1 Introduction

For electrostatic fields, we found from Coulomb’s law that there is a scalar function such that $\mathbf{E}=-\boldsymbol{\nabla}V$. For charge distributions that are finite in extent (e.g., not an infinite line or plane), we can compute $\mathbf{E}$ by first computing

$$\displaystyle V(\mathbf{r})=\frac{1}{4\pi\epsilon_o}\left(\int \frac{\rho(\mathbf{r}')}{\char"0509}d\tau' + \int \frac{\sigma(\mathbf{r}')}{\char"0509}dA' + \int \frac{\lambda(\mathbf{r}')}{\char"0509}dl'\right)$$

and then use $\mathbf{E}=-\boldsymbol{\nabla}V$. (In most problems, only one of the charge densities in the above equation was not zero; the equation above is the most general equation.) Given that the integrands are scalars, it is often easier to use this approach rather than computing $\mathbf{E}$ directly using Coulomb’s Law:

$$\displaystyle \mathbf{E}(\mathbf{r})=\frac{1}{4\pi\epsilon_o}\left(\int \frac{\hat{\textbf{\char"0509}}\phantom{^2}}{\char"0509^2}\rho(\mathbf{r}')d\tau' + \int \frac{\hat{\textbf{\char"0509}}\phantom{^2}}{\char"0509^2}\sigma(\mathbf{r}')dA' + \int \frac{\hat{\textbf{\char"0509}}\phantom{^2}}{\char"0509^2}\lambda(\mathbf{r}')dl'\right)$$

The reason that we can find $V$ for which $\mathbf{E}=-\boldsymbol{\nabla}V$ is electrostatic fields (field due to non–moving charges) have the property that $\boldsymbol{\nabla}\times \mathbf{E}=0$.

We can always compute $\mathbf{B}$ using the Biot–Savart law:

$$\displaystyle \mathbf{B}(\mathbf{r})=\frac{\mu_o}{4\pi}\left(\int d\mathbf{l}'\times\frac{\hat{\textbf{\char"0509}}\phantom{^2}}{\char"0509^2}J(\mathbf{r}')d\tau' + \int d\mathbf{l}'\times\frac{\hat{\textbf{\char"0509}}\phantom{^2}}{\char"0509^2}K(\mathbf{r}')dA' + \int d\mathbf{l}'\times\frac{\hat{\textbf{\char"0509}}\phantom{^2}}{\char"0509^2}I(\mathbf{r}')dl'\right)$$

(I’ve written the integrands in a slightly different way so that it looks more like the integrands for $\mathbf{E}$.)

From the Biot–Savart law, it can be shown that $\boldsymbol{\nabla}\times \mathbf{B}=\mu_o\mathbf{J}$. Because $\boldsymbol{\nabla}\times \mathbf{B}\ne 0$, we cannot in general find a scalar function $V_m$ such that $\mathbf{B}=-\nabla V_m$. ($V_m$ has some use; it can be used for solving boundary value problems in current–free regions and also when fitting measurements from current–free regions.)

However, we can find a vector function $\mathbf{A}$ for which $\mathbf{B}=\boldsymbol{\nabla}\times \mathbf{A}$. This vector is called the “vector potential,” and it applies to magnetic fields only.

A natural question is why we introduce this vector potential (see also 15-4 in the Feynman Lectures on Physics). One of the reasons that $V$ is useful is that it is a scalar, and so instead of discussing the vector function $\mathbf{E}$, we can discuss the scalar function $V$. Also, $V$ has a physical interpretation and use – for example, it can be used to find the work required to assemble a charge distribution. Neither of these is true for the vector potential $\mathbf{A}$. There are, however, several uses for $\mathbf{A}$. One is that the equations for finding it look similar to that for $V$:

$$\displaystyle \mathbf{A}(\mathbf{r})=\frac{\mu_o}{4\pi}\left(\int \frac{\mathbf{J}(\mathbf{r}')}{\char"0509}d\tau' + \int \frac{\mathbf{K}(\mathbf{r}')}{\char"0509}da' + \int \frac{\mathbf{I}(\mathbf{r}')}{\char"0509}dl'\right).$$

