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3. Magnetostatics
Charges give rise to electric fields. Current give rise to magnetic fields. In this section, we will study the magnetic fields induced by steady currents. This means that we are again looking for time independent solutions to the Maxwell equations. We will also restrict to situations in which the charge density vanishes, so ⇢ = 0. We can then set E = 0 and focus our attention only on the magnetic field. We’re left with two Maxwell equations to solve:
r ⇥ B = μ 0 J (3.1)
and
r · B = 0 (3.2)
If you fix the current density J, these equations have a unique solution. Our goal in this section is to find it.
Steady Currents
Before we solve (3.1) and (3.2), let’s pause to think about the kind of currents that we’re considering in this section. Because ⇢ = 0, there can’t be any net charge. But, of course, we still want charge to be moving! This means that we necessarily have both positive and negative charges which balance out at all points in space. Nonetheless, these charges can move so there is a current even though there is no net charge transport.
This may sound artificial, but in fact it’s exactly what happens in a typical wire. In that case, there is background of positive charge due to the lattice of ions in the metal. Meanwhile, the electrons are free to move. But they all move together so that at each point we still have ⇢ = 0. The continuity equation, which captures the conservation of electric charge, is
@⇢ @t
Since the charge density is unchanging (and, indeed, vanishing), we have
r · J = 0
Mathematically, this is just saying that if a current flows into some region of space, an equal current must flow out to avoid the build up of charge. Note that this is consistent with (3.1) since, for any vector field, r · (r ⇥ B) = 0.
3.1 Amp`ere’s Law
The first equation of magnetostatics,
r ⇥ B = μ 0 J (3.3)
is known as Amp`ere’s law. As with many of these vector dif-
J
S
C
Figure 25:
ferential equations, there is an equivalent form in terms of inte- grals. In this case, we choose some open surface S with boundary C = @S. Integrating (3.3) over the surface, we can use Stokes’ theorem to turn the integral of r ⇥ B into a line integral over the boundary C, Z
S
r ⇥ B · dS =
I
C
B · dr = μ (^0)
Z
S
J · dS
Recall that there’s an implicit orientation in these equations. The surface S comes with a normal vector ˆn which points away from S in one direction. The line integral around the boundary is then done in the right-handed sense, meaning that if you stick the thumb of your right hand in the direction ˆn then your fingers curl in the direction of the line integral.
The integral of the current density over the surface S is the same thing as the total current I that passes through S. Amp`ere’s law in integral form then reads I
C
B · dr = μ 0 I (3.4)
For most examples, this isn’t su cient to determine the form of the magnetic field; we’ll usually need to invoke (3.2) as well. However, there is one simple example where symmetry considerations mean that (3.4) is all we need...
3.1.1 A Long Straight Wire
Consider an infinite, straight wire carrying current I. We’ll take it to point in the ˆz direction. The symmetry of the problem is jumping up and down telling us that we need to use cylindrical polar coordinates, (r, ', z), where r =
p x 2 + y 2 is the radial distance away from the wire.
We take the open surface S to lie in the x y plane, centered on the wire. For the line integral in (3.4) to give something that doesn’t vanish, it’s clear that the magnetic field has to have some component that lies along the circumference of the disc.
B is oriented along the y direction. In fact, from the symmetry of the problem, it must look like
z
x
y (^) B
B
with B pointing in the ˆy direction when z > 0 and in the +ˆy direction when z < 0. We write
B = B(z)ˆy
with B(z) = B( z). We invoke Amp`ere’s law using the following open surface:
C
z
x
y
with length L in the y direction and extending to ±z. We have I
C
B · dr = LB(z) LB( z) = 2LB(z) = μ 0 KL
so we find that the magnetic field is constant above an infinite plane of surface current
B(z) = μ 0 K 2 z > 0
This is rather similar to the case of the electric field in the presence of an infinite plane of surface charge.
The analogy with electrostatics continues. The magnetic field is not continuous across a plane of surface current. We have
B(z! 0 +^ ) B(z! 0 ^ ) = μ 0 K
In fact, this is a general result that holds for any surface current K. We can prove this statement by using the same curve that we used in the Figure above and shrinking it
until it barely touches the surface on both sides. If the normal to the surface is ˆn and B (^) ± denotes the magnetic field on either side of the surface, then
ˆn ⇥ B| (^) + nˆ ⇥ B| (^) = μ 0 K (3.6)
Meanwhile, the magnetic field normal to the surface is continuous. (To see this, you can use a Gaussian pillbox, together with the other Maxwell equation r · B = 0).
