Magnetostatics

Now that we’ve spent a considerable amount of time covering the topic of electrostatics and the basic concepts of electric fields, we will now move to the topic of magnetostatics and the concepts of magnetic fields. Magnetostatics can informally be thought of as the magnetic analogue of electrostatics. Whereas in electrostatics we were interested in the behavior of electric fields due to static charges, in magnetostatics we’re interested in the behavior of magnetic fields due to static currents. We will approach the subject here by first defining the notion of electric current and current density, before proceeding to define the magnetic field through the empirical Biot-Savart law, as well as the empirical force law for magnetism, the Lorentz force law. We will then proceed to derive the field equations of magnetostatics and discuss their implications.

Current

If this were all there was to magnetism we wouldn’t have much more to say about the subject. However, it turns out that ferromagnets are not the only things that generate magnetic fields. It turns out electric currents also produce magnetic fields. Before discussing how this phenomenon arises though we first need to discuss the concept of electric current.

Definition

Up until now we’ve always assumed that charges were static. Now we will allow them to move around. This movement of charge is called electric current. By convention, if a set of positive charges are moving in some direction, we say there’s a current in that direction. Equivalently, if a set of negative charges are moving in some direction, we say there’s a current in the opposite direction. Informally, the current through a point is defined as the rate of change of charge through that point, or \[ I \equiv \frac{dq}{dt} \ . \] Current must have dimensions of charge per unit time. In Gaussian units, current is measured in units of esu per second, which is also called a statampere. In SI units, current is measured in units of Coulombs per second, also called an Ampere or an amp. The two units are related by \(1 \ \text{statamp} \approx 10^{-12} \ \text{A}\).

Current was perhaps first understood in the context of conducting wires. When a potential difference is created across a wire, for example by hooking the wire up to a battery, the charges inside the wire experience an electromotive force, causing the charges to move in the direction of lowest potential, thus creating an electric current through the wire.

If a wire has a line charge density \(\lambda\) and the charges inside the wire move with an average drift velocity \(\mathbf{v}\), the current \(I\) at any point in the wire is given by the product \[ I = \lambda |\mathbf{v}| \ . \] If we like, we can assign a direction to the current through the wire by making it a vector, \[ \mathbf{I} = \lambda \mathbf{v} \ . \] The current will always flow in the direction the positive charges are moving through the wire. Of course it’s really the negatively charged electrons that move through the wire, in the opposite direction of the current. This means if the drift velocity is in the \(\mathbf{v}\) direction, the electrons are really moving through the wire in the \(-\mathbf{v}\) direction. This convention, that the current flows opposite to the flowing electrons, is unfortunate, but it agrees with the convention practically always used in the subject.

So far we’ve only defined the current through a conducting wire, but we can generalize the definition to arbitrary moving charge distributions as well. Suppose some charge distribution with charge density \(\rho(\mathbf{x},t)\) is flowing through space with some flow velocity \(\mathbf{v}(\mathbf{x},t)\). Intuitively, we can think of this moving distribution as a bunch of wires of current flowing through space each with its own velocity. We define the current density \(\mathbf{J}(\mathbf{x},t)\) at each point in space by the product \[ \mathbf{J} \equiv \rho \mathbf{v} \ . \] Evidently, the current density has dimensions of charge times velocity, or equivalently current per unit area. In Gaussian units, the current density thus carries units of \(\text{statamp} / \text{cm}^2\). Unlike the charge density, the current density is a vector whose direction is the average direction of charge flow at any given point.

From the current density we can define a more general notion of current. Suppose some charge distribution is flowing through some surface \(\mathcal{S}\) with some given current density \(\mathbf{J}\). We define the current flowing through \(\mathcal{S}\) to be the flux integral \[ I \equiv \int_\mathcal{S} \mathbf{J} \cdot d\mathbf{a} \ . \] Notice that the current defined this way depends on the direction of the outward surface normal. If the direction of the outward normal is reversed, the current changes sign.

We can also define notions of current density for lower dimensional distributions analogously. Suppose a charge distribution with surface charge density \(\sigma(\mathbf{x},t)\) is flowing across some 2-dimensional surface with some flow velocity \(\mathbf{v}(\mathbf{x},t)\). Then we define the surface current density \(\mathbf{K}(\mathbf{x},t)\) at each point along the surface by the project \[ \mathbf{K} \equiv \sigma \mathbf{v} \ . \] As with the volume current density \(\mathbf{J}\), the surface current density \(\mathbf{K}\) is also a vector pointing in the direction of charge flow along the surface. The surface current density evidently carries dimensions of current per unit length.

From the surface current density we can define a notion of current flowing through some curve on a surface. If a surface \(\mathcal{S}\) carries some current density \(\mathbf{K}\) and \(\mathcal{C}\) is some curve on the surface, we define the current passing through the curve by \[ I \equiv \int_\mathcal{C} \mathbf{K} \cdot d\boldsymbol{\ell} \ . \] Finally, we’ll consider the 1-dimensional case. In this case there is no meaningful notion of line current density, since the current would just be the flow of line current density through a point. That is, line current density is the same thing as current for charge flowing along a curve. The closest thing to a line current density would be the current vector defined above for a wire. If a charge distribution with line density \(\lambda(\mathbf{x},t)\) is moving along some curve with flow velocity \(\mathbf{v}(\mathbf{x},t)\), we define the current vector \[ \mathbf{I} \equiv \lambda \mathbf{v} \ . \] From the current vector, we can easily see the current \(I\) at any point along the curve is just \(I = \lambda |\mathbf{v}|\), as we defined before.

See the figure below for a visual of these different types of current densities.

FIGURE (show current densities)

Conservation of Charge

One fundamental law of classical electromagnetism is that charge is conserved. That is, inside any closed surface the total charge \(Q_{\text{enc}}\) inside can’t spontaneously appear or disappear. It can only flow out of the surface with some current \(I_{\text{out}}\).

FIGURE (show charge and current through surface)

We can express conservation of charge mathematically then by writing \[ \frac{dQ_{\text{enc}}}{dt} = -I_{\text{out}} \ . \] The minus sign expresses the fact that a current flowing out of the closed surface decreases the amount of enclosed charge inside the enclosed region.

The statement of conservation of charge is stronger than this however. Not only is charged conserved globally, it’s also conserved locally. That is, no matter how small a closed surface we choose, charge must be conserved in the sense defined above. We can formulate the statement of local charge conservation most conveniently using a continuity equation.

To achieve this, we can express the enclosed charge as a volume integral over a charge density \(\rho(\mathbf{x},t)\) and the outward current as a closed surface integral over the current density \(\mathbf{J}(\mathbf{x},t)\), \[ Q_{\text{enc}} = \int_\mathcal{V} d^3 \mathbf{x} \ \rho(\mathbf{x},t) \quad , \quad I_{\text{out}} = \oint_\mathcal{S} \mathbf{J}(\mathbf{x},t) \cdot d\mathbf{a} \ . \] As usual, we denote the closed surface by \(\mathcal{S}\) and its enclosed volume by \(\mathcal{V}\). Plugging these integrals into the conservation law above, we then have \[ \frac{d}{dt} \int_\mathcal{V} d^3 \mathbf{x} \ \rho(\mathbf{x},t) = - \oint_\mathcal{S} \mathbf{J}(\mathbf{x},t) \cdot d\mathbf{a} \ . \] Now, we assume the closed surface \(\mathcal{S}\) is fixed, which means we can pull the time derivative inside the integral. We can then use the divergence on the right-hand side to express the current as a divergence. All together, we get \[ \int_\mathcal{V} d^3 \mathbf{x} \ \frac{\partial\rho(\mathbf{x},t)}{\partial t} = - \int_\mathcal{V} d^3 \mathbf{x} \ \nabla \cdot \mathbf{J}(\mathbf{x},t) \ . \] Since the volume \(\mathcal{V}\) is arbitrary, the only way this integral can hold is if the integrands equal. That is, if \[ \boxed{ \frac{\partial\rho}{\partial t} + \nabla \cdot \mathbf{J} = 0 } \ . \] This equation fully encapsulates the statement that charge is both locally and globally conserved. It’s common to refer to this equation as both conservation of charge and the continuity equation, even though we could express conservation of charge globally, and there can be continuity equations for other conserved quantities as well.

In this chapter, we will focus specifically on distributions in which \(\nabla \cdot \mathbf{J} = 0\). That is, we focus only on distributions where the current neither converges nor diverges at any point. This is called the steady current condition. The study of electromagnetism in the special scenario of steady current is called magnetostatics, the whole subject of this chapter.

FIGURE (show steady current flow, perhaps contrasting with non-steady current flow)

From the continuity equation, if the steady current condition holds, we must also have \[ \frac{\partial\rho}{\partial t} = 0 \ . \] That is, the charge distribution must be time-independent. That is, we must have \(\rho = \rho(\mathbf{x})\) and \(\mathbf{J} = \mathbf{J}(\mathbf{x})\), which means neither density is allowed to depend explicitly on time in magnetostatics. If they do, we violate the steady current condition.

Note fact that \(\rho = \rho(\mathbf{x})\) does not imply that charges aren’t moving around in space, only that they’re moving in such a way that \(\rho\) itself isn’t changing explicitly with time. The only way this can be true is if any charges flowing out a given region are being replaced by charges flowing back into the region at the same rate. Of course, this is an abstraction. Yet in many physical situations, for example when we have current moving through a wire, this is indeed approximately the case.

Though seemingly similar, this situation contrasts with electrostatics, where we also had \(\rho = \rho(\mathbf{x})\). In that setting, the charges were indeed assumed to be stationary. Charges placed at a point stayed there. They didn’t move around.

One immediately implication of all this is that point charges can never generate steady currents. To see why this is the case, suppose a point charge \(q\) is moving through space with some trajectory \(\mathbf{x}'(t)\). Then its charge density is \[ \rho(\mathbf{x},t) = q \delta\big(\mathbf{x} - \mathbf{x}'(t)\big) \ . \] This necessarily means that the charge density of a moving point charge must be an explicit function of time, which already violates the steady current condition.

An immediate implication of this is that, when enforcing the steady current condition, we must consider extended distributions of charge from the outset. We can’t start with moving point charges and build up steady currents from there. This is the major reason why magnetostatics is often much more cumbersome to introduce than electrostatics. In electrostatics, we could start by stating Coulomb’s law for a point charge, and from there extend to continuous distributions. In magnetostatics, we have to start immediately from continuous distributions and go from there.

Magnetic Fields

Magnetism as a qualitative subject is at least as old as electricity. In the days of Ancient Greece, Thales observed that rocks of iron ore, or lodestones, tended to attract small pieces of iron. He called this force magnetism, named after the ancient town of Magnesia, an area where lodestone was commonly found at the time. By the 11th century, the Chinese had discovered that iron needles could be used to create compasses for navigation, suggesting that the Earth itself had a magnetic force of its own. In the 16th century, Gilbert discovered was able to magnetic forces and electric forces must be distinctly different forces, though they appeared to be related to each other in a way that remained mysterious for a few hundred more years.

In the 19th century, it was discovered that electric current gives rise to magnetic fields. The first discovery in this direction was by Oersted, who noticed that if he placed a compass in the vicinity of a current carrying wire that the needle no longer pointed north. He noticed that as he moved the needle around the wire, it tended to point azimuthally to the wire, suggesting that somehow the current in the wire was altering the magnetic forces acting on the needle.

Not long after Oersted made his discovery, Biot and Savart were able to deduce a precise empirical relation between the magnetic field and the current flowing through a wire. They found that when a current was passed through a wire, a magnetic field strength was created due to the current. We will now present this law in more detail and derive its implications.

Biot-Savart Law

We’ll now present the Biot-Savart law, which provides a relation between the magnetic field \(\mathbf{B}\) and the current. In a modern formulation we can express this law as follows. Suppose an infinitesimal length of wire \(d\ell\) experiences a current \(\mathbf{I}\) at the source point \(\mathbf{x}'\). This will then give rise to an infinitesimal magnetic field \(d\mathbf{B}\) at the field point \(\mathbf{x}\) of the form

\[ d\mathbf{B} = k \mathbf{I}(\mathbf{x}') \times \frac{\mathbf{x} - \mathbf{x}'}{|\mathbf{x} - \mathbf{x}'|^3} \ . \] The constant \(k\) chosen depends on the choice of units. In Gaussian units we choose \(k \equiv 1/c\), where \(c\) is the speed of light. This strange choice of constant is needed to make sure the dimensions are consistent in the Gaussian system. Of course, one could argue that dividing by any characteristic velocity will work. Why exactly that velocity is the speed of light we’ll see in a future chapter when we study electrodynamics. Nevertheless, it does hint at a connection between electromagnetism and optics.

To find the magnetic field \(\mathbf{B}(\mathbf{x})\) generated by the full wire we need only integrate both sides along the length of the wire. If the wire lies on a curve \(\mathcal{C}\) in space, we have \[ \mathbf{B}(\mathbf{x}) = \frac{1}{c} \int_\mathcal{C} d\ell' \ \mathbf{I}(\mathbf{x}') \times \frac{\mathbf{x} - \mathbf{x}'}{|\mathbf{x} - \mathbf{x}'|^3} \ . \] This empirical result is known as the Biot-Savart Law. If we like, we can think of this as the definition of the magnetic field in magnetostatics. Just as we sometimes call the electric field an E-field for short, we sometimes call the magnetic field the B-field.