As a result, some of the techniques used in computing $V$ can be used in computing $\mathbf{A}$. In addition, some of the integrals found for $V$ can be re–used when they appear in integrals required to compute $\mathbf{A}$. Recall that the expansion of $1/\char"0509$ was used to find the multipole expansion of $V$ in Chapter 3.4 of Griffiths (so as to avoid having to integrate to find $V$ at points for which $r>r'$). The integrals for $\mathbf{A}$ have a $1/\char"0509$, and so we can use the multipole expansion to find an equation for $\mathbf{A}$ that allows us to avoid having to integrate to find $\mathbf{A}$ (and also $\mathbf{B}$ because it can be computed from $\mathbf{A}$).

A second reason that $\mathbf{A}$ is useful is that we can write $\nabla^2\mathbf{A}=-\mu_o\mathbf{J}$, which is the analog to the Poisson equation $\nabla^2V=-\rho/\epsilon_o$. Sometimes we can use the boundary value techniques for solving Poisson’s equation for solving $\nabla^2\mathbf{A}=-\mu_o\mathbf{J}$ (which is actually three equations, each which have the form of Poisson’s equation: $\nabla^2A_x=-\mu_oJ_x$, $\nabla^2A_y=-\mu_oJ_y$, $\nabla^2A_z=-\mu_oJ_z$.)

Finally, the vector potential appears as a natural quantity in the Lagrangian in classical mechanics and in many equations in quantum mechanics. This is discussed in detail in 15-4 in the Feynman Lectures on Physics.

2 Computing $\mathbf{B}$ given $\mathbf{A}$

Computing $\mathbf{B}$ given $\mathbf{A}$ requires only the evaluation of the curl of $\mathbf{A}$.

2.1 Example – $\mathbf{A}\sim \sin\theta \hat{\boldsymbol{\phi}}/{r^2}$

If $\displaystyle\mathbf{A}=\alpha \frac{\sin\theta \hat{\boldsymbol{\phi}}}{r^2}$, where $\alpha$ is a constant, find $\mathbf{B}$

Answer

Prior to computing the curl, it is useful to write out the scalar compoents of the vector to be curled, which are $A_r=0$, $A_\theta=0$, and $A_\phi=\alpha \sin\theta/r^2$.

The curl of a vector $\mathbf{A}$ in spherical coordinates is

$\boldsymbol{\nabla}\times \mathbf{A} = {\frac {1}{r\sin \theta }}\left({\frac {\partial }{\partial \theta }}\left(A_{\phi }\sin \theta \right) - {\frac {\partial A_{\theta }}{\partial \phi }}\right) {\hat {\mathbf {r} }} + {\frac {1}{r}}\left({\frac {1}{\sin \theta }}{\frac {\partial A_{r}}{\partial \phi }} - {\frac {\partial }{\partial r}}\left(rA_{\phi }\right)\right) {\hat {\boldsymbol {\theta }}} + {\frac {1}{r}}\left({\frac {\partial }{\partial r}}\left(rA_{\theta }\right) - {\frac {\partial A_{r}}{\partial \theta }}\right) {\hat {\boldsymbol {\phi }}}$

Of the six terms involving partial derivatives, only two are non–zero.

$\boldsymbol{\nabla}\times \mathbf{A} = {\frac {1}{r\sin \theta }}\left({\frac {\partial }{\partial \theta }}\left(A_{\phi }\sin \theta \right) - 0\right) {\hat {\mathbf {r} }} + {\frac {1}{r}}\left(0 - {\frac {\partial }{\partial r}}\left(rA_{\phi }\right)\right) {\hat {\boldsymbol {\theta }}} + {\frac {1}{r}}\left(0 - 0\right){\hat {\boldsymbol {\phi }}}$

Evaluation gives

$\mathbf{B}=\displaystyle\boldsymbol{\nabla}\times \mathbf{A} = \frac{\alpha}{r^3}\left(2\cos\theta{\hat {\mathbf {r} }}+\sin\theta\hat{\boldsymbol{\theta}}\right)$

This $\mathbf{B}$ has the same form as a magnetic dipole in spherical coordinates and unit vectors; we can conclude that this vector potential is that for a magnetic dipole for which the current system is a small loop of area $A$ centered on the origin in the $x$–$y$ plane if $\alpha = \mu_oIA/4\pi$.