When we looked at electric fields, we saw that the normal component was discontinu- ous in the presence of surface charge (2.9) while the tangential component is continuous. For magnetic fields, it’s the other way around: the tangential component is discontin- uous in the presence of surface currents.
A Solenoid
A solenoid consists of a surface current that travels around a cylin- B
z r
Figure 27:
der. It’s simplest to think of a single current-carrying wire winding many times around the outside of the cylinder. (Strictly speaking, the cross-sectional shape of the solenoid doesn’t have to be a circle – it can be anything. But we’ll stick with a circle here for simplicity). To make life easy, we’ll assume that the cylinder is infinitely long. This just means that we can neglect e↵ects due to the ends.
We’ll again use cylindrical polar coordinates, (r, ', z), with the axis of the cylinder along ˆz. By symmetry, we know that B will point along the z-axis. Its magnitude can depend only on the radial distance: B = B(r)ˆz. Once again, any magnetic field of this form immediately satisfies r · B = 0.
We solve Amp`ere’s law in di↵erential form. Anywhere other than
C
Figure 28:
the surface of the solenoid, we have J = 0 and
r ⇥ B = 0 )
dB dr = 0 ) B(r) = constant
Outside the solenoid, we must have B(r) = 0 since B(r) is constant and we know B(r)! 0 as r! 1. To figure out the magnetic field inside the solenoid, we turn to the integral form of Amp`ere’s law and consider the surface S, bounded by the curve C shown in the figure. Only the line that runs inside the solenoid contributes to the line integral. We have I
C
B · dr = BL = μ 0 IN L
3.2.1 Magnetic Monopoles
Above, we dispatched with the Maxwell equation r · B = 0 fairly quickly by writing B = r ⇥ A. But we never paused to think about what this equation is actually telling us. In fact, it has a very simple interpretation: it says that there are no magnetic charges. A point-like magnetic charge g would source the magnetic field, giving rise a 1 /r 2 fall-o↵
B = gˆr 4 ⇡r 2
An object with this behaviour is usually called a magnetic monopole. Maxwell’s equa- tions says that they don’t exist. And we have never found one in Nature.
However, we could ask: how robust is this conclusion? Are we sure that magnetic monopoles don’t exist? After all, it’s easy to adapt Maxwell’s equations to allow for presence of magnetic charges: we simply need to change (3.8) to read r · B = ⇢ (^) m where ⇢m is the magnetic charge distribution. Of course, this means that we no longer get to use the vector potential A. But is that such a big deal?
The twist comes when we turn to quantum mechanics. Because in quantum mechan- ics we’re obliged to use the vector potential A. Not only is the whole framework of electromagnetism in quantum mechanics based on writing things using A, but it turns out that there are experiments that actually detect certain properties of A that are lost when we compute B = r ⇥ A. I won’t explain the details here, but if you’re interested then look up the “Aharonov-Bohm e↵ect” in the lectures on Solid State Physics.
Monopoles After All?
To summarise, magnetic monopoles have never been observed. We have a law of physics (3.8) which says that they don’t exist. And when we turn to quantum mechanics we need to use the vector potential A which automatically means that (3.8) is true. It sounds like we should pretty much forget about magnetic monopoles, right?
Well, no. There are actually very good reasons to suspect that magnetic monopoles do exist. The most important part of the story is due to Dirac. He gave a beautiful argument which showed that it is in fact possible to introduce a vector potential A which allows for the presence of magnetic charge, but only if the magnetic charge g is related to the charge of the electron e by
ge = 2⇡~n n 2 Z (3.11)
This is known as the Dirac quantization condition.
Moreover, following work in the 1970s by ’t Hooft and Polyakov, we now realise that magnetic monopoles are ubiquitous in theories of particle physics. Our best current theory – the Standard Model – does not predict magnetic monopoles. But every theory that tries to go beyond the Standard Model, whether Grand Unified Theories, or String Theory or whatever, always ends up predicting that magnetic monopoles should exist. They’re one of the few predictions for new physics that nearly all theories agree upon.