Evidently, the magnetic field in Gaussian units carries the same units as electric field. Indeed, this is by design. Even though the electric field has units of \(\text{esu} / \text{cm}^2\), which means the magnetic field’s units are the same, for historical reasons though we give the unit of magnetic field a different name, the Gauss, with \(1 \ \text{G} = 1 \ \text{esu} / \text{cm}^2\) exactly. In SI units, the unit of magnetic field is called the Tesla, with \(1 \ \text{T} = 10^4 \ \text{G}\) exactly. Note though that the dimensions of magnetic field are different in the two unit systems, as discussed in the first chapter.

To get an idea for scales, the Earth’s magnetic field strength at the surface is about \(0.5 \ \text{G}\). A typical refrigerator magnet has a field strength of \(\sim 50 \ \text{G}\). An MRI machine might have a field strength of at least \(\sim 30 \ \text{kG}\), or \(\sim 3 \ \text{T}\). The strongest continuous magnet produced in a lab thus far has a field strength of \(\sim 450 \ \text{kG}\), or \(\sim 45 \ \text{T}\).

As written, the Biot-Savart law shows how to find the magnetic field due to a 1-dimensional wire. We can generalize the result immediately to 2-dimensional surfaces and 3-dimensional volumes in the usual way, by replacing a line density with a surface or volume density. The B-field due to a surface of current \(\mathcal{S}\) can be stated by replacing \(\mathbf{I} d\ell'\) with \(\mathbf{K} da'\) in the formula above to get \[ \mathbf{B}(\mathbf{x}) = \frac{1}{c} \int_\mathcal{S} da' \ \mathbf{K}(\mathbf{x}') \times \frac{\mathbf{x} - \mathbf{x}'}{|\mathbf{x} - \mathbf{x}'|^3} \ . \] Finally, the B-field due to a volume of current is given by replacing \(\mathbf{I} d\ell'\) with \(\mathbf{J} d^3\mathbf{x}'\) to get \[ \boxed{ \mathbf{B}(\mathbf{x}) = \frac{1}{c} \int d^3 \mathbf{x}' \ \mathbf{J}(\mathbf{x}') \times \frac{\mathbf{x} - \mathbf{x}'}{|\mathbf{x} - \mathbf{x}'|^3} } \ . \] It’s worth comparing the Biot-Savart law above to the integral form of Coulomb’s law in electrostatics. There, we had \[ \mathbf{E}(\mathbf{x}) = \int d^3 \mathbf{x}' \ \rho(\mathbf{x}') \frac{\mathbf{x} - \mathbf{x}'}{|\mathbf{x} - \mathbf{x}'|^3} \ . \] Notice that up to a constant the two laws look exactly the same, except with \(\rho(\mathbf{x}')\) replaced by the cross product \(\mathbf{J}(\mathbf{x}') \times\). The appearance of a cross product is perhaps the most important mathematical difference in the formulas between electrostatics and magnetostatics. An immediate implication of this is that the magnetic field never points in the direction of the current, but rather in a direction perpendicular to the current, given by the direction of the cross product \(\mathbf{J}(\mathbf{x}') \times (\mathbf{x} - \mathbf{x}')\).

The appearance of the cross product also makes calculating the magnetic field from the Biot-Savart law much more cumbersome to use than Coulomb’s law, which was itself already cumbersome. We’ll thus quickly move to finding better ways to calculate magnetic fields. But first we’ll show a relatively simple example of how to use the Biot-Savart law directly.

Example: Conducting loop of constant current

The canonical charge distribution in magnetostatics is the loop of constant current. Suppose a loop of radius \(R\) carries a constant current \(I\) in the counterclockwise direction. To achieve this, we’ll suppose the loop is centered at the origin in the \(xy\)-plane. Our goal will be to use the Biot-Savart law to calculate the B-field \(\mathbf{B}(z)\) along the \(z\)-axis.

FIGURE (loop of current)

With this setup, it’s not hard to see that \(\mathbf{x}' = R \mathbf{e}_\varrho\) and \(\mathbf{x} = z \mathbf{e}_z\), which means \[ \mathbf{x} - \mathbf{x}' = z \mathbf{e}_z - R \mathbf{e}_\varrho \ . \] Since the current is constant and flows counterclockwise along the loop, we have \(\mathbf{I} = I \mathbf{e}_\varphi\). This means we have \[ \mathbf{I} \times (\mathbf{x} - \mathbf{x}') = Iz (\mathbf{e}_\varphi \times \mathbf{e}_z) - IR (\mathbf{e}_\varphi \times \mathbf{e}_\varrho) = Iz \mathbf{e}_\varrho + IR \mathbf{e}_z \ . \] Now consider an infinitesimal element \(d\ell' = R d\varphi'\) of the loop. By the Biot-Savart law, we have \[ d\mathbf{B} = \frac{1}{c} d\ell' \ \mathbf{I} \times \frac{\mathbf{x} - \mathbf{x}'}{|\mathbf{x} - \mathbf{x}'|^3} = R d\varphi' \ \frac{I}{c} \bigg[ \frac{z}{(R^2 + z^2)^{3/2}} \mathbf{e}_\varrho + \frac{R}{(R^2 + z^2)^{3/2}} \mathbf{e}_z \bigg] \ . \] Integrating along the entire loop, we thus have \[ \mathbf{B}(z) = \frac{IR}{c} \int_0^{2\pi} d\varphi' \ \bigg[ \frac{z}{(R^2 + z^2)^{3/2}} \mathbf{e}_\varrho + \frac{R}{(R^2 + z^2)^{3/2}} \mathbf{e}_z \bigg] \ . \] At this point we need to be a bit careful, since cylindrical basis vectors are functions of position. It’s helpful to write \(\mathbf{e}_\varphi\) in terms of the Cartesian vectors to avoid this issue, \[ \mathbf{e}_\varphi = \cos\varphi' \mathbf{e}_x + \sin\varphi' \mathbf{e}_y \ . \] If we plug this into the first integral, we see it must vanish since the integrand is independent of \(\varphi'\) and \(\cos\phi'\) and \(\sin\varphi'\) are each \(2\pi\)-periodic functions. The second integral is also independent of \(\varphi'\). Since \(\mathbf{e}_z\) is constant, we finally get \[ \mathbf{B}(z) = \frac{2\pi IR^2}{c(R^2 + z^2)^{3/2}} \mathbf{e}_z \ . \] Thus, along the \(z\)-axis the B-field points only in the \(z\)-direction. This should make sense by symmetry. After all, the current along the loop is constant, which means any segment of field in the \(\varrho\)-direction must cancel with an opposite segment of field, leaving only the \(z\)-component to contribute anything.

Let’s now check the far field limit \(z \gg R\). In this limit, \((R^2 + z^2)^{-3/2} \approx 1/z^3\), and so \[ \mathbf{B}(z) \approx \frac{2\pi IR^2}{cz^3} \mathbf{e}_z \ . \]

Since the far field falls off like \(1/r^3\), in this limit the B-field seems to behave like a dipole field.

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In fact, the lowest non-vanishing moment of the B-field will always be at least a dipole field, as we’ll see later. There will never be a monopole field contribution. Said equivalently, there is no such thing as a magnetic monopole, as far as we can tell based on all experiments done up to this day. More advanced theories predict that they could exist, but experiments have never found them. Until found, we’re left to assume based on all experimental evidence that no magnetic monopoles exist in nature.

It’s tempting to ask whether we can use the Biot-Savart law to calculate the B-field generated by a moving point charge. If we naively attempted to apply the Biot-Savart law to a point charge \(q\) moving at constant velocity \(\mathbf{v}\), we’d find that \[ \mathbf{B}(\mathbf{x}) = q \frac{\mathbf{v}}{c} \times \frac{\mathbf{x} - \mathbf{v}t}{|\mathbf{x} - \mathbf{v}t|^3} \ . \] But this can’t be true, since the Biot-Savart law requires that B-field only be a function of position and not time. The only way the B-field can be time-independent is if the steady current condition is imposed, which means the Biot-Savart law implicitly assumes the steady current condition is true. We thus cannot use the Biot-Savart law on moving point charges. It only holds for continuous distributions of steady currents. We’ll revisit the magnetic field of a moving point charge in a later chapter.

Example: Infinite conducting wire of constant current

Consider now an infinitely long conducting wire of constant current \(I\). We’d again like to use the Biot-Savart law to find the B-field created due to this current-carrying wire.

We’ll assume the wire is oriented along the \(z\)-axis, which means \(\mathbf{I} = I \mathbf{e}_z\) and \(\mathbf{x}' = z' \mathbf{e}_z\). We’ll find it convenient for this problem to work in cylindrical coordinates, so that \(\mathbf{x} = \varrho \mathbf{e}_\varrho + z \mathbf{e}_z\). This means \[ \mathbf{x} - \mathbf{x}' = \varrho \mathbf{e}_\varrho + (z-z') \mathbf{e}_z \ . \] The cross product of the current vector with this separation vector is thus \[ \mathbf{I} \times (\mathbf{x} - \mathbf{x}') = I\varrho (\mathbf{e}_z \times \mathbf{e}_\varrho) = I\varrho \mathbf{e}_\varphi \ . \] The line element along the wire is clearly just \(d\ell' = dz'\), which means \[ d\mathbf{B} = \frac{1}{c} d\ell' \ \mathbf{I} \times \frac{\mathbf{x} - \mathbf{x}'}{|\mathbf{x} - \mathbf{x}'|^3} = \frac{I\varrho}{c} \mathbf{e}_\varphi dz' \ \frac{1}{\big(\varrho^2 + (z-z')^2\big)^{3/2}} \ . \] Notice we pulled \(\varrho\) and \(\mathbf{e}_\varphi\) out of the integral since neither depend on \(z'\). We now need to integrate this along the length of the wire, which is infinite, \[ \mathbf{B}(\mathbf{x}) = \frac{I\varrho}{c} \mathbf{e}_\varphi \int_{-\infty}^\infty \frac{dz'}{\big(\varrho^2 + (z-z')^2\big)^{3/2}} \ . \] Evaluating this integral can be done using elementary methods we’ve seen before in the course. In the end we simply get \[ \mathbf{B}(\mathbf{x}) = \frac{2I}{c\varrho} \mathbf{e}_\varphi \ . \] Evidently, the B-field of an infinite wire of constant current falls off like \(1/\varrho\), just like the E-field of an infinite wire of constant charge density. However, unlike the E-field, the B-field is directed azimuthally around the wire.

Lorentz Force Law

Also in the early 19th century, Ampere observed that if two parallel current carrying wires were placed near each other, they tended to attract or repel each other depending on whether the currents were aligned or anti-aligned. More precisely, suppose the first wire carries a current \(I_1\) and the second wire carries a current \(I_2\). If the wires are separated by a distance \(d\), Ampere found that the force per unit length \(f\) is given by \[ f = 2 k_m \frac{I_1 I_2}{d} \ . \] The constant of proportionality \(k_m\) depends on the choice of units. In Gaussian units we choose \(k_m \equiv 1/c^2\), where \(c\) is again the speed of light, which means the above law becomes \[ f = \frac{2 I_1 I_2}{c^2 d} \ . \] Later on in the century, Lorentz formulated a precise statement of the force on a moving point charge in a magnetic field. Suppose a point charge \(q\) is moving at a velocity \(\mathbf{v}\) in the presence of some external B-field \(\mathbf{B}\). This charge will feel a force due to the B-field given by the Lorentz Force Law, \[ \mathbf{F} = \alpha q \mathbf{v} \times \mathbf{B} \ . \] Here \(\alpha\) is yet another proportionality constant that depends on the choice of units. In the Gaussian system we choose \(\alpha = 1/c\), which again ensures \(\mathbf{E}\) and \(\mathbf{B}\) carry the same dimensions. In these units, we get \[ \boxed{ \mathbf{F} = q \frac{\mathbf{v}}{c} \times \mathbf{B} } \ . \] Evidently, the force due to a B-field will always be perpendicular to both the velocity of the charge as well as the B-field. This fact immediately implies something interesting, magnetic fields can only change the direction of a moving charge, not speed it up or slow it down. Said differently, magnetic forces do no work.

To see why this is true, consider the work done in moving a charge along some path \(\mathcal{C}\) in the presence of a B-field \(\mathbf{B}\). Plugging in the Lorentz force law and noting that \(d\boldsymbol{\ell} = \mathbf{v} dt\), the work \(W\) done in moving the charge along the path must be \[ W = \int_\mathcal{C} \mathbf{F} \cdot d\boldsymbol{\ell} = \int_\mathcal{C} \bigg(q \frac{\mathbf{v}}{c} \times \mathbf{B}\bigg) \cdot d\boldsymbol{\ell} = \frac{q}{c} \int dt \ \big(\mathbf{v} \times \mathbf{B}\big) \cdot \mathbf{v} = 0 \ . \] Thus, magnetic forces do no work. That is, they cannot change the kinetic energy of the moving charge. They can only deflect the moving charge in another direction.

