The magnetic field

Two neutral wires carrying currents push on each other, and nothing in the first six lessons predicts it.

The previous lesson ended there. The escape is to define a second field, and the definition has to be different in kind from the electric one, because the effect being described is different in kind: it depends on how fast the charge is moving, and it acts at right angles to the motion. A field defined as force per unit charge cannot express either of those. What follows assumes the vector geometry of Classical Mechanics, in particular the right hand rule for a cross product, and nothing else new.

Oersted's accident

The connection between electricity and magnetism was suspected for decades and demonstrated by Hans Christian Oersted in April 1820, during a lecture in Copenhagen. He placed a compass needle near a wire and closed a circuit, and the needle swung.

The reason this was startling, rather than merely new, is the direction it swung. Every force known in 1820 acted along the line joining two objects: gravity, Coulomb's force, contact forces, tension. The needle near the wire did not point towards the wire or away from it. It set itself across the wire, tangent to a circle drawn around it, and reversing the current reversed the deflection. Oersted's own report notes that the effect encircles the wire, and no force law of the existing kind can produce a circulation.

Within months André-Marie Ampère had shown that two currents attract or repel each other directly, with no magnet involved, and had proposed that all magnetism is due to circulating currents, which is essentially the modern view: the magnetism of iron comes from electrons whose spin and orbital motion constitute microscopic currents. Magnetism is not a separate substance. It is what electricity does when it moves.

Defining B by what it does

Since the effect is not a force per unit charge, the field cannot be defined as one. Instead it is defined by the whole force law it produces, and the law is the experimental result:

F = q v B \sin θ

where $θ$ is the angle between the velocity and the field, with the force perpendicular to both $v$ and $B$ , in the direction given by the right hand rule for $v \times B$ and reversed if the charge is negative. Compactly, $F = q v \times B$ .

Everything about $B$ is contained in that. Fire a test charge through a point in several directions: there is exactly one direction along which it feels no force at all, and that line is the direction of $B$ , with the sign fixed by the right hand rule applied to any other direction. Fire it perpendicular to that line and the force is maximal, at $F = q v B$ , which fixes the magnitude. The definition is operational and slightly awkward, and it is awkward because the phenomenon is.

The unit of $B$ is the tesla, one newton per ampere metre. It is a large unit. The Earth's field is about 50 μT, a refrigerator magnet a few millitesla, a clinical scanner 1.5 to 3 T, and the strongest steady laboratory field about 45 T. Fields above roughly 100 T can only be made in pulses that destroy the apparatus, because a field exerts a pressure on the currents making it, a point the tenth lesson quantifies. The older unit, the gauss, is $10^{- 4}$ T, and the Earth's field of half a gauss is why it survives in geophysics.

Magnetic force does no work

One consequence follows immediately and is worth isolating, because a great deal depends on it.

The force is always perpendicular to the velocity. Work is force times displacement along the direction of motion, and there is none. So

the magnetic force does no work, ever

and the speed of a charged particle in a magnetic field never changes. Only the direction does. Whatever a magnetic field is doing when a crane lifts scrap iron, it is not supplying energy directly through this force, and any explanation that says otherwise is wrong somewhere.

This is also why $B$ has no potential in the sense of the fourth lesson. A potential exists for a force that does path-independent work; a force that does no work at all cannot be got at that way.

Circular motion

Send a charge into a uniform field perpendicular to it. The force has constant magnitude $q v B$ , since $v$ never changes, and is always perpendicular to the motion. That is precisely the condition for uniform circular motion, so Newton's second law with the centripetal acceleration $v^{2} / r$ gives $q v B = m v^{2} / r$ and

r = \frac{m v}{q B}

The radius is proportional to the momentum. The period is $T = 2 π r / v = 2 π m / (q B)$ , so the cyclotron frequency

f = \frac{q B}{2 π m}

does not depend on the speed at all. A fast particle goes round a bigger circle in exactly the same time as a slow one, because the two effects cancel.

That cancellation is what Ernest Lawrence exploited in 1932. A cyclotron holds two hollow D-shaped electrodes in a uniform field, with an alternating voltage across the gap between them at the cyclotron frequency. A particle crosses the gap, gets a kick, spirals out to a larger radius, and arrives back at the gap in step with the alternating voltage because the period does not depend on how fast it is now going. Lawrence's first machines were centimetres across and reached a megaelectronvolt.