These days most theoretical physicists think that magnetic monopoles probably exist and there have been a number of experiments around the world designed to detect them. However, while theoretically monopoles seem like a good bet, their future observational status is far from certain. We don’t know how heavy magnetic monopoles will be, but all evidence suggests that producing monopoles is beyond the capabilities of our current (or, indeed, future) particle accelerators. Our only hope is to discover some that Nature made for us, presumably when the Universe was much younger. Unfortunately, here too things seem against us. Our best theories of cosmology, in particular inflation, suggest that any monopoles that were created back in the Big Bang have long ago been diluted. At a guess, there are probably only a few floating around our entire observable Universe. The chances of one falling into our laps seem slim. But I hope I’m wrong.
3.2.2 Gauge Transformations
The choice of A in (3.9) is far from unique: there are lots of di↵erent vector potentials A that all give rise to the same magnetic field B. This is because the curl of a gradient is automatically zero. This means that we can always add any vector potential of the form r for some function and the magnetic field remains the same,
A 0 = A + r ) r ⇥ A 0 = r ⇥ A
Such a change of A is called a gauge transformation. As we will see in Section 5.3.1, it is closely tied to the possible shifts of the electrostatic potential . Ultimately, such gauge transformations play a key role in theoretical physics. But, for now, we’re simply going to use this to our advantage. Because, by picking a cunning choice of , it’s possible to simplify our quest for the magnetic field.
Claim: We can always find a gauge transformation such that A 0 satisfies r·A 0 = 0. Making this choice is usually referred to as Coulomb gauge.
Proof: Suppose that we’ve found some A which gives us the magnetic field that we want, so r ⇥ A = B, but when we take the divergence we get some function r · A = (x). We instead choose A 0 = A + r which now has divergence
r · A 0 = r · A + r 2 = + r 2
It’s worth giving a word of warning at this point: the expression r 2 A is simple in Cartesian coordinates where, as we’ve seen above, it reduces to the Laplacian on each component. But, in other coordinate systems, this is no longer true. The Laplacian now also acts on the basis vectors such as ˆr and ˆ'. So in these other coordinate systems, r 2 A is a little more of a mess. (You should probably use the identity r 2 A = r ⇥ (r ⇥ A) + r(r · A) if you really want to compute in these other coordinate systems).
Anyway, if we stick to Cartesian coordinates then everything is simple. In fact, the resulting equations (3.13) are of exactly the same form that we had to solve in electrostatics. And, in analogy to (2.21), we know how to write down the most general solution using Green’s functions. It is
Ai (x) = μ (^0) 4 ⇡
Z
V
d 3 x 0 J (^) i (x 0 ) |x x 0 |
Or, if you’re feeling bold, you can revert back to vector notation and write
A(x) =
μ (^0) 4 ⇡
Z
V
d 3 x 0
J(x 0 ) |x x 0 |
where you’ve just got to remember that the vector index on A links up with that on J (and not on x or x 0 ).
Checking Coulomb Gauge
We’ve derived a solution to (3.12), but this is only a solution to Amp`ere’s equation (3.10) if the resulting A obeys the Coulomb gauge condition, r · A = 0. Let’s now check that it does. We have
r · A(x) = μ (^0) 4 ⇡
Z
V
d 3 x 0 r ·
J(x 0 ) |x x 0 |
where you need to remember that the index of r is dotted with the index of J, but the derivative in r is acting on x, not on x 0. We can write
r · A(x) = μ (^0) 4 ⇡
Z
V
d 3 x 0 J(x 0 ) · r
|x x 0 |
μ (^0) 4 ⇡
Z
V
d 3 x 0 J(x 0 ) · r 0
|x x 0 |
Here we’ve done something clever. Now our r 0 is di↵erentiating with respect to x 0. To get this, we’ve used the fact that if you di↵erentiate 1/|x x 0 | with respect to x then
you get the negative of the result from di↵erentiating with respect to x 0. But since r 0 sits inside an
R
d 3 x 0 integral, it’s ripe for integrating by parts. This gives
r · A(x) = μ (^0) 4 ⇡
Z
V
d 3 x 0
r 0 ·
J(x 0 ) |x x 0 |
r 0 · J(x 0 )
|x x 0 |
The second term vanishes because we’re dealing with steady currents obeying r·J = 0. The first term also vanishes if we take the current to be localised in some region of space, Vˆ ⇢ V so that J(x) = 0 on the boundary @V. We’ll assume that this is the case. We conclude that
r · A = 0
and (3.14) is indeed the general solution to the Maxwell equations (3.1) and (3.2) as we’d hoped.