The Lorentz force law is more often stated in the more general setting where both an external electric and magnetic fields may be present. Recall from electrostatics that the force on a point charge \(q\) due to an external E-field is given by \(\mathbf{F} = q \mathbf{E}\). By the principle of superposition then, the combined force on a moving charge will be the sum of electric and magnetic forces, \[ \boxed{ \mathbf{F} = q \bigg(\mathbf{E} + \frac{\mathbf{v}}{c} \times \mathbf{B}\bigg) } \ . \] Notice that we can immediately deduce something about scales from this equation. For a given field strength, the magnetic force on a charged particle will typically be much smaller than the electric force, unless the particle is moving at velocity that’s an appreciable fraction of the speed of light \(c\).

The Lorentz force law can easily be stated for continuous charge distributions as well. Consider an infinitesimal unit of charge \(dq = \rho d^3\mathbf{x}\). This must obey the above force law, giving rise to an infinitesimal force \(d\mathbf{F}\) given by \[ d\mathbf{F} = d^3\mathbf{x} \ \bigg[\rho\mathbf{E} + \rho \frac{\mathbf{v}}{c} \times \mathbf{B} \bigg] \ . \] Since \(\mathbf{J} = \rho \mathbf{v}\) is the current density, when we integrate over the entire distribution we get \[ \mathbf{F} = \int d^3\mathbf{x} \ \bigg[\rho\mathbf{E} + \frac{1}{c} \mathbf{J} \times \mathbf{B}\bigg] \ . \] This is the most general form of the Lorentz force law we can write down in electromagnetism. Remember that this is the force on a charge distribution due to external fields, not the force due to any fields the charge distribution itself may create. The issue of self fields is a delicate question in electromagnetism, one we’ve seen arise and will arise several times in the course.

As is typical in field theories, everything of interest is inside the integrand. The integrand is just the force per unit volume, which is called the force density \(\mathbf{f}\). For the Lorentz force, the force density is clearly \[ \boxed{ \mathbf{f} = \rho\mathbf{E} + \frac{1}{c} \mathbf{J} \times \mathbf{B} } \ . \] Clearly, when only B-fields are present, which we assume in magnetostatics, then the Lorentz force density reduces to \[ \mathbf{f} = \frac{1}{c} \mathbf{J} \times \mathbf{B} \ . \] The total Lorentz force \(\mathbf{F}\) for a continuous distribution is then the volume integral of \(\mathbf{f}\) over the entire distribution.

Example: Ampere’s Current Law

We’ll now derive Ampere’s current law between two current carrying wires. Suppose two parallel wires are separated by some distance \(d\), where the first wire carries a current \(I_1\) and the second wire carries a current \(I_2\). We mentioned above that the following result was found empirically, \[ f = \frac{2I_1 I_2}{c^2 d} \ , \] where \(f\) is the force per unit length between the two wires. It’s pretty trivial to derive this result using the Lorentz force law.

To do this it suffices to examine the force exerted by one wire on the other. We’ll focus on finding the force exerted on the second wire by the first. Suppose the two wires are oriented in the \(z\)-direction, with the first wire exactly along the \(z\)-axis and the second wire translated by \(d\) along the \(y\)-axis, as shown below.

FIGURE (show two current carrying wires)

Assuming both wires are sufficiently long that we can treat them as having infinite length. This means the first wire will produce a B-field of the form \[ \mathbf{B}(\mathbf{x}) = \frac{2I_1}{c\varrho} \mathbf{e}_{\varphi} \ . \] To find the force the first wire exerts on the second wire we only care about field points \(\mathbf{x} = z \mathbf{e}_z\) along the first wire. Along the first wire, \(\varrho = d\) and \(\mathbf{e}_\varphi = \mathbf{e}_x\), which means the B-field \(\mathbf{B}_1\) experienced by the second wire due to the first wire is \[ \mathbf{B}_1 = \frac{2I_1}{cd} \mathbf{e}_x \ . \] By the Lorentz force law, the force on the second wire due to the first wire will be given by integrating \[ d\mathbf{F} = \frac{1}{c} \mathbf{I}_2 \times \mathbf{B}_1 d\ell = \frac{1}{c} (I_2 \mathbf{e}_z) \times \bigg(\frac{2I_1}{cd} \mathbf{e}_x\bigg) d\ell = \frac{2I_1 I_2}{c^2 d} \mathbf{e}_y d\ell \ . \] Now, if we were interested in the total force exerted on the second wire, we’d need to integrate along the full length of the first wire, which would be infinite. But we’re only interested in the force per unit length \(\mathbf{f}\), which is just \[ \mathbf{f} = \frac{d\mathbf{F}}{d\ell} = \frac{2I_1 I_2}{c^2 d} \mathbf{e}_y \ . \] This is exactly Ampere’s current law. Notice that the force between the two wires is in the direction of separation between them. By Newton’s third law, the force felt by the first wire due to the second wire must be exactly the same. This means both wires are attracted are repelled by exactly the same force. The wires will attract if the two currents are opposite and repulse otherwise. This effect can be observed in the lab, though one would need to use pretty high currents to actually observe these forces by eye. For example, to create a force of \(1 \ \text{lbf} \approx 4 \cdot 10^5 \ \text{dyne}\) between two wires of length \(3 \ \text{ft} \approx 10^2 \ \text{cm}\) separated by a distance of \(6 \ \text{in} \approx 15 \ \text{cm}\), one would need to use a huge current of about \(2000\) amps!

A similar result can be derived for the force between two arbitrary wires each of constant current. If the first wire has a current \(\mathbf{I}_1\) along some curve \(\mathcal{C}_1\) and the second wire has a current \(\mathbf{I}_2\) along some curve \(\mathcal{C}_2\), then the force \(\mathbf{F}_{12}\) exerted on the first wire due to the second wire is given by \[ \mathbf{F}_{12} = - \frac{1}{c^2} \int_{\mathcal{C}_1} d\ell_1 \int_{\mathcal{C}_2} d\ell_2 \ \mathbf{I}_1 \cdot \mathbf{I}_2 \frac{\mathbf{x}_1 - \mathbf{x}_2}{|\mathbf{x}_1 - \mathbf{x}_2|^3} \ , \] where \(\mathbf{x}_1 - \mathbf{x}_2\) is the separation vector between any two source points along the two wires. This can easily be derived by using the Biot-Savart law on the second wire. By symmetry, it’s clear that the force \(\mathbf{F}_{21}\) on the second wire due to the first wire is equal and opposite, with \(\mathbf{F}_{21} = -\mathbf{F}_{12}\).

Field Equations

Just as it wasn’t convenient or illuminating in electrostatics to use Coulomb’s law to calculate and study electric fields, it’s not convenient in magnetostatics to use the Biot-Savart law to calculate or study magnetic fields. We’ll want to instead re-cast the Biot-Savart law as a set of field equations.

Divergence of \(\mathbf{B}\)

By the Biot-Savart law, the B-field created by a volume of steady current is given by \[ \mathbf{B}(\mathbf{x}) = \frac{1}{c} \int d^3 \mathbf{x}' \ \mathbf{J}(\mathbf{x}') \times \frac{\mathbf{x} - \mathbf{x}'}{|\mathbf{x} - \mathbf{x}'|^3} \ . \] Now, we know from electrostatics that the gradient with respect to \(\mathbf{x}\) is given by \[ \nabla \frac{1}{|\mathbf{x} - \mathbf{x}'|} = -\frac{\mathbf{x} - \mathbf{x}'}{|\mathbf{x} - \mathbf{x}'|^3} \ . \] Indeed, we used this fact to derive the scalar potential \(\phi(\mathbf{x})\). We also know from vector calculus that any curl of a scalar field \(f(\mathbf{x})\) times a vector \(\mathbf{v}\) can be rewritten in the form \[ \nabla \times \big(f(\mathbf{x})\mathbf{v}\big) = -\mathbf{v} \times \nabla f(\mathbf{x}) \ . \] If this isn’t obvious, just write the left-hand side in index notation and do the algebra. Together, these imply that \[ \mathbf{J}(\mathbf{x}') \times \frac{\mathbf{x} - \mathbf{x}'}{|\mathbf{x} - \mathbf{x}'|^3} = -\mathbf{J}(\mathbf{x}') \times \nabla \frac{1}{|\mathbf{x} - \mathbf{x}'|} = \nabla \times \frac{\mathbf{J}(\mathbf{x}')}{|\mathbf{x} - \mathbf{x}'|} \ . \] Plugging this into the Biot-Savart law and pulling the curl operator outside the integral, we thus have \[ \mathbf{B}(\mathbf{x}) = \nabla \times \frac{1}{c} \int d^3 \mathbf{x}' \ \frac{\mathbf{J}(\mathbf{x}')}{|\mathbf{x} - \mathbf{x}'|} \ . \] The argument inside the curl is evidently some vector field \(\mathbf{A}(\mathbf{x})\) given by the integral \[ \mathbf{A}(\mathbf{x}) \equiv \frac{1}{c} \int d^3 \mathbf{x}' \ \frac{\mathbf{J}(\mathbf{x}')}{|\mathbf{x} - \mathbf{x}'|} \ . \] We call this the vector potential. We’ll study this in more detail in the next section. For now, we’ve established that we can write \[ \mathbf{B} = \nabla \times \mathbf{A} \ . \] Since the divergence of a curl must vanish, the first field equation of magnetostatics immediately follows, \[ \boxed{ \nabla \cdot \mathbf{B} = 0 } \ . \] Given the similarity to Gauss’s law for electricity, this field equation is sometimes called Gauss’s law for magnetism.

Curl of \(\mathbf{B}\)

To derive the remaining field equation let’s now look at the curl of the B-field. Recall from vector calculus that \[ \nabla \times (\nabla \times \mathbf{A}) = \nabla (\nabla \cdot \mathbf{A}) - \nabla^2 \mathbf{A} \ . \] Clearly the left-hand side is just \(\nabla \times \mathbf{B}\), what we seek to find. Let’s now simplify the right-hand side, starting with the first term. Plugging in the integral for \(\mathbf{A}(\mathbf{x})\) and moving the divergence inside the integral, we have \[ \nabla (\nabla \cdot \mathbf{A}) = \nabla \int d^3 \mathbf{x}' \ \nabla \cdot \frac{\mathbf{J}(\mathbf{x}')}{|\mathbf{x} - \mathbf{x}'|} \ . \] To simplify this expression, notice first that the integrand can using the vector calculus identity \[ \nabla \cdot \frac{\mathbf{J}(\mathbf{x}')}{|\mathbf{x} - \mathbf{x}'|} = \mathbf{J}(\mathbf{x}') \cdot \nabla \frac{1}{|\mathbf{x} - \mathbf{x}'|} \] Next, we’ll use a useful fact that we’ll see a few times in the rest of this course. Namely, the gradient of \(1/|\mathbf{x} - \mathbf{x}'|\) with respect to \(\mathbf{x}\) is just minus its gradient with respect to \(\mathbf{x}'\), \[ \nabla \frac{1}{|\mathbf{x} - \mathbf{x}'|} = -\nabla' \frac{1}{|\mathbf{x} - \mathbf{x}'|} \ . \] If we apply this trick to the above integral and then integrate by parts to move the derivatives from \(1/|\mathbf{x} - \mathbf{x}'|\) to \(\mathbf{J}'(\mathbf{x}')\), we get \[ \int d^3 \mathbf{x}' \ \nabla \cdot \frac{\mathbf{J}(\mathbf{x}')}{|\mathbf{x} - \mathbf{x}'|} = -\int d^3 \mathbf{x}' \ \mathbf{J}(\mathbf{x}') \cdot \nabla' \frac{1}{|\mathbf{x} - \mathbf{x}'|} = \int d^3 \mathbf{x}' \ \frac{\nabla' \cdot \mathbf{J}(\mathbf{x}')}{|\mathbf{x} - \mathbf{x}'|} \ . \] Of course, for this to be true we require that the surface term vanishes as \(r' \to \infty\). This will be true so long as \(\mathbf{J}(\mathbf{x}')\) goes to zero at infinity faster than \(1/r'^2\), which will be the case at least for all localized current distributions.

Now, in magnetostatics we require that the steady current condition hold, which means we must have \(\nabla' \cdot \mathbf{J}(\mathbf{x}') = 0\). This means the integrand must vanish for all \(\mathbf{x}'\), which means the integral also must vanish, which means \(\nabla \cdot \mathbf{A} = 0\), which means that \(\nabla (\nabla \cdot \mathbf{A}) = 0\) as well.

We now need to simplify \(\nabla^2 \mathbf{A}\). Plugging in the integral for \(\mathbf{A}(\mathbf{x})\) and moving \(\nabla^2\) inside the integral, we have \[ \nabla^2 \mathbf{A} = \frac{1}{c} \int d^3 \mathbf{x}' \ \nabla^2 \frac{\mathbf{J}(\mathbf{x}')}{|\mathbf{x} - \mathbf{x}'|} \ . \] Since \(\nabla^2 \ 1/|\mathbf{x} - \mathbf{x}'| = -4\pi \delta(\mathbf{x} - \mathbf{x}')\) the integral is trivial, giving \(\nabla^2 \mathbf{A} = -4\pi/c \ \mathbf{J}\).

