11. Ch. 5, part 1

Lecture 11 - Basic Magnetism

There are no free magnetic charges in our part of the universe. The sources of magnetic fields, down to the atomic scale, are electrical currents. A small current loop appears from a distance as a magnetic dipole. Thus orbiting electrons give to an atom a magnetic dipole associated with its orbital angular momentum. In addition, subatomic particles also have magnetic dipoles. The magnetic dipole of nuclei comes in part from the proton currents within, and the magnetic dipole of a proton (and other baryons, and mesons) comes in part from quark currents within. In addition, the electron and the up and down quarks also have an “intrinsic” magnetic moment associated with their spin, although they are supposed to be point-like; the heavier leptons and quarks have one too. In a naive classical picture, the electron continually goes in circles around itself. The Dirac equation endows the electron with a spin S = \frac{1}{2}h̄ and a magnetic moment μ_{B} = eh̄/2mc. Both arise from quantum relativistic effects that are built into Dirac’s theory. Beyond that, quantum electrodynamics predicts that the intrinsic magnetic moment of the electron is actually μ = 1.00116 μ_{B}, in stunning agreement with experiment. Generally, the magnetic dipole moments of elementary particles are understood much better than their masses; but the last word has not been said.

To begin with, we discuss the magnetic field associated with steady currents, and we also postulate the existence of point-like magnetic dipoles that are of pure quantum origin. The “orbital” dipoles actually have atomic or nuclear sizes and can be described classically as tiny current loops, except that the associated angular momentum is quantized in units of h̄. The “intrinsic” moments of elementary particles may be truly point-like and can only be understood in a relativistic quantum theory.

11.1. Steady currents

Currents consist of a flow of particles and thus, classically, cannot really be steady. In quantum theory there are stationary states that have steady (time-independent) current densities. Thermally-averaged currents can also be steady, both classically and quantally.

The current density J is related to the charge density ρ by the continuity equation

\frac{∂ρ}{∂t} + ∇ · J = 0

In a steady state ρ cannot depend on t and consequently ∇ · J = 0, at least in the region under consideration. (There may be external current sources that are described as boundary conditions on J). A current field, or any vector field, that has zero divergence is said to be transverse. A reason for this name is that if we Fourier-expand

J(x) = ∫ exp(ik · x) J(k) d^{3}k/(2π)^{3}

we find that ∇ · J = 0 implies that k · J(k) = 0.

To find J within the region of interest, we also need an equation for ∇ × J or the equivalent. For particles streaming in empty space, we can find J(x) by solving the equations of motion of the individual particles, or the equivalent hydrodynamic equations for the particle stream. For currents in material media, we can use the simple local relation (Ohm’s law):

J(x) = σ^{(e)}(x)E(x)

where σ^{(e)} is the electrical conductivity. This relation holds in a conductor, which is a medium where a steady current flows indefinitely under the action of a steady field. We recall that in an insulator J = χ^{(e)}dE/dt as a consequence of J = dP/dt and P = χ^{(e)}E, so that we only have transient currents. Of course, there can be transient currents in a conductor as well; we will come back later to Ohm’s law and the polarizability in general time-dependent fields.

Here we note only that the equations

∇ × E = 0  ∇ · J = 0  J = σ^{(e)}E  (11.1)

can be used to find J and E in conductors in a similar way as

∇ × E = 0  ∇ · D = 4πρ  D = ϵ E

are used in to find D and E in insulators. To be complete, the set (11.1) must be supplemented in two ways:

• There must be sources, or boundary conditions that are in effect sources. Most commonly the normal component J_{n} is assigned on part of the outer boundary to describe current entering and exiting through “leads”.

• At the interface between different conductors J_{n} and E_{∥} are continuous, and correspondingly E_{n} and J_{∥} are discontinuous. The discontinuity of E_{n} means that a surface charge accumulates at the interface. There can be regions of empty space where σ^{(e)} = 0, hence J = 0; in these regions ∇ · J = 0 is replaced by ∇ · E = 4πρ.