The Magnetic Field
From the solution (3.14), it is simple to compute the magnetic field B = r ⇥ A. Again, we need to remember that the r acts on the x in (3.14) rather than the x 0. We find
B(x) = μ (^0) 4 ⇡
Z
V
d 3 x 0 J(x 0 ) ⇥ (x x 0 ) |x x 0 | 3
This is known as the Biot-Savart law. It describes the magnetic field due to a general current density.
There is a slight variation on (3.15) which more often goes by the name of the Biot- Savart law. This arises if the current is restricted to a thin wire which traces out a curve C. Then, for a current density J passing through a small volume V , we write J V = (JA) x where A is the cross-sectional area of the wire and x lies tangent to C. Assuming that the cross-sectional area is constant throughout the wire, the current I = JA is also constant. The Biot-Savart law becomes
B(x) = μ 0 I 4 ⇡
Z
C
dx 0 ⇥ (x x 0 ) |x x 0 | 3
This describes the magnetic field due to the current I in the wire.
An Example: The Straight Wire Revisited
Of course, we already derived the answer for a straight wire in (3.5) without using this fancy vector potential technology. Before proceeding, we should quickly check that the Biot-Savart law reproduces our earlier result. As before, we’ll work in cylindrical polar
There is an integral expression for the linking number, first written down by Gauss during his exploration of electromagnetism. The Biot-Savart formula (3.16) o↵ers a simple physics derivation of Gauss’ expression. Suppose that the curve C carries a current H I. This sets us a magnetic field everywhere in space. We will then compute
C 0 B·dx^
(^0) around another curve C. (If you want a justification for computing H C 0 B·dx^
0
then you can think of it as the work done when transporting a magnetic monopole of unit charge around C, but this interpretation isn’t necessary for what follows.) The Biot-Savart formula gives I
C 0
B(x 0 ) · dx 0 = μ 0 I 4 ⇡
I
C 0
dx 0 ·
I
C
dx ⇥ (x 0 x) |x x 0 | 3
where we’ve changed our conventions somewhat from (3.16): now x labels coordinates on C while x 0 labels coordinates on C 0.
Meanwhile, we can also use Stokes’ theorem, followed by Amp`ere’s law, to write I
C 0
B(x 0 ) · dx 0 =
Z
S 0
(r ⇥ B) · dS = μ (^0)
Z
S 0
J · dS
where S 0 is a surface bounded by C 0. The current is carried by the other curve, C, which pierces S 0 precisely n times, so that I
C 0
B(x 0 ) · dx 0 = μ (^0)
Z
S 0
J · dS = nμ 0 I
Comparing the two equations above, we arrive at Gauss’ double-line integral expression for the linking number n,
n =
I
C 0
dx 0 ·
I
C
dx ⇥ (x 0 x) |x x 0 | 3
Note that our final expression is symmetric in C and C 0 , even though these two curves played a rather di↵erent physical role in the original definition, with C carrying a current, and C 0 the path traced by some hypothetical monopole. To see that the expression is indeed symmetric, note that the triple product can be thought of as the determinant det(x 0 , x, x 0 x). Swapping x and x 0 changes the order of the first two vectors and changes the sign of the third, leaving the determinant una↵ected.
The formula (3.17) is rather pretty. It’s not at all obvious that the right-hand-side doesn’t change under (non-crossing) deformations of C and C 0 ; nor is it obvious that the right-hand-side must give an integer. Yet both are true, as the derivation above shows. This is the first time that ideas of topology sneak into physics. It’s not the last.
3.3 Magnetic Dipoles
We’ve seen that the Maxwell equations forbid magnetic monopoles with a long-range B ⇠ 1 /r 2 fall-o↵ (3.11). So what is the generic fall-o↵ for some distribution of currents which are localised in a region of space? In this section we will see that, if you’re standing suitably far from the currents, you’ll typically observe a dipole-like magnetic field.
