d\tau

We can exploit integration by parts to transfer the derivative to A\mathbf{A} and then use this product rule

∇⋅(A×B)=B⋅(∇×A)−A⋅(∇×B)\nabla \cdot(\mathbf{A}\times \mathbf{B})=\mathbf{B}\cdot(\nabla\times\mathbf{A})-\mathbf{A}\cdot(\nabla\times\mathbf{B})

and reversed

A⋅(∇×B)=B⋅B−∇⋅(A×B)\mathbf{A}\cdot(\nabla\times\mathbf{B})=\mathbf{B}\cdot \mathbf{B}-\nabla \cdot(\mathbf{A}\times \mathbf{B})

to find

W=12μ0[∫B2 dτ−∫∇⋅(A×B) dτ]==12μ0[∫VB2 dτ−∮S(A×B)⋅da]\begin{align} W&=\frac{1}{2\mu_{0}}\left[ \int B^{2}\ d\tau-\int \nabla \cdot(\mathbf{A}\times \mathbf{B})\ d\tau \right]= \\ & =\frac{1}{2\mu_{0}}\left[ \int_{\mathcal{V}} B^{2}\ d\tau-\oint_{\mathcal{S}} (\mathbf{A}\times \mathbf{B})\cdot d\mathbf{a} \right] \end{align}

where S\mathcal{S} is the bounding surface of V\mathcal{V}. V\mathcal{V} is the volume where the current is running through, but we might as well extend this to all space as any regione where the current is zero won't contribute anything anyway. As we increase the size of the volume, the volume integral gets larger and the surface integral gets smaller (we are making the surface further and further away from both A\mathbf{A} and B\mathbf{B} so it makes sense). In the limiting case of integration over all space, the volume integral is the only contribution and the surface integral vanishes, leaving us with

W=12μ0∫all spaceB2 dτ\boxed{W=\frac{1}{2\mu_{0}}\int _\text{all space}B^{2}\ d\tau}

Like many things in magnetism, this is a near carbon copy of the equivalent counterpart, the electric field energy, just with 1/μ01/\mu_{0} instead of ε0\varepsilon_{0} and BB instead of EE. Just like it, where the energy is located is beyond the scope of this formula. One could argue it's located in the field (as QFT wants) or that it's located in the current (as common sense might have it). What matters is the actual energy itself, which correctly given by any of the above formulas.