There is then a whole new set of boundary value problems dealing with current distribution in conductors, and an even wider set where both conductors and dielectrics are present. Not to mention “lossy” dielectrics where ϵ ≠ 1 and at the same time there is a finite conductivity σ^{(e)}. All these problems have practical importance, but Jackson mercifully ignores them. Some of the solutions obtained earlier can simply be transcribed. For instance, two-dimensional electrostatic problems become problems of conduction in a thin metal sheet or other conducting layer.

11.1.1. Conductance and resistance

The simplest and most common problem is that of a wire or rod of length d and constant cross section A. The current is I = JA and the electric field is E = V/d, where V is the potential difference between the ends of the wire. Thus I = (σ^{(e)}A/d) V and the conductance is σ^{(e)}A/d. This formula is analogous to C = ϵA/4πd for the capacitance of a parallel-plate capacitor. The inverse of the conductance is the resistance

R ≡ \frac{V}{I} = \frac{d}{σ^{(e)}A} = \frac{ρ^{(e)}d}{A}

where we have defined the resistivity ρ^{(e)} = 1/σ^{(e)}.

In the gaussian system, conductance has the dimensions of velocity and units cm/s; resistance has units s/cm. In the MKSA system the unit of resistance is the ohm

1 ohm = \frac{volt}{amp} = \frac{1}{9 · 10^{11}} \frac{s}{cm}

Resistivity is commonly expressed in ohm · cm rather than seconds, and conductivity in (ohm · cm)^{−1}. Representative values of σ^{(e)} in these units are 64 · 10^{4} for Cu and around 1 (but strongly dependent on doping) for Si, both at 0^{∘}C.

The product RC has dimensions of time in any system, and units of seconds in the MKSA and gaussian systems.

There is a perfect mathematical correspondence between conductance and capacitance, with J ↔ D and σ^{(e)} ↔ ϵ, but there is a major difference in the fact that J is confined to conductors, while D spills into insulators and the vacuum, where ϵ = 1. There is a conductance matrix corresponding to C_{ij} and all that. For instance:

• a spherical conductor of radius a in a medium of conductivity σ^{(e)} has conductanceto-infinity σ^{(e)}a

• the resistance between two perfectly conducting cylinders of radius a at opposite potentials, in a medium of conductivity σ^{(e)}, is

\frac{4}{σ^{(e)}} ln(d/a)

per unit length, where d > 2a is the distance between the cylinders. This result also applies to the resistance between two circular “leads” in a plate of thickness w, if we replace σ^{(e)} by σ^{(e)}w.

11.2. Currents and magnetic forces

Experimentally, the basic fact is that there is a force between two current-carrying wires. Ampère found that for two long parallel wires, separated by a distance d, the force per unit length of wire is

\frac{dF}{dl} = −\frac{2}{c^{2}} \frac{I_{1}I_{2}}{d}

where the - sign indicates that the force is attractive when the currents flow in the same direction. This is reminiscent of Coulomb’s law for the force between two wires carrying charge densities λ_{1} and λ_{2}

\frac{dF}{dl} = 2 \frac{λ_{1}λ_{2}}{d}

but the sign is opposite. The constant c has the dimensions of velocity and, amazingly, turns out to be the speed of light. If the current I is ascribed to a charge density λ moving with speed v, the ratio of the Ampère force to the Coulomb force is (v/c)^{2}. What Ampère discovered is a relativistic effect. It is not always a small effect compared to electric forces because the density of positive charge λ^{(+)} nearly compensates the density of negative charge −λ^{(−)} (we take both λ^{(−)} and λ^{(−)} as positive). Coulomb’s law involves the net charge λ^{(+)} − λ^{(−)}, while the current is I = λ^{(+)}v^{(+)} − λ^{(−)}v^{(−)}. For a steady current in a conductor, there is no net macroscopic charge, λ^{(+)} = λ^{(−)}, and only the electrons move, so that I = −λ^{(−)}v^{(−)}.