3.3.1 A Current Loop
We start with a specific, simple example. Consider
I
B
Figure 31:
a circular loop of wire C of radius R carrying a current I. We can guess what the magnetic field looks like simply by patching together our result for straight wires: it must roughly take the shape shown in the figure However, we can be more ac- curate. Here we restrict ourselves only to the mag- netic field far from the loop.
To compute the magnetic field far away, we won’t start with the Biot-Savart law but instead return to the original expression for A given in (3.14). We’re going to return to the notation in which a point in space is labelled as r rather than x. (This is more appropriate for long-distance distance fields which are essentially an expansion in r = |r|). The vector potential is then given by
A(r) = μ (^0) 4 ⇡
Z
V
d 3 r 0 J(r 0 ) |r r 0 |
Writing this in terms of the current I (rather than the current density J), we have
A(r) = μ 0 I 4 ⇡
I
C
dr 0 |r r 0 |
We want to ask what this looks like far from the loop. Just as we did for the electrostatic potential, we can Taylor expand the integrand using (2.22),
1 |r r 0 |
r
r · r 0 r 3
So that
A(r) = μ 0 I 4 ⇡
I
C
dr 0
r
r · r 0 r 3
This is our final, simple, answer for the long-range behaviour of the vector potential due to a current loop. It remains only to compute the magnetic field. A little algebra gives
B(r) = μ (^0) 4 ⇡
3(m · ˆr)ˆr m r 3
Now we see why m is called the magnetic dipole; this form of the magnetic field is exactly the same as the dipole electric field (2.19).
I stress that the B field due to a current loop and E field due to two charges don’t look the same close up. But they have identical “dipole” long-range fall-o↵s.
3.3.2 General Current Distributions
We can now perform the same kind of expansion for a general current distribution J localised within some region of space. We use the Taylor expansion (2.22) in the general form of the vector potential (3.14),
A (^) i (r) = μ (^0) 4 ⇡
Z
d 3 r 0 J (^) i (r 0 ) |r r 0 |
μ (^0) 4 ⇡
Z
d 3 r 0
J (^) i (r 0 ) r
J (^) i (r 0 ) (r · r 0 ) r 3
where we’re using a combination of vector and index notation to help remember how the indices on the left and right-hand sides match up.
The first term above vanishes. Heuristically, this is because currents can’t stop and end, they have to go around in loops. This means that the contribution from one part must be cancelled by the current somewhere else. To see this mathematically, we use the slightly odd identity
@ (^) j (J (^) j r (^) i ) = (@ (^) j J (^) j ) r (^) i + J (^) i = J (^) i (3.22)
where the last equality follows from the continuity condition r · J = 0. Using this, we see that the first term in (3.21) is a total derivative (of @/@r (^0) i rather than @/@ri ) which vanishes if we take the integral over R 3 and keep the current localised within some interior region.
For the second term in (3.21) we use a similar trick, now with the identity
@ (^) j (J (^) j r (^) i r (^) k ) = (@ (^) j J (^) j )r (^) i r (^) k + J (^) i r (^) k + J (^) k r (^) i = J (^) i r (^) k + J (^) k r (^) i
Because J in (3.21) is a function of r 0 , we actually need to apply this trick to the J (^) i r (^0) j terms in the expression. We once again abandon the boundary term to infinity.
Dropping the argument of J, we can use the identity above to write the relevant piece of the second term as Z d 3 r 0 J (^) i r (^) j r (^) j^0 =
Z
d 3 r 0 r (^) j 2 (J (^) i r (^) j^0 J (^) j r (^0) i ) =
Z
d 3 r 0
(J (^) i (r · r 0 ) r (^) i^0 (J · r))
But now this is in a form that is ripe for the vector product identity a ⇥ (b ⇥ c) = b(a · c) c(a · b). This means that we can rewrite this term as Z d 3 r 0 J (r · r 0 ) =
r ⇥
Z
d 3 r 0 J ⇥ r 0 (3.23)
With this in hand, we see that the long distance fall-o↵ of any current distribution again takes the dipole form (3.19)
A(r) = μ (^0) 4 ⇡
m ⇥ r r 3
now with the magnetic dipole moment given by the integral,
m =
Z
d 3 r 0 r 0 ⇥ J(r 0 ) (3.24)
Just as in the electric case, the multipole expansion continues to higher terms. This time you need to use vector spherical harmonics. Just as in the electric case, if you want further details then look in Jackson.