2.2 Problem – $\mathbf{A}\sim \hat{\boldsymbol{\phi}}$

If $\mathbf{A}= \alpha\hat{\boldsymbol{\phi}}$, where $\alpha$ is a constant, what is $\mathbf{B}$?

Answer

The curl of a vector $\mathbf{A}$ in spherical coordinates is

$\displaystyle\boldsymbol{\nabla}\times \mathbf{A} = {\frac {1}{r\sin \theta }}\left({\frac {\partial }{\partial \theta }}\left(A_{\phi }\sin \theta \right) - {\frac {\partial A_{\theta }}{\partial \phi }}\right) {\hat {\mathbf {r} }} + {\frac {1}{r}}\left({\frac {1}{\sin \theta }}{\frac {\partial A_{r}}{\partial \phi }} - {\frac {\partial }{\partial r}}\left(rA_{\phi }\right)\right) {\hat {\boldsymbol {\theta }}} + {\frac {1}{r}}\left({\frac {\partial }{\partial r}}\left(rA_{\theta }\right) - {\frac {\partial A_{r}}{\partial \theta }}\right) {\hat {\boldsymbol {\phi }}}$

In this problem, $A_r=A_\theta=0$ and $A_\phi=\alpha$.

Of the six terms involving partial derivatives, only two are non–zero, so

$\displaystyle\boldsymbol{\nabla}\times \mathbf{A} = {\frac {1}{r\sin \theta }}\left({\frac {\partial }{\partial \theta }}\left(A_{\phi }\sin \theta \right) - 0\right) {\hat {\mathbf {r} }} + {\frac {1}{r}}\left(0 - {\frac {\partial }{\partial r}}\left(rA_{\phi }\right)\right) {\hat {\boldsymbol {\theta }}} + {\frac {1}{r}}\left( 0 - 0\right) {\hat {\boldsymbol {\phi }}}$.

Evaluation gives

$\displaystyle\mathbf{B}= \frac{\alpha}{r\tan\theta}\hat {\mathbf {r} } - \frac{\alpha}{r}\hat{\boldsymbol {\theta }}$

One could also do the same calculation in cylindrical coordinates using

$\displaystyle\boldsymbol{\nabla}\times \mathbf{A} = \left({\frac {1}{s }}{\frac {\partial A_{z}}{\partial \phi }}-{\frac {\partial A_{\phi }}{\partial z}}\right) {\hat {\boldsymbol {s }}} + \left({\frac {\partial A_{s}}{\partial z}}-{\frac {\partial A_{z}}{\partial s}}\right) {\hat {\boldsymbol {\phi }}} + {\frac {1}{s}}\left({\frac {\partial \left(s A_{\phi }\right)}{\partial s}}-{\frac {\partial A_{s}}{\partial \phi }}\right) {\hat {\mathbf {z} }}$

In this case,

$\displaystyle\mathbf{B}=\frac{\alpha}{s}\mathbf{\hat{z}}$.

Using $\mathbf{\hat{z}} = \cos\theta \hat {\mathbf {r} } - \sin\theta \hat {\boldsymbol {\theta} }$ and $s=r\sin\theta$, this is

$\displaystyle\mathbf{B}=\frac{\alpha}{r\sin\theta}\left(\cos\theta \hat {\mathbf {r} } - \sin\theta \hat {\boldsymbol {\theta} }\right)=\frac{\alpha}{r\tan\theta}\hat {\mathbf {r} } - \frac{\alpha}{r}\hat{\boldsymbol {\theta }}$,

which is the same as that found using the curl in spherical coordinates.

2.3 Problem – $\mathbf{A}\sim |y|\mathbf{\hat{z}}$

If $\mathbf{A} = -\alpha y \mathbf{\hat{x}}$ for $y\gt 0$ and $\mathbf{A} = +\alpha y \mathbf{\hat{x}}$ for $y\lt 0$, where $\alpha$ is a constant,

1. find $\mathbf{B}$ and

2. plot $B_x(y)$.