Thus, in magnetostatics, the curl of the B-field is evidently proportional to the current density, with \[ \boxed{ \nabla \times \mathbf{B} = \frac{4\pi}{c} \mathbf{J} } \ . \] This result is known as Ampere’s Law in differential form. We’ll study it in more detail shortly.

Summarizing what we’ve found, we’ve established the following two field equations of magnetostatics, \[ \nabla \cdot \mathbf{B} = 0 \quad , \quad \nabla \times \mathbf{B} = \frac{4\pi}{c} \mathbf{J} \ . \] So what do these tell us? Let’s compare these with the field equations we found for electrostatics, \[ \nabla \cdot \mathbf{E} = 4\pi \rho \quad , \quad \nabla \times \mathbf{E} = 0 \ . \] In electrostatics, it appears that it’s the charge distribution that gives rise to the E-field. There are no currents present. Meanwhile, in magnetostatics it appears it’s the current distribution that gives rise to the B-field. Evidently, in classical electromagnetism it’s only electric currents that create B-fields. Magnetic charges do not exist, or equivalently, magnetic monopoles do not exist. It’s entirely possible that magnetic monopoles may exist, in which case we’d need to modify the theory of electromagnetism, but we’ve never found them. Until that happens, we take these equations as reality.

Ampere’s Law

In electrostatics, we were able to use the integral form of Gauss’s law to easily find the electric field of several charge distributions with high degrees of symmetry. It turns out we can do something similar with Ampere’s law, but to do that we first need to convert the differential form of Ampere’s law into its integral form.

To do this, we’ll choose some arbitrary surface \(\mathcal{S}\) and integrate both sides of Ampere’s law over this surface, \[ \int_{\mathcal{S}} (\nabla \times \mathbf{B}) \cdot d\boldsymbol{\ell} = \frac{4\pi}{c} \int_{\mathcal{S}} \mathbf{J} \cdot d\mathbf{a} \ . \] Now, notice that the path integral on the right-hand side is just the total enclosed current \(I_\text{enc}\) flowing through the closed loop, \[ I_\text{enc} = \int_{\mathcal{S}} \mathbf{J} \cdot d\mathbf{a} \ . \] The integral on the left-hand side can be rewritten as a surface integral using Stokes’ theorem. Suppose \(\mathcal{C}\) is the closed loop that bounds the surface \(\mathcal{S}\) chosen above. Then by Stokes’ theorem we have \[ \int_{\mathcal{S}} (\nabla \times \mathbf{B}) \cdot d\mathbf{a} = \oint_{\mathcal{C}} \mathbf{B} \cdot d\boldsymbol{\ell} \ . \] Putting these together, we have the integral form of Ampere’s law, \[ \boxed{ \oint_{\mathcal{C}} \mathbf{B} \cdot d\boldsymbol{\ell} = \frac{4\pi}{c} I_\text{enc} } \ . \] In words, the integral form of Ampere’s law says that the total circulation of magnetic field through any closed loop is proportional to the enclosed current flowing through that loop.

The power of this form of Ampere’s law lies in the fact that for a given current distribution, Ampere’s law must be true for any closed loop \(\mathcal{C}\) we choose. This means if we need to calculate the B-field of an enclosed current distribution then we can choose any closed loop we like to do so. We call any such choice of closed loop an Amperian loop. Indeed, this is very similar to Gauss’s law, which says we’re allowed to choose any closed surface, or Gaussian surface, we like to calculate the electric field of an enclosed charge distribution.

Suppose for a given current distribution we know that it has a certain type of symmetry that enables us to choose an Amperian loop such that the magnetic field strength along this loop is constant. If the field is constant along some section of the loop with length \(L\), then we’d have \[ \oint_{\mathcal{C}} \mathbf{B} \cdot d\boldsymbol{\ell} = |\mathbf{B}| \oint_{\mathcal{C}} d\ell = |\mathbf{B}| L \ . \] This means the magnetic field strength \(|\mathbf{B}|\) along this loop is evidently just \[ |\mathbf{B}| = \frac{4\pi}{c} \frac{I_\text{enc}}{L} \ . \] Of course, for this to work we need to easily be able to identify the symmetry of a given current distribution and find an Amperian loop that matches this symmetry. Just as with Gauss’s law, there are only a handful of distributions in which this is really possible. Those are the infinite wire, the infinite sheet, and the solenoid. Let’s look at each of these one by one.

Infinite conducting wire of current

Consider again the infinite conducting wire of uniform current \(I\). We calculated the B-field of this wire before using the Biot-Savart law. Now we’ll do the same using Ampere’s law. We’ll again suppose the wire is oriented along the \(z\)-axis.

FIGURE

From symmetry it’s clear that the B-field of the wire should only be a function of the cylindrical radius \(\varrho\), and the field should point in the \(\varphi\)-direction by the right-hand rule. We thus need only to find the field strength \(B(\varrho)\), where \[ \mathbf{B}(\mathbf{x}) = B(\varrho) \mathbf{e}_\varphi \ . \] Now, \(B(\varrho)\) is clearly constant along any counterclockwise circular loop of radius \(\varrho\) around the wire. We’ll thus choose these circular loops as the Amperian loop. Inside such a loop the enclosed current is clearly just the current flowing through the wire, meaning \(I_\text{enc} = I\). Since such a loop has a length \(L = 2\pi \varrho\), we evidently have \[ B(\varrho) = \frac{4\pi}{c} \frac{I_\text{enc}}{L} = \frac{4\pi}{c} \frac{I}{2\pi \varrho} \ . \] Thus, the B-field of the infinite wire of current is just \[ \mathbf{B}(\mathbf{x}) = \frac{2I}{c\varrho} \mathbf{e}_\varphi \ , \] which is the same result we found before using the Biot-Savart law directly. Just as we saw with the infinite wire of charge in electrostatics, the field doesn’t fall off like \(1/r^2\) due to the assumption that the wire is infinite. Unlike with electrostatics though, the B-field points azimuthally, while the E-field points radially.

Infinite conducting sheet of current

Consider now an infinite conducting sheet of uniform surface current density \(K\). The B-field of this sheet can also be easily calculated using Ampere’s law.

We’ll suppose the sheet is oriented in the \(xy\)-plane, with the current flowing in the \(x\)-direction, so \(\mathbf{K} = K \mathbf{e}_x\). From symmetry, it’s clear that the magnetic field of the sheet can only be a function of \(z\), and the field should point in the negative \(y\)-direction by the right-hand rule. We thus need only to find the field strength \(B(z)\), where \[ \mathbf{B}(\mathbf{x}) = -B(z) \mathbf{e}_y \ . \] It’s also clear from symmetry that we must have \(B(-z) = -B(z)\), meaning the field changes direction crossing the \(z\)-axis.

Now, \(B(z)\) is clearly constant along any section of loop parallel to the \(xy\)-plane. We’ll thus define an Amperian loop \(\mathcal{C}\) as follows.

• \(\mathcal{C}_1\): A directed line segment of length \(L\) from \((0,-L/2,z)\) to \((0,L/2,z)\).
• \(\mathcal{C}_2\): A directed line segment of length \(2|z|\) from \((0,L/2,z)\) to \((0,L/2,-z)\).
• \(\mathcal{C}_3\): A directed line segment of length \(L\) from \((0,L/2,-z)\) to \((0,-L/2,-z)\).
• \(\mathcal{C}_4\): A directed line segment of length \(2|z|\) from \((0,-L/2,-z)\) to \((0,-L/2,z)\).

See the figure below for a visual of the loop along with the flow of current along the sheet.

FIGURE (show sheet of current with loop)

Then, the circulation integral over the entire loop \(\mathcal{C}\) is just the circulation integral over each of the four segments, \[ \int_\mathcal{C} \mathbf{B} \cdot d\boldsymbol{\ell} = \int_{\mathcal{C}_1} \mathbf{B} \cdot d\boldsymbol{\ell} + \int_{\mathcal{C}_2} \mathbf{B} \cdot d\boldsymbol{\ell} + \int_{\mathcal{C}_3} \mathbf{B} \cdot d\boldsymbol{\ell} + \int_{\mathcal{C}_4} \mathbf{B} \cdot d\boldsymbol{\ell} \ . \] Now, notice that the integrals over the vertical segments \(\mathcal{C}_2\) and \(\mathcal{C}_4\) must be equal and opposite since the field depends only on \(z\) and not \(x\) and \(y\). This means only the horizontal segments contribute anything, giving \(B(z) L\) over \(\mathcal{C}_1\) and \(B(-z) L\) over \(\mathcal{C}_3\). Thus, \[ \int_\mathcal{C} \mathbf{B} \cdot d\boldsymbol{\ell} = L B(z) + L B(-z) = 2L B(z) \ . \] The enclosed current through the loop is just the current flowing through the line segment joining \((0,-L/2,0)\) to \((0,L/2,0)\). Since this segment has length \(L\) and the current density through it is a constant \(K\), the enclosed current is just \(I_\text{enc} = K L\).

Putting these together, by Ampere’s law we have \[ B(z) = \frac{2\pi K}{c} \ , \] which means the magnetic field of the infinite sheet of current is evidently \[ \mathbf{B}(\mathbf{x}) = \begin{cases} 2\pi K/c \ \mathbf{e}_y & z > 0 \ , \\ -2\pi K/c \ \mathbf{e}_y & z < 0 \ . \end{cases} \] Thus, the B-field strength is constant at any point above or below the sheet, just as the electric field of an infinite sheet of charge is constant. Since \(B(-z) = -B(z)\), the field strength experiences a discontinuity when crossing the sheet. The B-field strength changes discontinuously by exactly \(\Delta B = 4\pi K/c\). We saw this behavior in electrostatics as well, where the E-field through an infinite sheet of charge changes by a discontinuous amount \(\Delta E = 4\pi \sigma\) when crossing the sheet.

Solenoid

The final current distribution we’ll consider is the solenoid. A solenoid is a tightly wound coil of conducting wire through which a current is pumped, so tightly wound that the current always points azimuthally along the coil. Provided the coil is wound tightly enough, the B-field inside the coil will be constant and point cylindrically outward from the coil, while outside the wire the field will more or less vanish. To see why, suppose we have an infinitely long coil of wire oriented along the \(z\)-axis. We’ll assume that each coil is a loop of radius \(a\), and that there are exactly \(n\) coils per unit length of wire. Through this coil we imagine some constant current \(I\) is flowing azimuthally.

Now, if the current is moving azimuthally, there can be no azimuthal component to the field since the B-field is always perpendicular to the current. But there also can’t be a radial direction to the field either. To see why, suppose we choose a B-field point inside the solenoid. Since reversing the direction of the current through the solenoid shouldn’t alter the radial current of the field due to symmetry. This means the B- field of the solenoid must point in the \(z\)-direction. Moreover, the field strength cannot depend on \(\varphi\) or \(z\) since the solenoid is cylindrically symmetric. We thus must conclude that \[ \mathbf{B}(\mathbf{x}) = B(\varrho) \mathbf{e}_z \ . \] Let’s now look more closely at the \(\varrho\)-dependence of the field strength. To analyze this dependence we’ll need to consider two cases, field points lying either inside or outside the radius of the solenoid.

FIGURE (solenoid with two loops, inner and outer rectangles)

To that end let’s first suppose we have an Amperian loop lying completely outside the solenoid consisting of a rectangle of height \(\ell\) oriented along the \(z\)-axis, as shown in the above figure. Since there is no enclosed current flowing through this loop, and since for each horizontal segment of loop we have \(\mathbf{B} \cdot d\boldsymbol{\ell} = 0\), by Ampere’s law we must have \[ \int_\mathcal{C} \mathbf{B} \cdot d\boldsymbol{\ell} = B(a) \ell - B(b) \ell = 0 \ . \] This means \(B(a) = B(b)\), which means the field strength also doesn’t depend on \(\varrho\), and hence must be constant. Now, outside of the solenoid we expect the field should go to zero at infinity as it would for any wire. But since the B-field is independent of \(\varrho\), the only way this can be true is if \(B = 0\) outside the solenoid, i.e. when \(\varrho \geq a\).

By the same argument, the field also must be constant inside the solenoid, when \(\varrho < a\), but it need not vanish. To find its value inside the solenoid, we’ll take the same Amperian loop defined above, but translate it inward so the inner vertical segment lies inside the solenoid. If we now look at the circulation integral again, we find that only the vertical segment inside the solenoid contributes anything to the circulation since the outer vertical segment must vanish, giving \[ \int_\mathcal{C} \mathbf{B} \cdot d\boldsymbol{\ell} = B \ell \ . \] But now this new loop contains an enclosed current consisting of the product of the current \(I\) with the number of coils \(n \ell\) passing through the enclosed surface of the loop, \(I_\text{enc} = n\ell I\). By Ampere’s law, inside the wire we thus have \[ B = \frac{4\pi}{c} n I \ . \] In summary then, the B-field of an infinitely long solenoid of radius \(a\) is thus \[ \mathbf{B}(\mathbf{x}) = \begin{cases} 4\pi n I/c \ \mathbf{e}_z & \varrho < a \ , \\ \mathbf{0} & \varrho \geq a \ . \end{cases} \] Evidently, the solenoid is able to produce a uniform B-field inside the coils. In this sense, the solenoid is to magnetostatics what the parallel plate capacitor is to electrostatics, a simple device for producing a uniform field inside a closed volume. This makes the solenoid a very useful and practical means for generating magnetic fields in the lab and in many engineering devices.

