Magnetic Effects of Current and Magnetism

1Biot–Savart law and the straight wire

Every short length of a current-carrying wire seeds the space around it. Biot–Savart says an element I dl at distance r contributes dB = (μ₀/4π)(I dl sin θ)/r², where θ is the angle between the element and the line drawn from it to the point, and μ₀/4π = 10⁻⁷ T m A⁻¹ in SI units. Its direction is that of dl × r, perpendicular to both, so the field wraps around the wire instead of pointing away from it as an electric field would.

Add up every element of a long straight wire and the answer is B = μ₀I/2πr. Notice what happened to the powers: an inverse square for each element, but only an inverse first power for the finished wire, because the further away you stand the longer the stretch of wire that is still close enough to matter.

Figure. The element and the line to P meet at 60° here, so this element gives sin 60° of its best. Slide the element far up or far down the wire and θ closes toward 0° or opens toward 180°; either way sin θ dies and so does the contribution, which is why only the stretch of wire near the foot of the perpendicular really counts.

How it works

1. Chop the wire upTake one element I dl and write its contribution dB = (μ₀/4π)(I dl sin θ)/r².
2. Kill the parallel bitsWhere the element points straight at the field point, sin θ = 0 and that element contributes nothing at all.
3. Add along the wireFor a straight wire every element sends dB the same way through the page, so the magnitudes simply add and give μ₀I/2πr.

2The loop, the arc, and the 2πr trap

Bend the same wire into a circle and every element sits at the same distance R from the centre and square-on to it, so sin θ = 1 throughout and the integral collapses to B = μ₀I/2R. Bend it only part of the way, into an arc subtending φ radians, and you get that same field scaled by the fraction of a full turn actually present: B = (μ₀I/2R)(φ/2π).

The two standard results look alike and are not. A straight wire gives μ₀I/2πr; a loop's centre gives μ₀I/2R. At the same distance the loop's centre field is π times the straight wire's, and mistaking one for the other is the single commonest slip in this chapter.

Figure. The field at the centre is strictly proportional to the angle the arc subtends, because every element of the arc lies at the same distance R and contributes equally. Read a quarter turn off the line and you have a quarter of the full loop's field; there is no further geometry to do. Nothing on this line describes a straight wire, whose field falls off with distance instead.

How it works

1. Name the shape firstA loop or an arc puts 2R underneath; a straight wire puts 2πr. Decide which before substituting anything.
2. Take the fractionAn arc of φ radians gives φ/2π of the full loop's centre field: a semicircle a half, a quadrant a quarter.
3. Ignore the radial leadsStraight wires running in along a radius point directly at the centre, so sin θ = 0 and they add nothing to it.

3Ampère's circuital law

Ampère's law is the magnetic counterpart of Gauss's: walk once round any closed loop and ∮B·dl = μ₀I_enc. Currents threading the loop count; currents outside it do not, however strong their field on the loop may be. And like Gauss's law it is always true but only sometimes useful — you need a path on which B has a single constant magnitude and lies either along the path or square across it, which in practice means a circle round a wire, a rectangle straddling a solenoid's wall, or a circle inside a toroid.

Inside a solid wire of radius R carrying a uniform current density, only the fraction (r/R)² of the current is enclosed, so B = μ₀Ir/2πR². The field therefore climbs in a straight line from zero on the axis to μ₀I/2πR at the surface, and only then starts falling as 1/r.

Figure. Field against distance from the axis of a solid wire carrying a uniform current. Inside, the enclosed current grows as r² while the path length grows only as r, so B rises along a straight line; outside, the whole current is enclosed and B falls as 1/r. The two marked points sit at the same height because half the radius inside and twice it outside both give half the peak — the worked example read off the curve.

How it works

1. Let symmetry pick the pathChoose a loop on which you already know B has one size all the way round — the field's own circular path around a wire, for instance.
2. Drop the useless sectionsAnywhere B is perpendicular to dl the dot product is zero, so those stretches leave the integral untouched.
3. Count only what threadsB × (the length that actually counted) = μ₀ × (the current passing through the loop), and B falls straight out.

4Solenoid and toroid

Wind the wire into a long close-packed helix and the turns reinforce one another along the axis while cancelling outside, leaving B = μ₀nI inside, with n the number of turns per metre. That field is uniform in a strong sense: it does not depend on where in the cross-section you stand, nor on how wide the solenoid is — only on how tightly it is wound and how hard it is driven. Well away from the ends, at least; at each end the field has dropped to half its interior value.

Bend the solenoid round until it closes into a ring and you have a toroid, with B = μ₀NI/2πr inside it, where N is the total number of turns and r the distance from the ring's centre. So the field is slightly stronger against the inner wall than the outer, and outside the ring altogether there is no field at all.

Figure. The solenoid seen from the side. All three interior arrows are drawn the same length because the field really is uniform in there — moving off the axis or widening the coil changes nothing. Outside the wall the turns' contributions cancel and almost nothing is left.

5The Lorentz force and circular motion

A charge moving through a magnetic field feels F = q(v × B): sideways to its own motion, sideways to the field, and exactly zero if it moves along the field. Being always perpendicular to v, this force does no work — the speed and the kinetic energy of a charged particle in a purely magnetic field never change. What it can do is bend the path, and for v perpendicular to a uniform B it bends it into a circle, with qvB supplying the mv²/r, so r = mv/qB and one turn takes T = 2πm/qB.