Ampère also found that two long current-carrying wires at right angles exert no force on each other, quite unlike the Coulomb case. By fiddling with two current loops 1 and 2 he found that in general

F_{on 1 by 2} = −\frac{I_{1}I_{2}}{c^{2}} ∮ ∮ \frac{x_{1} − x_{2}}{|x_{1} − x_{2}|^{3}} dl_{1}·dl_{2}  (11.2)

It is not obvious how this action-at-a-distance law can be written in terms of a force field. A clue came from earlier observations by Biot and Savart who had seen that a compass needle is deflected by a nearby current. ————————————

In the MKSA system the Ampère and Coulomb force between parallel wires are

Ampère: (\frac{dF}{dl} )_{mag} = −2 \frac{μ_{0}}{4π} \frac{I_{1}I_{2}}{d}

Coulomb: (\frac{dF}{dl} )_{el} = 2 \frac{1}{4πϵ_{0}} \frac{λ_{1}λ_{2}}{d}

The ratio of these forces must be the same in the MKSA and gaussian systems, since charges cancel out in the combination I_{1}I_{2}/λ_{1}λ_{2}. Thus

ϵ_{0}μ_{0} = \frac{1}{c^{2}}

\frac{μ_{0}}{4π} = \frac{1}{4πϵ_{0}c^{2}} = \frac{9 · 10^{9}}{(3 · 10^{8})^{2}} \begin{matrix}nt · s & 2 \\ coul & 2 \end{matrix} = 10^{−7} \frac{nt}{amp^{2}}

————————————

11.3. The magnetic field

It was known since Gilbert that two magnetized bars, or needles, exert on each other forces and torques entirely analogous to those between two electric dipoles. That suggests that a magnetized thin bar or needle has a magnetic dipole moment m and generates a magnetic field

B(x) = −∇ (\frac{m · x}{|x|^{3}})

(when the bar is at x = 0). And the energy of a dipole m in the field B is −B · m. These relations can be used to define the standard unit dipole and the units of B, in the same way as for Coulomb’s law. However, these definitions have now been replaced by others , as discussed below.

Comparing the forces exerted by currents with those exerted by magnetic bars, Biot and Savart found that at a distance R from a long straight wire carrying a current I there is a magnetic field of magnitude

B = \frac{2I}{cR}

and direction given by the right-hand rule (Jackson Fig. 5.2). Vectorially one can write B = (I × R) /cR^{2}.

For a general current loop the Biot-Savart law is

B(x_{1}) = \frac{I_{2}}{c} ∮ \frac{dl_{2}× (x_{1} − x_{2})}{|x_{1} − x_{2}|^{3}}  (11.3)

and this suggests that one could write the Ampère force law as

F_{on 1 by 2} = \frac{I_{1}}{c} ∮ dl_{1}×B(x_{1})

or more explicitly

F_{on 1 by 2} = \frac{I_{1}I_{2}}{c^{2}} ∮ ∮ dl_{1}×\frac{dl_{2}× (x_{1} − x_{2})}{|x_{1} − x_{2}|^{3}}

It is a simple matter to get back eq. (11.2) from this by using

dl_{1}× [dl_{2}× (x_{1} − x_{2})] = − (dl_{1}·dl_{2}) (x_{1} − x_{2}) + dl_{2} [dl_{1} · (x_{1} − x_{2})]

and noting that

∮ dl_{1} · \frac{x_{1} − x_{2}}{|x_{1} − x_{2}|^{3}} = − ∮ dl_{1} · ∇\frac{1}{|x_{1} − x_{2}|} = 0

More generally, the magnetic force acting on a current distribution J is

F = \frac{1}{c} ∫ J(x) × B(x) d^{3}x

11.3.1. Units and dimensions

In the gaussian system, the dimensions of B are M^{1/2}L^{−1/2}T^{−1}, same as those of E, but the unit of B is called the gauss (not the statvolt/cm). In the MKSA system, the dimensions of B are MT^{−1}C^{−1} and the unit of B is the tesla

1 tesla = 10^{4} gauss = 10 kilogauss

The Earth’s field on its surface is around 0.5 gauss, but interstellar fields are of the order of 10^{−5} gauss. A typical laboratory field is a few kgauss; the strongest available continuous field is 60 tesla. Neutron stars are believed to have fields of several Megagauss.