As we saw with the infinite sheet of current, the B-field changes discontinuously when crossing outside the solenoid. In fact, since \(K = nI\), the field strength also changes by exactly \(\Delta B = 4\pi K/c\). This is due to the unstated assumption that the coils are infinitely thin, which means we can effectively think of a solenoid as a hollow cylinder with surface density \(K = nI\).

Vector Potential

In electrostatics, we saw that we could always write the E-field as the gradient of the scalar potential. Something similar exists in magnetostatics. We cannot write the B-field as the gradient of a scalar potential, but we can write it as the curl of a vector potential. Indeed, the need for a vector potential leads to the many differences between electricity and magnetism.

Definition

In the previous section, we showed that we can express the B-field in terms of the curl of a vector potential, \[ \boxed{ \mathbf{B} = \nabla \times \mathbf{A} } \ , \] where the vector potential \(\mathbf{A}(\mathbf{x})\) is given by the integral \[ \boxed{ \mathbf{A}(\mathbf{x}) = \frac{1}{c} \int d^3 \mathbf{x}' \ \frac{\mathbf{J}(\mathbf{x}')}{|\mathbf{x} - \mathbf{x}'|} } \ . \] Notice that this is a vector integral, meaning that there will be three component integrals to calculate. Also, notice that \(\mathbf{A}\) will always point in the same direction as the current density \(\mathbf{J}\), not in the direction of the field itself.

Let’s compare this to the equivalent statement in electrostatics. In that setting, we found we could express the electric field as the gradient of a scalar potential, \[ \mathbf{E} = -\nabla \phi \ , \] where the scalar potential \(\phi(\mathbf{x})\) could be expressed via the integral \[ \phi(\mathbf{x}) = \int d^3 \mathbf{x}' \ \frac{\rho(\mathbf{x}')}{|\mathbf{x} - \mathbf{x}'|} \ . \] Notice the similarities and differences between these two settings. In electrostatics, the electric field is the gradient of a scalar potential, while in magnetostatics the magnetic field is the curl of a vector potential. In a sense, the differences between the two are what gives rise to the differences in behavior we see between electric and magnetic fields. But on another hand the two expressions for the potentials look quite similar, except with \(\rho\) replaced by \(\mathbf{J}/c\).

Since \(\rho\) and \(\mathbf{J}/c\) have the same dimensions, evidently the scalar potential and the vector potential must also have the same dimensions in Gaussian units, or units of statvolts. In practice though, we tend to express the units of vector potential in \(\text{G} \cdot \text{cm}\), which is equivalent to statvolts. In SI units, the vector potential and scalar potential unfortunately have different dimensions. In those units, the scalar potential has units of volts, while the vector potential has units of \(\text{T} \cdot \text{m}\).

Gauge Transformations

In electrostatics, we saw that we could add any constant to the scalar potential and it wouldn’t effect the E-field, which is the essential physical quantity of interest since the E-field is what gives rise to forces. A similar concept exists in magnetostatics, except it’s more subtle. What physically matters is the B-field, since it’s the B-field that gives rise to forces.

Suppose we have a vector potential \(\mathbf{A}\) whose magnetic field is \(\mathbf{B} = \nabla \times \mathbf{A}\). What operations can we do to \(\mathbf{A}\) so that \(\mathbf{B}\) itself doesn’t change? Clearly we can add any constant vector to \(\mathbf{A}\), but that’s not all. Generally speaking we can add anything we like to \(\mathbf{A}\) whose curl vanishes. The most general thing we can add would be the gradient of some scalar field. Suppose we transform \(\mathbf{A}\) by adding to it the gradient of some scalar field \(\chi(\mathbf{x})\), \[ \mathbf{A}' = \mathbf{A} + \nabla \chi \ . \] Since the curl of a gradient must always vanish we have \[ \mathbf{B} = \nabla \times \mathbf{A} = \nabla \times \mathbf{A}' \ . \] We call such a transformation \(\mathbf{A} \to \mathbf{A}'\) a gauge transformation, and the scalar field \(\chi(\mathbf{x})\) a gauge potential. Evidently, there is no physical information contained in the gauge potential. We’re free to choose any \(\chi\) we like without changing \(\mathbf{B}\).

An interesting consequence of this is that two vector potentials may look very different from each other, yet still give rise to the same B-field. When this happens, we say \(\mathbf{A}\) and \(\mathbf{A}'\) are gauge equivalent.

Since we’re free to choose or fix whatever gauge we wish, we might as well choose a gauge that simplifies our theory. Typically, when fixing a gauge potential in a given theory, for example in magnetostatics, we don’t specify a gauge potential directly like this. Instead, we specify a condition on the divergence of the vector potential. To see why this is equivalent, remember that \(\mathbf{A}\) is a vector field. From the Helmholtz theorem, we know we can uniquely specify any vector field in terms of its divergence and its curl. With the vector potential we specify the curl by requiring that \(\mathbf{B} = \nabla \times \mathbf{A}\), but we still have freedom to choose \(\nabla \cdot \mathbf{A}\).

Specifying the divergence of \(\mathbf{A}\) turns out to be equivalent to specifying a condition on the Laplacian of the gauge potential. Indeed, if \(\mathbf{A}' = \mathbf{A} + \nabla \chi\), then \(\nabla \cdot \mathbf{A}'\) must satisfy \[ \nabla \cdot \mathbf{A}' = \nabla \cdot \mathbf{A} + \nabla^2 \chi \ . \] If we now fix a gauge by requiring that \(\nabla \cdot \mathbf{A}' = \nabla \cdot \mathbf{A}\), what we’re left with is \(\nabla^2 \chi = 0\). Thus, fixing a gauge with the divergence of the vector potential is equivalent to requiring that any valid gauge potential satisfy Laplace’s equation.

In magnetostatics, we know that we can always just choose the vector potential given by the integral \[ \mathbf{A}(\mathbf{x}) = \frac{1}{c} \int d^3 \mathbf{x}' \ \frac{\mathbf{J}(\mathbf{x}')}{|\mathbf{x} - \mathbf{x}'|} \ . \] It’s thus natural in magnetostatics to fix a gauge whose divergence is the same as the divergence of this integral. But we showed in the previous section that the divergence of this integral is zero for any localized current distribution. This means the natural choice of gauge is one where the divergence of the vector potential is always zero. In magnetostatics, we thus choose \[ \boxed{ \nabla \cdot \mathbf{A} = 0 } \ . \] We call this choice of gauge the Coulomb gauge. We’ll see more on why the Coulomb gauge is a convenient choice for magnetostatics in the next section.

The gauge invariance of the vector potential makes it much more complicated to interpret physically what exactly the vector potential is. This wasn’t the case in electrostatics, where we found we could interpret the scalar potential naturally as the potential energy per unit charge of a distribution. The vector potential is much more subtle. Sometimes the vector potential can be thought of as a kind of momentum per unit charge, but not always. Despite its difficulty to physically interpret though, the vector potential still has great theoretical importance in the theory of electromagnetism, as we’ll see.

Poisson’s Equation

In electrostatics, we saw that we could combine the two first order field equations for the E-field into a single second order field equation for the scalar potential, \[ \begin{cases} \nabla \cdot \mathbf{E} = 4\pi\rho \ , \\ \nabla \times \mathbf{E} = \mathbf{0} \end{cases} \quad \Longrightarrow \quad \nabla^2 \phi = -4\pi\rho \ . \] In magnetostatics, we can also combine the two first order field equations for the B-field into a single second order field equation for the vector potential. Starting with the two field equations for the B-field, we have \[ \begin{cases} \nabla \cdot \mathbf{B} = 0 \ , \\ \nabla \times \mathbf{B} = \frac{4\pi}{c} \mathbf{J} \ . \end{cases} \] By writing \(\mathbf{B} = \nabla \times \mathbf{A}\), we automatically get the first field equation for free. All we need to do now is plug the vector potential into Ampere’s law. To do that, we’ll use the identity \[ \nabla \times \mathbf{B} = \nabla \times (\nabla \times \mathbf{A}) = \nabla(\nabla \cdot \mathbf{A}) - \nabla^2 \mathbf{A} \ . \] The left-hand side is just \(\nabla \times \mathbf{B}\), which by Ampere’s law is \(4\pi/c \ \mathbf{J}\). This means in general we have \[ \nabla^2 \mathbf{A} - \nabla(\nabla \cdot \mathbf{A}) = -\frac{4\pi}{c} \mathbf{J} \ . \] Now we can see why specifying the divergence of \(\mathbf{A}\) is a sensible way to fix a gauge, and why the Coulomb potential is so natural in magnetostatics. For if we fix \(\nabla(\nabla \cdot \mathbf{A})\), then the above equation reduces to \[ \boxed{ \nabla^2 \mathbf{A} = -\frac{4\pi}{c} \mathbf{J} } \ . \] This is just a Poisson equation in each component of \(\mathbf{A}\), which is analogous to what we saw in electrostatics where we had \[ \nabla^2 \phi = -4\pi\rho \ . \] In particular, this means the Green’s function for the vector potential (in the Coulomb gauge) is the same as the Green’s function for the scalar potential in electrostatics, \[ G(\mathbf{x} - \mathbf{x}') = \frac{1}{|\mathbf{x} - \mathbf{x}'|} \ . \] Any particular solution to Poisson’s equation will be the convolution of the Green’s function with \(\mathbf{J}/c\), which of course gives back the integral we derived before, \[ \mathbf{A}(\mathbf{x}) = \frac{1}{c} \int d^3 \mathbf{x}' \ \frac{\mathbf{J}(\mathbf{x}')}{|\mathbf{x} - \mathbf{x}'|} \ . \] For a given boundary value problem, the general solution will be this particular solution summed with any solution to the vector Laplace’s equation \(\nabla^2 \mathbf{A} = 0\) subject to any given boundary conditions.

Unfortunately, unlike Poisson’s equation for the scalar potential, solving Poisson’s equation directly for the vector potential is rarely useful as a means to solve a given problem. This is due to the fact that we now have a coupled vector differential equation to solve, which is much more difficult. The coupling comes from the fact that any solution must also satisfy the Coulomb gauge condition \(\nabla \cdot \mathbf{A} = 0\). Even worse, the boundary conditions for magnetostatics problems are often much less trivial. For instance, the B-field on the surface of a conductor need not vanish, nor is it physically realistic to fix a constant vector potential on the surface of a given material. We’ll cover some ways to calculate the vector potential using other methods below.

Surface Boundary Conditions

We saw in a few earlier examples in the chapter that the magnetic field across a surface of current changes discontinuously by an amount \(4\pi / c \ |\mathbf{K}|\). We will now prove this fact more generally. Specifically, we’ll show that only the tangential part of the B-field changes discontinuously when crossing a surface of current, while the normal part of the B-field remains continuous.

FIGURE (surface of current with normal and tangential B-field)

Suppose \(\mathcal{S}\) is some smooth surface of current with a surface current density \(\mathbf{K}(\mathbf{x})\), where \(\mathbf{x}\) is some point on the surface. The magnetic field \(\mathbf{B}(\mathbf{x})\) at this point can be decomposed into two vectors, one vector \(\mathbf{B}^\perp(\mathbf{x})\) normal to the surface, and another vector \(\mathbf{B}^\parallel(\mathbf{x})\) tangential to the surface, \[ \mathbf{B}(\mathbf{x}) = \mathbf{B}^\perp(\mathbf{x}) + \mathbf{B}^\parallel(\mathbf{x}) \ . \] Let’s first show the normal vector is continuous across the surface. Since \(\nabla \cdot \mathbf{B} = 0\), by the divergence theorem the surface integral of the B-field through any Gaussian surface must always satisfy \[ \oint \mathbf{B} \cdot d\mathbf{a} = 0 \ . \] We can now apply Gauss’s law exactly the same way we did with the E-field before. Choose an infinitesimally high and very thin pillbox with top and bottom areas \(\delta A\) as the Gaussian surface, where \(\delta A\) is so small that any deviations in the curvature of the surface \(\mathcal{S}\) are negligible inside the Gaussian surface. If \(B_+\) is the value of the B-field above the surface and \(B_-\) is the value of the B-field below the surface, by Gauss’s Law we must have \[ (B_+^\perp - B_-^\perp) \delta A \approx \int \mathbf{B} \cdot d\mathbf{a} = 0 \ . \] Assuming \(\delta A\) is infinitesimal, we thus have \(B_+^\perp(\mathbf{x}) = B_-^\perp(\mathbf{x})\). That is, the normal component of the B-field remains continuous across the surface of current.