3.4 Magnetic Forces
We’ve seen that a current produces a magnetic field. But a current is simply moving charge. And we know from the Lorentz force law that a charge q moving with velocity v will experience a force
F = qv ⇥ B
This means that if a second current is placed somewhere in the neighbourhood of the first, then they will exert a force on one another. Our goal in this section is to figure out this force.
3.4.1 Force Between Currents
Let’s start simple. Take two parallel wires carrying currents I 1 and I 2 respectively. We’ll place them a distance d apart in the x direction.
Now we place a second current distribution J 2 in this magnetic field. It experiences a force per unit area given by (1.3), so the total force is
F =
Z
d 3 r J 2 (r) ⇥ B(r) (3.26)
Again, if the current J 2 is restricted to lie on a curve C 2 , then this volume integral can be replaced by the line integral
F = I 2
I
C (^2)
dr ⇥ B(r)
and the force can now be expressed as a double line integral,
F =
μ (^0) 4 ⇡
I 1 I 2
I
C (^1)
I
C (^2)
dr 2 ⇥
dr 1 ⇥ r 2 r (^1) |r 2 r 1 | 3
In general, this integral will be quite tricky to perform. However, if the currents are localised, and well-separated, there is a somewhat better approach where the force can be expressed purely in terms of the dipole moment of the current.
3.4.2 Force and Energy for a Dipole
We start by asking a slightly di↵erent question. We’ll forget about the second current and just focus on the first: call it J(r). We’ll place this current distribution in a magnetic field B(r) and ask: what force does it feel?
In general, there will be two kinds of forces. There will be a force on the centre of mass of the current distribution, which will make it move. There will also be a torque on the current distribution, which will want to make it re-orient itself with respect to the magnetic field. Here we’re going to focus on the former. Rather remarkably, we’ll see that we get the answer to the latter for free!
The Lorentz force experienced by the current distribution is
F =
Z
V
d 3 r J(r) ⇥ B(r)
We’re going to assume that the current is localised in some small region r = R and that the magnetic field B varies only slowly in this region. This allows us to Taylor expand
B(r) = B(R) + (r · r)B(R) +...
We then get the expression for the force
F = B(R) ⇥
Z
V
d 3 r J(r) +
Z
V
d 3 r J(r) ⇥ [(r · r)B(R)] +...
The first term vanishes because the currents have to go around in loops; we’ve already seen a proof of this following equation (3.21). We’re going to do some fiddly manipula- tions with the second term. To help us remember that the derivative r is acting on B, which is then evaluated at R, we’ll introduce a dummy variable r 0 and write the force as
F =
Z
V
d 3 r J(r) ⇥ [(r · r 0 )B(r 0 )]
r 0 =R
Now we want to play around with this. First, using the fact that r ⇥ B = 0 in the vicinity of the second current, we’re going to show, that we can rewrite the integrand as
J(r) ⇥ [(r · r 0 )B(r 0 )] = r 0 ⇥ [(r · B(r 0 ))J(r)]
To see why this is true, it’s simplest to rewrite it in index notation. After shu✏ing a couple of indices, what we want to show is:
✏ (^) ijk J (^) j (r) r (^) l @ (^) l^0 B (^) k (r 0 ) = ✏ (^) ijk J (^) j (r) r (^) l @ (^0) k B (^) l (r 0 )
Or, subtracting one from the other,
✏ (^) ijk J (^) j (r) r (^) l (@ (^) l^0 B (^) k (r 0 ) @ (^) k^0 B (^) l (r 0 )) = 0
But the terms in the brackets are the components of r ⇥ B and so vanish. So our result is true and we can rewrite the force (3.27) as
F = r 0 ⇥
Z
V
d 3 r (r · B(r 0 )) J(r)
(^) r 0 =R
Now we need to manipulate this a little more. We make use of the identity (3.23) where we replace the constant vector by B. Thus, up to some relabelling, (3.23) is the same as Z
V
d 3 r (B · r)J =
B ⇥
Z
V
d 3 r J ⇥ r = B ⇥ m
where m is the magnetic dipole moment of the current distribution. Suddenly, our expression for the force is looking much nicer: it reads
F = r ⇥ (B ⇥ m)