11.4. Differential equations for the B field

The basic equation (11.3) for the magnetic field for any current distribution J(x) is

B(x) = \frac{1}{c} ∫ J(x′) × \frac{x − x′}{|x − x′|^{3}} d^{3}x′

B(x) = \frac{1}{c}∇× ∫ \frac{J(x′)}{|x − x′|} d^{3}x′

which are the analog of

E(x) = ∫ ρ(x′)\frac{x − x′}{|x − x′|^{3}} d^{3}x′

E(x) = −∇ ∫ \frac{ρ(x′)}{|x − x′|} d^{3}x′

It follows that the analog (so to speak) of ∇ × E = 0 is

∇ · B = 0

and (less obviously) the analog of ∇ · E = 4πρ is

∇ × B = \frac{4π}{c} J

This is called Ampère’s law. By using Stokes’ theorem it can be transformed into an integral relation, which is the analog of Gauss’s law (see Jackson Fig. 5.4)

∮_{C} B·dl =\frac{4π}{c} ∫_{S} J · n da = \frac{4π}{c} I

This is Ampère’s law, in integrated form. There is some possible confusion in these names, considering that there is also an Ampère’s law for the force between currents.

—————————— Derivation of Ampère’s law

∇× (∇× ∫ \frac{J(x′)}{|x − x′|} d^{3}x′) = ∇ (∇· ∫ \frac{J(x′)}{|x − x′|} d^{3}x′) − ∇^{2} ∫ \frac{J(x′)}{|x − x′|} d^{3}x′

∇· ∫ \frac{J(x′)}{|x − x′|} d^{3}x′ = ∫ J(x′) · ∇\frac{1}{|x − x′|} d^{3}x′

Replace ∇\frac{1}{|x − x′|} with − ∇′ \frac{1}{|x − x′|} and integrate by parts

∫ J(x′) · ∇\frac{1}{|x − x′|} d^{3}x′ = ∫ \frac{∇′ · J(x′)}{|x − x′|} d^{3}x′ = 0

Hence we are left with

∇× (∇× ∫ \frac{J(x′)}{|x − x′|} d^{3}x′) = − ∫ J(x′)∇^{2} \frac{1}{|x − x′|} d^{3}x′ = 4πJ(x′)

———————————

11.5. The vector potential

Just as the equation ∇ × E = 0 has the general solution E = −∇Φ, so the equation ∇ · B = 0 has the general solution

B = ∇ × A

where A is the vector potential. The analog of ∇^{2}Φ = −4πρ is

∇× (∇ × A) = \frac{4π}{c} J

or

∇ (∇ · A) − ∇^{2}A = \frac{4π}{c} J  (11.4)

This equation can be further simplified (see next section) but in any case A, being a vector, is not as convenient to work with as Φ. Boundary value problems in magnetostatics are harder than those in electrostatics.

The equations for the B field in a current-free region

∇ · B = 0
∇ × B = 0

are the same as those for the E field in a charge-free region. Thus in many problems we can introduce a magnetic scalar potential by

B = −∇Φ_{M}

and reduce the problem to solving

∇^{2}Φ_{M} = 0

subject to the appropriate boundary conditions. This is generally much easier than solving for A. However, some authors regard the use of Φ_{M} as wrong in principle and even immoral.