That period contains no v and no r. A faster particle simply sweeps a proportionally larger circle in the same time, which is exactly what lets a cyclotron push a particle to its final energy with a fixed-frequency supply.

Figure. One instant of the motion. The force stands at a right angle to the velocity, which is why it turns the particle without hurrying it, and it points straight at the centre of the circle about to be traced. Double the speed and the bracket marked r simply doubles; the time to get round is untouched.

How it works

1. Check the angle firstOnly the part of v across B produces any force; motion along the field is left completely alone.
2. Balance the circleSet qvB = mv²/r to get r = mv/qB — the radius grows with momentum and shrinks with the field.
3. Read the period offT = 2πr/v = 2πm/qB, which loses both v and r and depends only on the particle and the field.

6Force on a current-carrying wire

A wire is a pipe full of moving charges, so it inherits their force: F = I(L × B), of size BIL sin θ with θ the angle between the wire and the field. Lay the wire along the field and the force vanishes; lay it across and the force is BIL, perpendicular to both the wire and the field.

Since each wire makes a field and each feels the other's, two long parallel wires a distance d apart push on each other with F/L = μ₀I₁I₂/2πd — attracting when the currents run the same way and repelling when they oppose, which is the exact reverse of what two like charges do. A magnetic force still does no work on the individual charges; when a rod on rails accelerates, the energy is coming from the source driving the current, not from the field.

Figure. Both currents run up the page and each wire sits in the field of the other, so the two forces come out equal, opposite and inward: the pair pulls together. Reverse either current and both arrows turn outward, exactly the same size. The fields doing the pushing point straight through the page and are named rather than drawn.

How it works

1. Take the perpendicular partOnly the component of the wire lying across the field counts, so the magnitude is BIL sin θ.
2. Point with the right handFingers along the current, curl them into B, and the thumb gives F — square to both.
3. Use the other wire's fieldIn a two-wire problem the field acting on each wire is the other's, never its own: a wire feels no force from the field it makes.

7The current loop as a magnetic dipole

A coil of N turns and area A carrying current I behaves in a magnetic field exactly as an electric dipole behaves in an electric field. Its magnetic moment is M = NIA, directed along the coil's normal — curl the right hand's fingers along the current and the thumb points along M. In a uniform field the forces on opposite sides are equal and opposite, so there is no net force, but they act along different lines and leave a couple τ = MB sin θ that turns M toward B.

The stored energy is U = −MB cos θ: least at θ = 0°, where the coil has settled and will oscillate about that position if nudged, and greatest at θ = 180°, where it is balanced but poised to flip.

Figure. The coil is seen edge-on, tilted so that its moment makes a true 30° with the field. The two BIL arrows are equal and opposite, so the coil does not drift anywhere; because they act along different lines they twist it, clockwise here, until M lies along B. At that point the two forces pull head-on against each other and everything is still.

How it works

1. Find M and its directionM = NIA along the coil's normal; fingers curled along the current, thumb along M.
2. Measure θ from M, not the coilThe θ in MB sin θ is the angle between M and B. If the question gives the angle to the coil's plane, take it from 90° first.
3. Torque and energy peak apartτ is greatest at 90°, where U is zero; U is greatest at 180°, where τ is zero.

8The moving coil galvanometer

Hang a coil of N turns and area A in a magnetic field, pass a current through it, and the couple NIAB twists it until the suspension's own restoring torque kφ matches it: φ = (NAB/k)I. The deflection is proportional to the current, which is the whole reason the instrument can carry a scale. The field is made radial by curved pole pieces and a soft-iron core, so that at every angle the coil's plane lies along B and the sin θ in the torque stays pinned at 1 — without that trick the divisions would crowd together at large deflections.

Current sensitivity is φ/I = NAB/k. Voltage sensitivity is φ/V = NAB/kR, and the two do not improve together: winding on more turns raises NAB but drags the coil's resistance R up with it, so a galvanometer made more sensitive to current may read voltage no better than it did before.

Figure. Radial field makes deflecting torque NIAB proportional to I, and the suspension restores with kφ, so the needle angle is linear in current — that linearity is the whole instrument.

How it works

1. The current makes a coupleThe coil is a magnetic dipole of moment NIA, so the field twists it with a torque NIAB.
2. The suspension answers backThe fibre resists with kφ, and the pointer settles exactly where NIAB = kφ.
3. Read off the sensitivityφ/I = NAB/k: more turns, more area, a stronger field or a weaker fibre all make it twitchier.

9Magnetic materials

Put matter inside a field and its own electrons answer back. The magnetisation M is the dipole moment per unit volume, the magnetising field is H, and the susceptibility is defined by M = χH. The total field is then B = μ₀(H + M) = μ₀(1 + χ)H, so the permeability is μ = μ₀(1 + χ) and the relative permeability μ_r is simply 1 + χ. That one number sorts every substance into three classes.

Diamagnets have χ small and negative: the moments induced in them oppose the applied field, they are feebly pushed toward the weaker part of a non-uniform field, and heating them changes nothing. Paramagnets have χ small and positive and are feebly pulled toward the stronger part, their alignment fighting thermal agitation, so χ falls off as 1/T. Ferromagnets have χ enormous and positive because whole domains of atoms align together; they keep some magnetisation after the field is removed, and above the Curie temperature the domains break up and they behave as paramagnets.

Figure. Susceptibility spans about nine decades across the three classes. A log axis is the only honest single chart; the sign (χ < 0 for diamagnets) lives in the comparison table beside the magnitudes.