Let’s now deal with the tangential vector. This time we’ll apply Ampere’s law. For the Amperian loop we choose a square with infinitesimal side lengths \(\delta\ell\), oriented with the face of the perpendicular to the direction of \(\mathbf{K}\) as shown below.

FIGURE (show amperian loop)

When we calculate the circulation integral, only the sides parallel to the surface contribute. Since \(I_{\text{enc}} \approx |\mathbf{K}| \delta\ell\), by Ampere’s law we thus have \[ (B_+^\parallel - B_-^\parallel) \delta\ell \approx \int \mathbf{B} \cdot d\boldsymbol{\ell} = \frac{4\pi}{c} |\mathbf{K}| \delta\ell \ . \] Assuming \(\delta\ell\) is infinitesimal, we thus have \[ B_+^\parallel(\mathbf{x}) - B_-^\parallel(\mathbf{x}) = \frac{4\pi}{c} |\mathbf{K}(\mathbf{x})| \ . \] That is, the tangential part of the B-field changes discontinuously across the surface by an amount \(\frac{4\pi}{c} |\mathbf{K}|\).

We can combine these results together into a single equation. Notice both conditions are equivalent to the statement that \[ \mathbf{B}_+(\mathbf{x}) - \mathbf{B}_-(\mathbf{x}) = \frac{4\pi}{c} \mathbf{K}(\mathbf{x}) \times \mathbf{n} \ . \] In a similar manner, one can show that the vector potential \(\mathbf{A}(\mathbf{x})\) across the surface is always continuous, both its tangential and normal parts. That is, for the vector potential we have \[ \mathbf{A}_+(\mathbf{x}) = \mathbf{A}_-(\mathbf{x}) \ . \] That the tangential part remains continuous follows from the fact that \(\mathbf{B} = \nabla \times \mathbf{A}\). The normal part is only continuous in the Coulomb gauge, since we require \(\nabla \cdot \mathbf{A} = 0\) for this to be true.

We can also express the discontinuity of the B-field across the surface using the normal derivative of the vector potential. We’ll define the normal derivative of the vector potential in terms of its components by \[ \frac{\partial A_i}{\partial n} \equiv \partial_j A_i n_j \ . \] One can then show that the B-field relations above are equivalent to the statement that \[ \frac{\partial}{\partial n} \mathbf{A}_+(\mathbf{x}) - \frac{\partial}{\partial n} \mathbf{A}_-(\mathbf{x}) = -\frac{4\pi}{c} \mathbf{K}(\mathbf{x}) \ . \] Together, we thus have the following boundary conditions for the vector potential across a surface of current, \[ \begin{align*} \mathbf{A}_+(\mathbf{x}) - \mathbf{A}_-(\mathbf{x}) &= \mathbf{0} \ , \\ \frac{\partial}{\partial n} \mathbf{A}_+(\mathbf{x}) - \frac{\partial}{\partial n} \mathbf{A}_-(\mathbf{x}) &= -\frac{4\pi}{c} \mathbf{K}(\mathbf{x}) \ . \end{align*} \] While these boundary conditions for the vector potential are nice to know, we rarely use them in practice since we rarely solve the vector Poisson equation for reasons described in the previous section.

Calculating Vector Potentials

Unlike in electrostatics, it’s typically much less helpful to solve problems in magnetostatics by first calculating the potential and then using that to calculate the B-field. The reason for this is that the integral for \(\mathbf{A}(\mathbf{x})\) is still a vector equation, which can often be as difficult to solve as finding \(\mathbf{B}(\mathbf{x})\) from the Biot-Savart law directly. Nevertheless, when an expression for the vector potential is needed, there are a few ways to calculate them. One way is to use the integral directly, \[ \mathbf{A}(\mathbf{x}) = \frac{1}{c} \int d^3 \mathbf{x}' \ \frac{\mathbf{J}(\mathbf{x}')}{|\mathbf{x} - \mathbf{x}'|} \ . \] If the magnetic field is already known, we can calculate the vector potential another way. In magnetostatics, we can write \[ \nabla^2 \mathbf{A} = -\nabla \times \mathbf{B} \ . \] Notice this equation looks like the Poisson equation \(\nabla^2 \phi = -4\pi\rho\) in electrostatics, except with three components. This means \(\nabla^2 \mathbf{A}\) must have the same Green’s function as we had in electrostatics, namely \(1/|\mathbf{x} - \mathbf{x}'|\). We know then that the solution of the equation \(\nabla^2 \mathbf{A} = -4\pi \mathbf{F}\) can be expressed as the convolution of \(\mathbf{F}\) with the Green’s function. If we set \(\mathbf{F} = 1/4\pi \ \nabla \times \mathbf{B}\), we thus immediate get that \[ \mathbf{A}(\mathbf{x}) = \frac{1}{4\pi} \int d^3\mathbf{x}' \ \frac{\nabla' \times \mathbf{B}(\mathbf{x}')}{|\mathbf{x} - \mathbf{x}'|} \ . \] This provides a way to calculate the vector potential of some current distribution if we already know the magnetic field. Note that this formula implicitly assumes that the current distribution is localized. If it’s infinite we need to do something else to find \(\mathbf{A}\). In these cases, we can instead calculate the circulation of \(\mathbf{A}\) around some Amperian loop. By Stoke’s theorem, we have \[ \oint_\mathcal{C} \mathbf{A} \cdot d\boldsymbol{\ell} = \int_\mathcal{S} (\nabla \times \mathbf{A}) \cdot d\mathbf{a} = \int_\mathcal{S} \mathbf{B} \cdot d\mathbf{a} \ . \] The integral on the right-hand side is just the magnetic flux through the surface \(\mathcal{S}\), which we’ll denote \(\Phi_B\), so we have \[ \oint_\mathcal{C} \mathbf{A} \cdot d\boldsymbol{\ell} = \Phi_B \ . \] Notice how similar this expression looks to Ampere’s law, except with \(\mathbf{B}\) replaced by \(\mathbf{A}\) and \(4\pi/c \ \mathbf{J}\) replaced by \(\Phi_B\). This means if the symmetry of a problem permits, we can often use the same tricks we used with Ampere’s law to find \(\mathbf{A}\).

Example: Vector potential of a uniform magnetic field

Suppose some current distribution gives rise to a uniform magnetic field \(\mathbf{B}\). We’d like to find the vector potential \(\mathbf{A}\). First, it’s clear that no localized current distribution can give rise to a constant magnetic field \(\mathbf{B}\), since the field would go to zero at infinity for any localized current distribution. We thus need to use the modified Ampere’s law to find \(\mathbf{A}\), \[ \oint_\mathcal{C} \mathbf{A} \cdot d\boldsymbol{\ell} = \Phi_B \ . \] Suppose the magnetic field is oriented in the positive \(z\)-direction, with \(\mathbf{B} = B \mathbf{e}_z\). This means \(\mathbf{A}\) must then be oriented in the \(xy\)-plane. In fact, it’s not too hard to see that \(\mathbf{A}\) must be azimuthal, with \(\mathbf{A} = A \mathbf{e}_\varphi\).

With this in mind, consider as an Amperian loop a counterclockwise circle of radius \(\varrho\) centered in the \(xy\)-plane. Along this loop the vector potential must be constant, and \(\Phi_B = \pi\varrho^2 B\). Thus, we get \[ \oint_\mathcal{C} \mathbf{A} \cdot d\boldsymbol{\ell} = 2\pi \varrho A = \pi\varrho^2 B \ . \] This means we have \[ \mathbf{A}(\mathbf{x}) = \frac{1}{2} \varrho B \mathbf{e}_\varphi = -\frac{1}{2} (\varrho\mathbf{e}_\varrho \times B\mathbf{e}_z) = -\frac{1}{2} \mathbf{x} \times \mathbf{B} \ . \] The vector potential of a uniform B-field can thus be expressed in the form \[ \mathbf{A}(\mathbf{x}) = -\frac{1}{2} \mathbf{x} \times \mathbf{B} \ . \] One can easily verify this gives the answer by confirming that \(\mathbf{B} = \nabla \times \mathbf{A}\). This is perhaps one of the few vector potentials worth actually remembering. It’s the magnetic analogue of the similar statement that \(\phi(\mathbf{x}) = -\mathbf{x} \cdot \mathbf{E}\) for a uniform E-field.

Example: Vector potential of a rotating charged sphere

Suppose a sphere of radius \(R\) is rotating with some constant angular velocity \(\boldsymbol{\omega}\) about some axis. On the sphere is a uniform surface charge density \(\sigma\). We’d like to calculate the vector potential outside this rotating sphere, when \(r > R\).

FIGURE (rotating sphere)

To calculate the vector potential, we need to evaluate the vector integral \[ \mathbf{A}(\mathbf{x}) = \frac{1}{c} \int da' \ \frac{\mathbf{K}(\mathbf{x}')}{|\mathbf{x} - \mathbf{x}'|} \ . \] Now, the area element in this case is \(da' = R^2 \sin\theta' d\theta'd\varphi'\), and since \(|\mathbf{x}'| = R\) we have \[ |\mathbf{x} - \mathbf{x}'| = \sqrt{r^2 + R^2 - 2 \mathbf{x} \cdot \mathbf{x}'} \ . \] By definition \(\mathbf{K}(\mathbf{x}') = \sigma \mathbf{v}'\). Since the sphere is rigid and rotating with the same angular velocity \(\boldsymbol{\omega}\), the velocity \(\mathbf{v}'\) is given by \[ \mathbf{v}' = \boldsymbol{\omega} \times \mathbf{x}' \ . \] Since the surface charge density \(\sigma\) is constant on the sphere, we thus have \[ \mathbf{A}(\mathbf{x}) = \frac{R^2\sigma}{c} \int_0^\pi \sin\theta' d\theta' \int_0^{2\pi} d\varphi' \ \frac{\boldsymbol{\omega} \times\mathbf{x}'}{\sqrt{r^2 + R^2 - 2 \mathbf{x} \cdot \mathbf{x}'}} \ . \] At this point we need to choose how we’d like to align the coordinate axes to perform the integration. While it seems natural to align axes such that the axis of rotation is the \(z\)-axis, it turns out to be much easier to evaluate the integral if we let \(\mathbf{x}\) lie on the \(z\)-axis instead, so that \[ \mathbf{x} \cdot \mathbf{x}' = rR \cos\theta' \ . \] But this still leaves a choice as to how to align the axes in the \(xy\)-plane. It’ll be most convenient to orient the \(xy\)-plane such that the axis of rotation lies in the \(xz\)-plane. If \(\boldsymbol{\omega}\) lies at a constant angle \(\gamma\) from the \(z\)-axis, we have \[ \boldsymbol{\omega} \times \mathbf{x}' = R\omega[-\cos\gamma\sin\theta'\sin\varphi' \mathbf{e}_x + (\cos\gamma\sin\theta'\cos\varphi' - \sin\gamma\cos\theta') \mathbf{e}_y + \sin\gamma\sin\theta'\sin\varphi' \mathbf{e}_z] \ . \] Now, notice that each component of \(\mathbf{A}\) containing a \(\sin\varphi'\) or \(\cos\varphi'\) will vanish when integrated over \(\varphi'\). The only component without such a term will be \(A_y\), which upon integrating over \(\varphi'\) reduces to \[ A_y(\mathbf{x}) = -\frac{2\pi R^3\omega\sigma}{c} \sin\gamma \int_0^\pi d\theta' \ \frac{\sin\theta'\cos\theta'}{\sqrt{r^2 + R^2 - 2 \cos\theta'}} \ . \] The remaining integral over \(\theta'\) can be done by substituting \(u = \cos\theta'\). For \(r > R\), the result will be \(2R/3r^2\). We thus have \[ A_y(\mathbf{x}) = -\frac{4\pi R^4\sigma}{3cr^2} \omega\sin\gamma \ . \] Now, notice that in our choice of coordinates we have \(\boldsymbol{\omega} \times \mathbf{x} = -\omega r \sin\gamma \mathbf{e}_y\). This means we can write the vector potential as \[ \mathbf{A}(\mathbf{x}) = \frac{4\pi R^4 \sigma}{3cr^3} \boldsymbol{\omega} \times \mathbf{x} \ . \] In this notation we can now easily rotate the coordinate axes if we like so that \(\boldsymbol{\omega}\) is along the \(z\)-axis. Then we have \[ \mathbf{A}(\mathbf{x}) = \frac{4\pi R^4 \sigma\omega}{3cr^2} \sin\theta\mathbf{e}_\varphi \ . \] Thus, the vector potential of a rotation sphere moves in the direction the sphere is rotating, and falls off like \(1/r^2\) outside the sphere. In fact, this is exactly the behavior of a magnetic dipole field, which we’ll study in the next section. From the multipole expansion we’ll soon derive, the magnetic dipole moment \(\mathbf{m}\) for this rotating sphere can easily be identified as \[ \mathbf{m} = \frac{4\pi R^4 \sigma}{3c} \boldsymbol{\omega} \ . \] One can also show the vector potential inside the sphere is the same, except multiplied by \(r^3/R^3\). This means the magnetic field inside the sphere will be constant and pointed along the axis of rotation.