11.5.1. Gauge invariance

The vector potential is not unique, because for any Λ(x) we obtain the same B from A and from

A′ = A + ∇Λ  (11.5)

This freedom of adding an arbitrary gradient to A is called gauge invariance. It has a deep meaning that can only be fully appreciated in quantum theory. It allows us to simplify the equation for A by imposing one additional condition that can in principle be satisfied by an appropriate choice of Λ. For instance, in the infinite domain the exact expression

B(x) = \frac{1}{c}∇× ∫ \frac{J(x′)}{|x − x′|} d^{3}x′

suggests taking for A the particular choice

A_{C}(x) = \frac{1}{c} ∫ \frac{J(x′)}{|x − x′|} d^{3}x′  (11.6)

which is the direct analog of the Coulomb potential,

Φ(x) = ∫ \frac{ρ(x′)}{|x − x′|} d^{3}x′

It is easily checked that the explicit solution (11.6) obeys the condition

∇ · A_{C} = 0

More generally, gauge invariance allows us to impose ∇ · A_{C} = 0 as an additional condition, thus restricting A to the Coulomb gauge (which in the time-independent case is the same as another popular gauge, the Lorentz gauge). In this gauge eq. (11.4) reduces to

∇^{2}A_{C} = − \frac{4π}{c} J  (11.7)

which is the close analog of the Poisson equation ∇^{2}Φ = −4πρ. In general the choice of a suitable gauge condition can be advantageous. For instance, in Cartesian coordinates ∇^{2}A = − (4π/c) J gives three separate Poisson equations for A_{x}, A_{y}, A_{z}. One can get general solutions for each component and then impose ∇ · A = 0. Often there are shortcuts.

11.5.2. Two-dimensional problems

When J has only the component J_{z}(x, y) (and also, as we will see later, when the boundaries of magnetic media are parallel to the z axis), the equations for A reduce to

(\frac{∂^{2}}{∂x^{2}} + \frac{∂^{2}}{∂y^{2}} ) A_{z} = −4πJ_{z}

We are automatically in the Coulomb gauge, since ∇ · A = ∂A_{z}/∂z = 0. We drop the subscript _{z} and use z to denote x + iy, since complex variable methods are very useful. The field components are

B_{x} = \frac{∂A}{∂y} B_{y} = −\frac{∂A}{∂x}

We can also introduce the scalar magnetic potential Φ_{M} in terms of which

B_{x} = −\frac{∂Φ_{M}}{∂x} B_{y} = −\frac{∂Φ_{M}}{∂y}

We look for a complex potential F (z) that has singularities corresponding to the points where J ≠ 0. The above equations are consistent with putting

F = A + iΦ_{M}

or −iF = Φ_{M} − iA. As in the electric case,

B_{x} − iB_{y} = −\frac{d}{dz} (−iF) = i\frac{dF}{dz}

This is now more conveniently written

B_{y} + iB_{x} = −\frac{dF}{dz}

As an example, for a line current F = −2(I/c) ln z, and A = −2(I/c) ln ρ, while Φ_{M} = 2Iϕ/c. This gives

−\frac{dF}{dz} = \frac{2I}{cz} = \frac{2I}{c} \frac{x − iy}{ρ^{2}}

B_{y} = \frac{2I}{c} \frac{x}{ρ^{2}} = \frac{2I}{c} \frac{cos θ}{ρ} B_{x} = −\frac{2I}{c} \frac{y}{ρ^{2}} = −\frac{2I}{c} \frac{sin θ}{ρ}

and of course B = 2I/cρ, in agreement with Biot and Savart. We note that Φ_{M} is multivalued as one goes around the origin and must be “cut” on an arbitrary line from the origin to ∞, for instance along the negative x axis. This unphysical “cut” disturbs some people. It means that Φ_{M} is the potential of a dipole layer terminating on the current line, except that the dipole layer is not really there. We will see later that in general the Φ_{M} of a current loop is the potential of a fictitious dipole layer bounded by the loop.

In any case, the lines of constant A are the streamlines of B. In our example they circle the origin without any break. Other two-dimensional examples are left as exercises.