3. Based on the plot for 2., what is the current system that produced this $\mathbf{B}$?

Answer

$\boldsymbol{\nabla}\times \mathbf{A} = \left(\frac{\partial A_z }{\partial y} - \frac{\partial A_y }{\partial z} \right)\mathbf{\hat{x}} + \left(\frac{\partial A_x }{\partial z} - \frac{\partial A_z }{\partial x} \right)\mathbf{\hat{y}} + \left(\frac{\partial A_y }{\partial x} - \frac{\partial A_x }{\partial y} \right)\mathbf{\hat{z}}$

$A_x=0$, $A_y=0$, and $A_z=\mp \alpha y$

Inserting zeros for terms involving partial derivative of $A_x$ and $A_y$ gives

$\boldsymbol{\nabla}\times \mathbf{A} = \left(\frac{\partial A_z }{\partial y} - 0 \right)\mathbf{\hat{x}} + \left(0 - \frac{\partial A_z }{\partial x} \right)\mathbf{\hat{y}} + \left(0 + 0\right)\mathbf{\hat{z}}$

Evaluation gives

$\mathbf{B} = \boldsymbol{\nabla}\times \mathbf{A} = \mp\alpha\mathbf{\hat{x}}$

This is the field of a sheet of current in the $x$–$y$ plane with current flowing in the $+\mathbf{\hat{z}}$ direction. (To see this, draw out the bound current sheets and see HW #11.1.

2.4 Dipole Vector Potential

The general equation for the vector potential for a magnetic dipole is

$$\mathbf{A}=\frac{\mu_o}{4\pi}\frac{\mathbf{m}\times\hat{\mathbf{r}}}{r^2}$$

If $\mathbf{m}=m_o\mathbf{\hat{x}}$, find $\mathbf{B}$ in cartesian coordinates with cartesian unit vectors. Note that you can check your answer using equation 5.89 of Griffiths 4th edition:

$$\mathbf{B} = \frac{\mu_o}{4\pi}\frac{1}{r^3}\left[3(\mathbf{m}\boldsymbol{\cdot}\hat{\mathbf{r}})\hat{\mathbf{r}} - \mathbf{m}\right]$$

Answer

Method 1

In Magnetic Dipoles, the equation for a with dipole $\mathbf{m}=I\pi b^2\mathbf{\hat{z}}$ was given as

$\displaystyle\mathbf{B}=\frac{\mu_o}{4\pi}\frac{I\pi b^2}{r^5}\left(3xz\hat{\mathbf{x}}+3yz\hat{\mathbf{y}}+(3z^2-r^2)\hat{\mathbf{z}}\right)$

Swapping $z$ and $x$ and replacing $I\pi b^2$ with $m_o$ gives the answer to this problem:

$\displaystyle\mathbf{B}=\frac{\mu_o}{4\pi}\frac{m_o}{r^5}\left[(3x^2-r^2)\hat{\mathbf{x}}+3xy\hat{\mathbf{y}}+3xz\hat{\mathbf{z}}\right]$

Method 2

Using $\mathbf{\hat{r}}=\mathbf{r}/r$ and $\mathbf{r}=x\mathbf{\hat{x}}+y\mathbf{\hat{y}}+z\mathbf{\hat{z}}$,

$\displaystyle \mathbf{A}=\frac{\mu_om_o}{4\pi}\frac{\mathbf{\hat{x}}\times \mathbf{\hat{r}}}{r^3}=\frac{\mu_om_o}{4\pi}\frac{\mathbf{\hat{x}}\times (x\mathbf{\hat{x}}+y\mathbf{\hat{y}}+z\mathbf{\hat{z}})}{r^3}=\frac{\mu_om_o}{4\pi}\left(\frac{-z\mathbf{\hat{y}}+y\mathbf{\hat{z}}}{r^3}\right)$

Written fully in cartesian coordinates, this is

$\displaystyle \mathbf{A}=\frac{\mu_om_o}{4\pi}\left(\frac{-z\mathbf{\hat{y}}+y\mathbf{\hat{z}}}{\left(x^2+y^2+z^2\right)^{3/2}}\right)$

Using this and

$\displaystyle\boldsymbol{\nabla}\times \mathbf{A} = \left(\frac{\partial A_z }{\partial y} - \frac{\partial A_y }{\partial z} \right)\mathbf{\hat{x}} + \left(\frac{\partial A_x }{\partial z} - \frac{\partial A_z }{\partial x} \right)\mathbf{\hat{y}} + \left(\frac{\partial A_y }{\partial x} - \frac{\partial A_x }{\partial y} \right)\mathbf{\hat{z}}$

will give the same answer as given above.