The scheme fails at high energy, and the failure is instructive. As the particle approaches the speed of light its effective mass rises, the period lengthens, and it drifts out of step with the driving voltage. The synchrocyclotron sweeps the driving frequency to follow it, and the synchrotron instead ramps the magnetic field to hold the radius fixed, which is why large accelerators are rings of fixed radius rather than spirals. An electron cyclotron frequency of 28.0 MHz per millitesla is a useful figure to carry, and a proton in 1.5 T circulates at 22.9 MHz.

Example. A proton moves at $2.0 \times 10^{6}$ m/s perpendicular to a 0.40 T field. What is the radius of its path, and how long does one orbit take?

$r = m v / (q B) = (1.673 \times 10^{- 27}) (2.0 \times 10^{6}) / [(1.602 \times 10^{- 19}) (0.40)] = 5.22 \times 10^{- 2}$ m, about 5.2 cm. The period is $T = 2 π m / (q B) = 2 π (1.673 \times 10^{- 27}) / [(1.602 \times 10^{- 19}) (0.40)] = 1.64 \times 10^{- 7}$ s, and it would be the same for a proton of any speed in that field.

Now you. An electron moves at $5.0 \times 10^{6}$ m/s perpendicular to a 0.25 T field. What is the radius of its path?

Answer

$r = m v / (q B) = (9.109 \times 10^{- 31}) (5.0 \times 10^{6}) / [(1.602 \times 10^{- 19}) (0.25)] = 1.14 \times 10^{- 4}$ m, about a tenth of a millimetre. The electron's small mass makes it curl very tightly.

Selecting and weighing

Cross an electric and a magnetic field at right angles and send charges through the region. The electric force $q E$ is fixed; the magnetic force $q v B$ grows with speed. Arrange them to oppose and there is exactly one speed at which they cancel:

v = \frac{E}{B}

and the charge, whatever it is and whatever its mass, passes through undeflected. Everything else is swept aside. This is a velocity selector, and its output does not depend on $q$ or $m$ , which is what makes it useful as a first stage.

Feed that beam into a second region with a uniform field $B'$ and the particles curve with $r = m v / (q B')$ , so measuring where a particle lands measures its mass to charge ratio:

\frac{m}{q} = \frac{B' r}{v}

That is a mass spectrometer, and it is how isotopes were discovered. J. J. Thomson found in 1913 that neon gave two parabolas, at masses 20 and 22, and Francis Aston built the instrument properly and by 1922 had identified over 200 isotopes with it. The same principle, refined, is now standard for identifying molecules by their exact mass.

Thomson's earlier and more famous use of crossed fields, in 1897, was the other way round: he measured $m / q$ for cathode rays, found it a thousand times smaller than for any ion, and concluded that the particles were far lighter than atoms. That was the discovery of the electron, and it came out of the force law of this lesson.

Example. A velocity selector uses $E = 2.4 \times 10^{5}$ V/m and $B = 0.30$ T. Ions of charge $+ e$ emerge and enter a 1.00 T field, curving on a radius of 33.2 cm. What is their mass, in atomic mass units?

The selected speed is $v = E / B = 2.4 \times 10^{5} / 0.30 = 8.0 \times 10^{5}$ m/s. Then $m = q B' r / v = (1.602 \times 10^{- 19}) (1.00) (0.332) / (8.0 \times 10^{5}) = 6.65 \times 10^{- 26}$ kg. Dividing by the atomic mass unit, $1.6605 \times 10^{- 27}$ kg, gives 40.0 u, so this is argon-40, which makes up 99.6 per cent of atmospheric argon.

Now you. In the same instrument, another singly charged ion curves on a radius of 29.0 cm. What is its mass in atomic mass units?

Answer

$m = q B' r / v = (1.602 \times 10^{- 19}) (1.00) (0.290) / (8.0 \times 10^{5}) = 5.81 \times 10^{- 26}$ kg, which divided by $1.6605 \times 10^{- 27}$ is 35.0 u: chlorine-35.

Force on a wire, and torque on a loop

A current is charge in motion, so a current-carrying wire in a field feels the sum of the forces on its carriers. Take a straight segment of length $L$ and cross section $A$ , with $n$ carriers per cubic metre each of charge $q$ drifting at $v_{d}$ . The segment contains $n A L$ carriers, each feeling $q v_{d} B \sin θ$ , so the total is $n A L q v_{d} B \sin θ$ . But $n q v_{d} A$ is exactly the current, from the previous lesson, so

F = I L B \sin θ

with the direction given by the right hand rule applied to the current direction and the field. Ten amperes through 20 cm of wire across a 0.50 T field gives 1.0 N, which is easy to feel and is the whole basis of the electric motor.