Multipole Expansion

We saw that it is generally much more difficult to calculate the vector potential than it was the scalar potential in electrostatics. For this reason, it’s even more useful in magnetostatics than it was in electrostatics to find approximate solutions for the vector potential. We already know that one useful way to do that is to do a multipole expansion, which provides good approximations to the potential when far away from the source distribution. We’ll derive an expansion of this type for magnetostatics as well.

Derivation

Recall that in magnetostatics we can express the vector potential via the integral \[ \mathbf{A}(\mathbf{x}) = \frac{1}{c} \int d^3 \mathbf{x}' \ \frac{\mathbf{J}(\mathbf{x}')}{|\mathbf{x} - \mathbf{x}'|} \ . \] Since the Green’s function inside this integral is the same as the one in electrostatics, deriving this expansion will be fairly simple. Recall from the derivation of the multipole expansion in electrostatics that we can expand the Green’s function as \[ \frac{1}{|\mathbf{x} - \mathbf{x}'|} = \frac{1}{r} + \frac{\mathbf{x} \cdot \mathbf{x}'}{r^3} + \cdots \ . \] We only expanded up to the dipole term, but of course there will be a quadrupole term after that, and so on. Up to the dipole term then, each component \(A_i\) of the vector potential \(\mathbf{A}\) can be expanded in index notation as \[ A_i(\mathbf{x}) = \frac{1}{r} \int \frac{d^3 \mathbf{x}'}{c} \ J_i(\mathbf{x}') + \frac{x_j}{r^3} \int \frac{d^3 \mathbf{x}'}{c} \ x_j' J_i(\mathbf{x}') + \cdots \ . \] Notice that neither of these integrals depends on the field point \(\mathbf{x}\), only the source point \(\mathbf{x}'\). Similar to what we did with the multipole expansion in electrostatics, we call these integrals magnetic multipole moments. The first integral is evidently the magnetic monopole moment, and the second integral is evidently the magnetic dipole moment. If we included more terms, the next term would be the magnetic quadrupole moment, and so on.

We can calculate what those moments are most easily by first proving the following identity. For any two scalar fields \(f(\mathbf{x})\) and \(g(\mathbf{x})\) and any localized current density \(\mathbf{J}(\mathbf{x})\) in the Coulomb gauge, we must have \[ \int d^3 \mathbf{x} \ (f \mathbf{J} \cdot \nabla g + g \mathbf{J} \cdot \nabla f) = 0 \ . \] To prove this we’ll use integration by parts on the second term. Assuming the current is localized it must be zero at infinity, which means when we use integration by parts on the second term, the surface term will vanish leaving us with \[ \int d^3 \mathbf{x} \ g \mathbf{J} \cdot \nabla f = -\int d^3 \mathbf{x} \ (\nabla \cdot g \mathbf{J}) f \ . \] Finally, since \(\nabla \cdot g \mathbf{J} = \nabla g \cdot \mathbf{J} + g \nabla \cdot \mathbf{J}\) and \(\nabla \cdot \mathbf{J} = 0\) in the Coulomb gauge, the identity is proved.

For our uses below it’ll be helpful to write the above identity in index notation in terms of the source point \(\mathbf{x}'\), \[ \int d^3 \mathbf{x}' \ (f J_i \partial_i' g + g J_i \partial_i' f) = 0 \ . \] First, observe if we set \(f=1\) and \(g = x_i\), the second term vanishes and the first term reduces \(J_i\). We thus get \[ \int d^3 \mathbf{x}' \ \mathbf{J}(\mathbf{x}') = 0 \ . \] But these are just the components of the magnetic monopole moment. Thus, the vector potential has no monopole moment, and hence no monopole potential or field. This is a formal proof if we like of the statement that magnetic monopoles do not exist. It all follows directly from the fact that \(\nabla \cdot \mathbf{B} = 0\), which is what allows us to define the vector potential in the first place.

Magnetic Dipole Moment

Let’s now look at the of the dipole potential, which we can write in vector form as \[ \mathbf{A}^{(1)}(\mathbf{x}) = \frac{1}{cr^3} \int d^3\mathbf{x}' \ (\mathbf{x} \cdot \mathbf{x}') \mathbf{J}(\mathbf{x}') \ . \] We’ll find it most convenient to express this in index notation, \[ A_i^{(1)} = \frac{x_j}{cr^3} \int d^3 \mathbf{x}' \ x_j' J_i \ . \] If we instead set \(f = -x_i/2\) and \(g = -x_j/2\) in the above identity we get \[ -\frac{1}{2} \int d^3 \mathbf{x}' \ (x_i' J_j + x_j' J_i) = 0 \ . \]

Since this integral vanishes, we can add it to the dipole integral without changing its value. If we do that, we get \[ \int d^3 \mathbf{x}' \ x_j' J_i = \int d^3 \mathbf{x}' \ \bigg(x_j' J_i - \frac{1}{2} (x_i' J_j - x_j' J_i)\bigg) = -\frac{1}{2} \int d^3 \mathbf{x}' \ (x_i' J_j - x_j' J_i) \ . \] This integral is identical to the original one, just written differently. If we now add it to the integral in the dipole potential, we get \[ A_i^{(1)} = -\frac{x_j}{2cr^3} \int d^3 \mathbf{x}' \ (x_i' J_j - x_j' J_i) \ . \] Now, recall that any expression of the form \(x_i' J_j - x_j' J_i\) can be related to the cross product \(\mathbf{x}' \times \mathbf{J}\) by \[ x_i' J_j - x_j' J_i = \varepsilon_{ijk} (\mathbf{x}' \times \mathbf{J})_k \ . \] This means the dipole potential can also be rewritten as \[ A_i^{(1)} = -\frac{1}{2cr^3} \varepsilon_{ijk} x_j \int d^3 \mathbf{x}' \ (\mathbf{x}' \times \mathbf{J})_k \ . \] Finally, observe that \(\varepsilon_{ijk} x_j (\mathbf{x}' \times \mathbf{J})_k\) is just the \(i\)^th component of the triple product \(\mathbf{x} \times (\mathbf{x}' \times \mathbf{J})\). In vector notation, the dipole potential can thus evidently be written as \[ \mathbf{A}^{(1)}(\mathbf{x}) = \frac{\mathbf{x}}{r^3} \times \frac{1}{2c} \int d^3 \mathbf{x}' \ (\mathbf{J} \times \mathbf{x}') \ . \] From this, we can see that the magnetic dipole moment, usually denoted \(\mathbf{m}\), has the form

\[ \boxed{ \mathbf{m} \equiv \frac{1}{2c} \int d^3\mathbf{x}' \ \mathbf{x}' \times \mathbf{J}(\mathbf{x}') } \ . \]

From this definition it’s clear that the magnetic dipole moment has dimensions of charge times distance in Gaussian units, or units of \(\text{esu} \cdot \text{cm}\), the same dimension and units that the electric dipole moment has. Typically, the magnetic monopole moment is expressed in \(\text{erg} / \text{G}\) in Gaussian units, which is equivalent. In SI units, the magnetic monopole moment carries units of \(\text{A} \cdot \text{m}^2\).

In general, calculating the magnetic dipole moment can be challenging. But there’s one important case where it isn’t difficult to calculate, namely a planar loop of current.

Example: Dipole moment of a planar loop of constant current

Suppose a planar current loop with interior area \(A\) carries a constant current \(I\). We want to calculate the magnetic monopole moment of this current loop. Since the loop is 1-dimensional, the dipole moment will be given by

\[ \mathbf{m} = \frac{1}{2c} \oint_\mathcal{C} d\ell' \ \mathbf{x}' \times \mathbf{I}(\mathbf{x}') \ . \] We’ll orient the loop to lie in the \(xy\)-plane, which means both \(\mathbf{x}'\) and \(\mathbf{I}(\mathbf{x}')\) are also in the \(xy\)-plane.

Since the current along the loop is constant and \(d\mathbf{I}(\mathbf{x}') = I d\boldsymbol{\ell}'\), we can write \[ \mathbf{m} = \frac{I}{2c} \oint_\mathcal{C} \mathbf{x}' \times d\boldsymbol{\ell}' \ . \] Now, observe that \(\mathbf{x}' \times d\boldsymbol{\ell}'\) is just twice the area element \(d\mathbf{a}'\) inside the loop. Since the area \(A\) of the loop is given, we get \[ \mathbf{m} = \frac{I}{c} \int_\mathcal{S} d\mathbf{a}' = \frac{IA}{c} \mathbf{e}_z \ . \] This is the correct dipole moment so long as the loop lies in the \(xy\)-plane. In general, for a surface normal \(\mathbf{n}\) we’d have \[ \boxed{ \mathbf{m} = \frac{IA}{c} \mathbf{n} } \ . \] Thus, the dipole moment of a planar loop of current is proportional to the product of its current and its area, and points in the direction normal to the loop.

Example: Dipole moment of an orbiting point charge

Suppose a charged particle with mass \(m\) and charge \(q\) is orbiting about some center with velocity \(\mathbf{v}\). For instance, this could be an electron orbiting about the nucleus of an atom. We’d like to estimate the magnetic dipole moment of this orbiting charge.

If we model the orbiting charge as a planar loop of current, its magnetic dipole moment is given by the previous example, \[ \mathbf{m} = \frac{IA}{c} \mathbf{n} \ . \] We can estimate the current \(I\) as the charge per period \(\tau\) of orbit, \[ I = \frac{q}{\tau} \ . \] Note that this strictly speaking isn’t a steady current. It’s only on average a steady current, where we imagine the period \(\tau\) is much much smaller than the time period of observation. For small particles this is typically the case. For example, an electron orbiting a hydrogen atom has an orbital period of only \(\tau \sim 10^{-15} \ \text{s}\).

To calculate the area of orbit, we recall that an orbiting particle satisfies Kepler’s second law, which says \[ \frac{da}{dt} = \frac{|\mathbf{L}|}{2m} \ , \] where \(\mathbf{L} = |\mathbf{L}| \mathbf{n}\) is the particle’s angular momentum vector. Since angular momentum is conserved, \(\mathbf{L}\) will be constant. If we integrate \(da\) over a full period \(\tau\) we thus get the full area of the orbit, \[ A = \frac{|\mathbf{L}|}{2m} \tau \ . \] Thus, the magnetic moment of an orbiting point charge is evidently \[ \mathbf{m} = \frac{q}{2mc} \mathbf{L} \ . \] That is, the magnetic dipole moment of an orbiting point charge is proportional to its angular momentum, and points in the direction of the angular momentum vector. Note that this formula only holds in the non-relativistic limit \(|\mathbf{v}| \ll c\), since the classical mechanics assumptions we made need to be modified to include special relativity when \(|\mathbf{v}| \sim c\).

The proportionality constant \(\gamma \equiv q/2mc\) is usually given a name, the gyromagnetic ratio. The gyromagnetic ratio depends only on the mass and charge of the orbiting particle, not its position or velocity.

We see a version of this formula arise in quantum mechanics, where among other things we’re interested in the magnetic dipole moment of a spinning electron. Assuming a spinning particle has a spin angular momentum \(\mathbf{S}\), the above formula suggests that the magnetic dipole moment of a spinning electron might be \[ \mathbf{m} = -\frac{e}{2m_m c} \mathbf{S} \ , \] where \(-e\) is the charge of the electron and \(m_e\) is the mass of the electron. In fact, what we find in quantum mechanics is that the magnetic dipole moment of a spinning electron is approximately twice this value, \[ \mathbf{m} = -\frac{e}{m_m c} \mathbf{S} \ . \] The extra factor of two comes from relativistic quantum mechanics, which we won’t go into. In this setting, we typically define a different proportionality constant called the Bohr magneton \(\mu_B\) defined by \[ \mu_B \equiv \frac{e \hbar}{2 m_e c} \approx 9.274 \cdot 10^{-21} \ \text{erg} / \text{G} \ . \] Notice the inclusion of Planck’s constant \(\hbar\) in the Bohr magneton. This is because in quantum mechanics the spin is quantized in integer units of \(\hbar\), which means \(\mathbf{S} / \hbar\) is a dimensionless vector, hence \(\mu_B\) has the units of a magnetic dipole moment. We’d thus write the magnetic dipole moment of a spinning electron in quantum mechanics as \[ \mathbf{m} = -2 \mu_B \frac{\mathbf{S}}{\hbar} \ . \] At any rate, it’s interesting we could get this far by treating an orbiting charged particle as a classical loop of current.

Example: Dipole moment of a rotating disk of uniform charge

Suppose we have a disk with radius \(R\) and uniform surface charge density \(\sigma\) that’s made to rotate about its center with a constant angular velocity \(\boldsymbol{\omega}\) normal to the disk.