3 Computing $\mathbf{A}$

There are two general types of problems related to computing $\mathbf{A}$

1. if $\mathbf{B}$ is known, and

2. if $\mathbf{B}$ is not known

If $\mathbf{B}$ is known, we can sometimes (a.) find $\mathbf{A}$ by guessing what $\mathbf{A}$ will give $\mathbf{B}=\boldsymbol{\nabla}\times \mathbf{A}$, (b.) use $\mathbf{B}=\boldsymbol{\nabla}\times \mathbf{A}$, and (c.) and additional information, or using $\oint \mathbf{A}\boldsymbol{\cdot} d\mathbf{l}=\int \mathbf{B}\boldsymbol{\cdot} d\mathbf{l}$. As an example of (c.), see Example 5.12 of Griffiths.

If $\mathbf{B}$ is not known, we can always use direct integration to find $\mathbf{A}$. Often the integral encountered will be one that was previously used to find $V$ using direct integration. Most generally, the integral will not have a closed–form solution, and we will write the integrant of $\mathbf{A}$ as a power series as was done in the monopole expansion for $V$.

3.1 Example – Infinitely Long and Straight Wire

Suppose that we know $\mathbf{B}$ for an infinitely long wire along the $z$–axis with $\mathbf{I}=I\mathbf{\hat{z}}$. It can be argued using symmetry that for a certain choice of integration path for the line integral that

$\displaystyle\oint \mathbf{A}\boldsymbol{\cdot} d\mathbf{l}=\int \mathbf{B}\boldsymbol{\cdot} d\mathbf{a}$

simplifies to

$\displaystyle A_z(s)L-A(s_o)L = L\int_{s_o}^{s} B_{\phi} ds'$

Using $\displaystyle B_\phi = \frac{\mu_o}{2\pi}\frac{I}{s}$ gives

$\displaystyle A_z(s)=A_z(s_o)-\frac{\mu_oI}{2\pi}\ln\frac{s}{s_o}$

Suppose $\mathbf{B}$ was not known. We can solve for $\mathbf{A}$ by analogy. For a finite wire along the $z$ axis that is centered on the origin and has a length $L$, the vector potential is

$\displaystyle \mathbf{A}=\int \frac{\mathbf{I}(\mathbf{r}')}{\char"0509}dl'$

Here, $dl'=dz'$ (the $dl'$ here is the differential length along the wire; in $\oint \mathbf{A}\boldsymbol{\cdot} d\mathbf{l}$ used above, $dl$ is a path of integration). Using $\mathbf{I}=I\mathbf{\hat{z}}$, we have

$\displaystyle A_z=\frac{\mu_o}{4\pi}\int_{-L/2}^{L/2} \frac{I}{\char"0509}dz'$

We have encountered an integral like this before. If, instead of a wire, we have a uniformly charged line of charge density $\lambda$, then

$\displaystyle V=\frac{1}{4\pi\epsilon_o}\int_{-L/2}^{L/2}\frac{\lambda}{\char"0509}dz'$

From this, we can conclude that we can take the answer for $V$ and replace $\lambda/\epsilon_o$ with $\mu_oI$ to get $A_z$.

For an infinitly long wire, we can’t actually use the above equation to find $V$. We can only compute the potential difference $V(s)-V(s_o)=-\int_{s_o}^{s}\mathbf{E}\boldsymbol{\cdot} d\mathbf{\hat{s}}$. For an infinite wire, $\mathbf{E}=(\lambda/2\pi\epsilon_o)\mathbf{\hat{s}}/s$ and integration gives $V(s)-V(s_o)=-(\lambda/2\pi\epsilon_o)\ln(s/s_o)$. Using the replacement of $\lambda/\epsilon_o$ with $\mu_oI$ gives

$\displaystyle A_z(s)=A_z(s_o)-\frac{\mu_oI}{2\pi}\ln\frac{s}{s_o}$

as found earlier.