Now bend the wire into a rectangular loop of sides $a$ and $b$ , carrying current $I$ , sitting in a uniform field with the loop's plane containing $B$ . The two sides of length $a$ that run perpendicular to the field feel forces $I a B$ , in opposite directions since the current runs opposite ways in them, and they are separated by $b$ . The net force is zero, exactly as for the electric dipole in the second lesson, and the net torque is not:

τ = I a b B = I A B

with $A = a b$ the area. For $N$ turns it is $N$ times larger, and for a general angle a factor $\sin θ$ appears, with $θ$ measured between the field and the normal to the loop. Defining the magnetic moment $μ = N I A$ , a vector along the loop's normal by the right hand rule,

τ = μ B \sin θ

which is the same form as the electric dipole's $τ = p E \sin θ$ , and for the same reason: at a distance, a current loop and a bar magnet are indistinguishable, and both are magnetic dipoles. This is Ampère's insight made quantitative.

Everything rotational in electrical engineering is here. A motor is a coil in a field, with a commutator that reverses the current twice per revolution so the torque never changes sign. A moving-coil galvanometer balances this torque against a spring, so the deflection reads the current. A compass needle is a magnetic moment aligning with the Earth's field, and Oersted's needle was reporting the field of his wire in exactly this way.

Example. A coil of 50 turns and area 4.0 cm² carries 0.20 A in a 0.10 T field, with its plane parallel to the field. What is the torque on it?

The magnetic moment is $μ = N I A = (50) (0.20) (4.0 \times 10^{- 4}) = 4.0 \times 10^{- 3}$ A m². With the plane parallel to the field, the normal is perpendicular to it, so $\sin θ = 1$ and the torque is maximal: $τ = μ B = (4.0 \times 10^{- 3}) (0.10) = 4.0 \times 10^{- 4}$ N m.

Now you. A coil of 120 turns and area 6.0 cm² carries 0.15 A in a 0.25 T field, with its plane parallel to the field. What is the torque?

Answer

$μ = N I A = (120) (0.15) (6.0 \times 10^{- 4}) = 1.08 \times 10^{- 2}$ A m², and $τ = μ B = (1.08 \times 10^{- 2}) (0.25) = 2.7 \times 10^{- 3}$ N m.

The Hall effect, and the sign of the carriers

Nothing so far distinguishes positive charges drifting one way from negative charges drifting the other, since both give the same current. Edwin Hall found the experiment that does, in 1879, while a graduate student.

Pass a current along a flat strip and apply a field perpendicular to its face. The carriers, whatever their sign, are pushed sideways by $q v \times B$ , and they pile up on one edge until the transverse electric field they create cancels the magnetic push. A voltage then appears across the strip, and its sign depends on the sign of the carriers, because positive carriers moving one way and negative carriers moving the other are deflected to the same edge, arriving there with opposite charge.

Balancing the two forces gives $q E_{H} = q v_{d} B$ , and with $E_{H} = V_{H} / w$ across a strip of width $w$ and $I = n q v_{d} w t$ for thickness $t$ , the width cancels:

V_{H} = \frac{I B}{n q t}

which also measures $n$ , the carrier density, an otherwise awkward quantity. In copper the effect is minute: 5 A through a strip 0.10 mm thick in a 1.0 T field gives $V_{H} = 3.7$ μV. In a semiconductor with $n$ around $10^{21}$ m⁻³, eight orders of magnitude fewer carriers, the same numbers give hundreds of volts in principle and a comfortably measurable signal in practice, which is why Hall sensors are made of semiconductors.

The results were a puzzle for fifty years. In some metals, notably zinc and cadmium, the Hall voltage comes out with the sign for positive carriers, which is impossible if conduction is by electrons. The resolution needed quantum band theory and the concept of a hole, a missing electron in a nearly full band that responds to fields exactly as a positive particle would. Hall's experiment was correct and unexplainable for half a century, which is a useful reminder that a clean measurement can outrun the theory available to interpret it.

What is still missing

The force law now covers everything a magnetic field does to a moving charge, a current, or a loop. It says nothing whatsoever about what produces the field.

That is a genuine asymmetry with the electric case, where Coulomb's law gave the source and the force in one statement. Here the source law has to be found separately, and the previous lesson's observation, that two parallel currents attract, is the experiment that will fix it. Doing so is next, and it will turn up an equation with the same structure as Gauss's law and one flat contradiction with it.