FIGURE (rotating disk)

Since the disk is 2-dimensional, the magnetic monopole moment will be given by the integral \[ \mathbf{m} = \frac{1}{2c} \int_\mathcal{S} da' \ \mathbf{x}' \times \mathbf{K}(\mathbf{x}') \ . \] Assuming the disk is centered in the \(xy\)-plane, then each point \(\mathbf{x}'\) in the disk will rotate with a velocity \[ \mathbf{v}' = \boldsymbol{\omega} \times \mathbf{x}' = \omega \varrho' \mathbf{e}_z \times \mathbf{e}_\varrho = \omega \varrho' \mathbf{e}_\varphi \ . \] Since the surface charge density on the disk is uniform, this means the surface current density \(\mathbf{K}(\mathbf{x}')\) will be \[ \mathbf{K}(\mathbf{x}') = \sigma \mathbf{v}' = \sigma \omega \varrho' \mathbf{e}_\varphi \ . \] This means \(\mathbf{x}' \times \mathbf{K}(\mathbf{x}') = \sigma\omega\varrho'^2 \mathbf{e}_z\), and so the magnetic monopole moment will be given by the integral \[ \mathbf{m} = \frac{\sigma\omega}{2c} \int_0^R \varrho' d\varrho' \int_0^{2\pi} d\varphi' \ \varrho'^2 \mathbf{e}_z = \frac{\pi R^4 \sigma\omega}{4c} \mathbf{e}_z \ . \] Incidentally, this same result could be found in a different way by making use of the result we found for the planar current loop in the previous example. The idea is to break the disk up into a bunch of differential rings of charge. For a given ring of radius \(\varrho'\), there will be a planar loop with differential current \(dI = K d\varrho' = \sigma\omega \varrho' d\varrho'\). Then the ring has a differential dipole moment \[ d\mathbf{m} = \frac{dI}{c} (\pi \varrho'^2) \mathbf{e}_z \ . \] Integrating \(d\mathbf{m}\) over the whole disk will then give the same answer obtained above.

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Up to the dipole term, the vector potential can evidently be expanded as \[ \mathbf{A}(\mathbf{x}) = \frac{\mathbf{m} \times \mathbf{x}}{r^3} + \cdots \ . \]

We could keep proceed to find the higher order magnetic multipole terms as well. The next component would be the quadrupole term, falling off like \(1/r^3\). But these higher moments are generally less useful in magnetism than in electricity, and far more complicated to write down, so we’ll stop at the dipole term.

Let’s compare this multipole expansion with the multipole expansion we found for the scalar potential in electrostatics, \[ \phi(\mathbf{x}) = \frac{q}{r} + \frac{\mathbf{p} \cdot \mathbf{x}}{r^3} + \cdots \ . \] First, as we noted before, the vector potential has no monopole moment. Second, the magnetic dipole term involves a cross product with the dipole moment, instead of a dot product with the dipole moment. This partly reflects the obvious fact that the vector potential is a vector, not a scalar. But it also reflects the fact that the electric dipole term is strongest in the direction of the dipole vector, while the magnetic dipole term is strongest in directions perpendicular to the dipole vector.

By taking the curl of the vector potential and manipulating terms, one can show that the B-field of the dipole potential can be expressed in the form \[ \mathbf{B}^{(1)}(\mathbf{x}) = \frac{3(\mathbf{m} \cdot \mathbf{x}) \mathbf{x} - r^2 \mathbf{m}}{r^5} \ . \] Interestingly, the dipole field has the same form for both the electric and magnetic fields. Indeed, this is why they’re called dipoles in both cases. The fields look exactly the same, even if the potentials do not. In particular, this means in analogy to electrostatics we could express the dipole field as the gradient of a scalar pseudo-potential \(\psi\), since \[ \mathbf{B}^{(1)} = -\nabla \frac{\mathbf{m} \cdot \mathbf{x}}{r^3} \equiv -\nabla \psi \ . \] It may seem strange that we can express the B-field as the gradient of a scalar potential. We know we can’t do this in general. The reason we can in this case is because we’re only considering field points outside the current distribution, and outside the current distribution we must have \(\nabla \times \mathbf{B} = \mathbf{0}\), which means \(\mathbf{B} = -\nabla \psi\) for some scalar field \(\psi\). In this special case then we can evidently introduce a scalar potential, even though in general we can’t. This is why we call it a pseudo-potential.

Just like the lowest non-vanishing moment of the scalar potential is independent of the choice of origin, the same is true of the vector potential. Since the magnetic monopole moment always vanishes, this means the magnetic dipole moment will always be independent of the choice of origin. If the dipole moment vanishes, the quadrupole will be independent of origin, and so on.

Currents in External Magnetic Fields

Thus far we’ve studied the multipole expansion of a localized current distribution in free space, assuming no external fields are present. Similar to what we did in electrostatics, we now want to ask how the potential energy of localized current distribution depends on the presence of an external B-field, particular its multipole moments. We’ll assume the external B-field is strong enough that we can neglect any self fields created by the current distribution of interest.

Suppose we have a localized current distribution \(\mathbf{J}(\mathbf{x}')\) about the origin, and in the background we’ll suppose there is an external B-field \(\mathbf{B}_{\text{ext}}(\mathbf{x})\) acting on this distribution. Assuming we can Taylor expand this field about the origin, we get \[ \mathbf{B}_{\text{ext}}(\mathbf{x}) = \mathbf{B}_{\text{ext}}(\mathbf{0}) + (\mathbf{x} \cdot \nabla) \mathbf{B}_{\text{ext}}(\mathbf{x}) \bigg|_{\mathbf{x} = \mathbf{0}} + \cdots \ . \] Now, this external field will create a force on the current distribution given by the Lorentz force law, \[ \mathbf{F} = \frac{1}{c} \int d^3\mathbf{x}' \ \mathbf{J}(\mathbf{x}') \times \mathbf{B}_{\text{ext}}(\mathbf{x}) \ , \] where \(\mathcal{V}\) is the enclosed volume of the current distribution. If we plug in the Taylor expansion above into the force law and group terms, we have to leading order \[ \begin{align*} \mathbf{F} &= \frac{1}{c} \int d^3\mathbf{x}' \ \mathbf{J}(\mathbf{x}') \times \bigg[\mathbf{B}_{\text{ext}}(\mathbf{0}) + (\mathbf{x} \cdot \nabla) \mathbf{B}_{\text{ext}}(\mathbf{x}) \bigg|_{\mathbf{x} = \mathbf{0}} + \cdots \bigg] \\ &\approx -\frac{1}{c} \mathbf{B}_{\text{ext}}(\mathbf{0}) \times \int d^3\mathbf{x}' \ \mathbf{J}(\mathbf{x}') + \frac{1}{c} \int d^3\mathbf{x}' \ \mathbf{J}(\mathbf{x}') \times(\mathbf{x} \cdot \nabla) \mathbf{B}_{\text{ext}}(\mathbf{x}) \bigg|_{\mathbf{x} = \mathbf{0}} + \cdots \ . \end{align*} \] Now, the first term will vanish, since if \(\mathbf{J}\) is a localized steady current we’ve already shown that \[ \int d^3\mathbf{x}' \ \mathbf{J}(\mathbf{x}') = \mathbf{0} \ . \] We’ll thus focus on the second term. To simplify this term it’ll be helpful to rewrite the force in index notation as \[ F_i \approx \frac{1}{c} \varepsilon_{ijk} \int d^3\mathbf{x}' \ J_j (x_\ell \partial_\ell) B_k = \varepsilon_{ijk} \bigg(\frac{1}{c} \int d^3\mathbf{x}' \ J_j x_\ell\bigg) \partial_\ell B_k \ , \] where \(B_k\) is the \(k\)^th component of \(\mathbf{B}_{\text{ext}}(\mathbf{x}) \big|_{\mathbf{x} = \mathbf{0}}\). Now, we’ve studied the integral in parenthesis already and showed that \[ \frac{1}{c} \int d^3\mathbf{x}' \ J_j x_\ell = \varepsilon_{\ell j n} m_n \ , \] where \(m_n\) is the \(n\)^th component of the magnetic dipole moment \(\mathbf{m}\). Plugging this in and using the identity to convert the contraction of two Levi-Civita symbols into delta functions, we have \[ \begin{align*} F_i &\approx \varepsilon_{ijk} \varepsilon_{\ell j n} m_n \partial_\ell B_k \\ &\approx \varepsilon_{ikj} \varepsilon_{j \ell n} m_n \partial_\ell B_k \\ &\approx (\delta_{i\ell} \delta_{kn} - \delta_{in} \delta_{k\ell}) m_n \partial_\ell B_k \\ &\approx m_k \partial_i B_k - m_i \partial_k B_k \ . \end{align*} \] Notice that the second term is proportional to \(\nabla \cdot \mathbf{B}_\text{ext} = 0\), so the second term vanishes. Since the dipole moment \(\mathbf{m}\) doesn’t depend on \(\mathbf{x}\), we can finally pull the derivative out of the product to get \[ \mathbf{F} \approx \nabla \big(\mathbf{m} \cdot \mathbf{B}_\text{ext}(0)\big) \ . \] We now want to convert this into an expansion for the potential energy \(U\). Since \(\mathbf{F} = -\nabla U\), we immediately get \[ U \approx - \mathbf{m} \cdot \mathbf{B}_\text{ext}(0) \ . \] Thus, to first order the potential energy of a current distribution in the presence of an external B-field is just the negative dot product of the magnetic dipole moment with the external field evaluated at the origin.

We can also calculate the torque exerted on the current distribution in the presence of an external B-field. By definition, the torque at each source point \(\mathbf{x}'\) is given by \(d\mathbf{N} = d(\mathbf{x}' \times \mathbf{F})\), which means the total torque is \[ \mathbf{N} = \frac{1}{c} \int d^3\mathbf{x}' \ \mathbf{x}' \times [\mathbf{J}(\mathbf{x}') \times \mathbf{B}_{\text{ext}}(\mathbf{x})] \ . \] If we again substitute in the Taylor expansion for \(\mathbf{B}_\text{ext}(\mathbf{x})\), to leading order we have \[ \mathbf{N} \approx \frac{1}{c} \int d^3\mathbf{x}' \ \mathbf{x}' \times [\mathbf{J}(\mathbf{x}') \times \mathbf{B}_{\text{ext}}(\mathbf{0})] \ . \] This time we only kept the first leading term, since as we’ll see this term will not vanish. We can simplify the triple product inside the integrand using the BAC-CAB rule, \[ \mathbf{x}' \times [\mathbf{J} \times \mathbf{B}_{\text{ext}}(\mathbf{0})] = (\mathbf{x}' \cdot \mathbf{B}_\text{ext}(0)) \mathbf{J} - (\mathbf{x}' \cdot \mathbf{J}) \mathbf{B}_\text{ext}(0) \ , \] which means we have \[ \mathbf{N} \approx \frac{1}{c} \int d^3\mathbf{x}' \ \big[(\mathbf{x}' \cdot \mathbf{B}_\text{ext}(0)) \mathbf{J} - (\mathbf{x}' \cdot \mathbf{J}) \mathbf{B}_\text{ext}(0) \big] \ . \]

• First term simplifies by prior argument, second term vanishes by prior argument. See Jackson.

\[ \mathbf{N} \approx \mathbf{m} \times \mathbf{B}_\text{ext}(0) \ . \]

Spherical Multipole Expansion

Since the Green’s function for the vector potential is the same as the one for the scalar potential, we can immediately write down the spherical multipole expansion for the vector potential in magnetostatics. Recall that we could expand the Green’s function as \[ \frac{1}{|\mathbf{x} - \mathbf{x}'|} = \frac{1}{r} \sum_{\ell=0}^\infty \frac{4\pi}{2\ell+1} \bigg(\frac{r'}{r}\bigg)^\ell \sum_{m=-\ell}^\ell Y_{\ell m}(\theta,\varphi) Y_{\ell m}^*(\theta',\varphi') \ . \] If we plug this expansion into the integral for the vector potential and group together the source-dependent terms, we get \[ \mathbf{A}(\mathbf{x}) = \frac{1}{c} \sum_{\ell=0}^\infty \sum_{m=-\ell}^\ell \frac{4\pi}{2\ell+1} \frac{Y_{\ell m}(\theta,\varphi)}{r^{\ell+1}} \int d^3 \mathbf{x}' \ \mathbf{J}(\mathbf{x}') r'^\ell Y_{\ell m}^*(\theta',\varphi') \ . \] The terms inside the integral depend only on the source current distribution. These are the magnetic spherical multipole moments, which we’ll denote by \(\mathbf{m}_{\ell m}\), \[ \mathbf{m}_{\ell m} \equiv \int d^3 \mathbf{x}' \ \mathbf{J}(\mathbf{x}') r'^\ell Y_{\ell m}^*(\theta',\varphi') \ . \] Unlike the spherical multipole moments \(q_{\ell m}\) in electrostatics, which are scalars, the spherical multipole moments \(\mathbf{m}_{\ell m}\) in magnetostatics are vectors, and hence will have three components per moment.

We can thus formally express the magnetic spherical multipole expansion of the vector potential via the sum \[ \mathbf{A}(\mathbf{x}) = \frac{1}{c} \sum_{\ell=0}^\infty \sum_{m=-\ell}^\ell \frac{4\pi}{2\ell+1} \frac{Y_{\ell m}(\theta,\varphi)}{r^{\ell+1}} \mathbf{m}_{\ell m} \ . \] One can show that the \(\ell=0,1\) expansion terms give the same results we found above in the Cartesian expansion.