3.2 Problem – $\mathbf{A}$ due to a finite wire segment

For a segment of wire that extends from $-L/2$ to $L/2$ on the $z$–axis with a current that flows in the $+z$ direction, find $\mathbf{A}(x)$.

Answer

In general,

$\displaystyle \mathbf{A}(\mathbf{r})=\frac{\mu_o}{4\pi}\int \frac{\mathbf{I}(\mathbf{r}')}{\char"0509}dl'$

For this problem,

$\displaystyle \mathbf{A}(x)=\frac{\mu_o}{4\pi}\int_{-L/2}^{L/2}\frac{I\mathbf{\hat{z}}}{\sqrt{x^2+z'^2}}dz'$

Let $u=xz'$, then

$\displaystyle\mathbf{A}=\mathbf{\hat{z}}\frac{\mu_oI}{4\pi}\frac{x}{|x|}\int_{-L/2x}^{L/2x} \frac{du}{\sqrt{1+u^2}}$

The integral can be evaluated using an integral table.

$\displaystyle\int_{u_f}^{u_o}\frac{du}{\sqrt{1+u^2}}=\tanh^{-1}\left(\frac{u}{\sqrt{1 + u^2}}\right)\Bigg|_{u_f}^{u_o}$

3.3 Example – Magnetic Dipole $\mathbf{A}$ Derivation

Starting with

$$\mathbf{A} = \frac{\mu_o}{4\pi}\int \frac{\mathbf{I}dl'}{\char"0509}$$

set up the integral that must be solved to find $\mathbf{A}(x,y,z)$ due to a loop with current $I$ of radius $b$ that lies in the $x–y$ plane and is centered on the origin.

Answer

$\mathbf{I}=I\hat{\boldsymbol{\phi}}$

$dl'= bd\phi'$

$\mathbf{r}'=b\hat{\mathbf{s}}$

$\mathbf{r}=x\hat{\mathbf{x}}+y\hat{\mathbf{y}}+z\hat{\mathbf{z}}$

To do the integration, we need to convert non-cartesian unit vectors to cartesian unit vectors.

$\mathbf{r}'=b\hat{\mathbf{s}}=b\cos\phi'\hat{\mathbf{x}}+b\sin\phi'\hat{\mathbf{y}}$

$\mathbf{I}=I(-\sin\phi'\hat{\mathbf{x}}+\cos\phi'\hat{\mathbf{y}})$

$\char"0509=|\mathbf{r}-\mathbf{r}'|=\sqrt{(x-b\cos\phi')^2+(y-b\sin\phi')^2+z^2}$

Substitution of these three equations into

$\displaystyle \mathbf{A} = \frac{\mu_o}{4\pi}\int \frac{\mathbf{I}dl'}{\char"0509}$

gives

$\displaystyle\mathbf{A} = \frac{\mu_oIb}{4\pi}\int_0^{2\pi} \frac{d\phi'(-\sin\phi'\hat{\mathbf{x}}+\cos\phi'\hat{\mathbf{y}})}{\sqrt{(x-b\cos\phi')^2+(y-b\sin\phi')^2+z^2}}$

This equation would be the answer to a “set up the integral problem”.

3.4 Problem

1. Use the binomial expansion on $1/\sqrt{(x-b\cos\phi')^2+(y-b\sin\phi')^2+z^2}$ to show that to leading order in $1/r$

$$\mathbf{A}(x,y,z) \simeq \frac{\mu_o}{4\pi}\frac{I\pi b^2}{r^2}\left(-\frac{y}{r}\hat{\mathbf{x}}+\frac{x}{r}\hat{\mathbf{y}}\right)$$

where $r^2=x^2+y^2+z^2$.

2. Use the above equation for $\mathbf{A}$ to compute $\mathbf{B}$.

3. What is $\boldsymbol{\nabla}\times\mathbf{B}$ at $r=2b$? (You can answer this without doing any partial derivatives.)