Magnetic Fields
For seven chapters the electric field has been the whole story. Now a second field enters — the magnetic field \(\vec{B}\) — and with it a force that behaves quite unlike any we have met. The magnetic force on a charge depends on how fast the charge moves and in which direction, it always acts sideways, and it never speeds the charge up. This chapter defines \(\vec{B}\) through that force \(\vec{F}_B = q\vec{v}\times\vec{B}\), traces the circular and helical paths a charge follows in a field, and then carries the same force over to a current-carrying wire and a current loop — the seed of the electric motor.
- How the magnetic field \(\vec{B}\) is defined by the force on a moving charge, \(\vec{F}_{B} = q\vec{v}\times\vec{B}\), with magnitude \(F_{B} = |q|vB\sin\phi\), measured in teslas.
- Why the magnetic force is always perpendicular to the velocity, so it changes a particle's direction but never its speed or kinetic energy.
- Crossed fields: the speed selector \(v = E/B\), Thomson's discovery of the electron, and the Hall effect that reveals the sign and density of charge carriers.
- Circular motion of radius \(r = \dfrac{mv}{|q|B}\) and period \(T = \dfrac{2\pi m}{|q|B}\) — independent of speed — plus helical paths and the cyclotron resonance condition.
- The force on a current-carrying wire \(\vec{F}_{B} = i\vec{L}\times\vec{B}\), the torque on a coil \(\tau = NiAB\sin\theta\), and the magnetic dipole moment \(\vec{\mu} = Ni\vec{A}\) with energy \(U = -\vec{\mu}\cdot\vec{B}\).
What Is Physics?
One central goal of physics is to understand how an electric field produces a force on a charged object. A closely related goal is to understand how a magnetic field produces a force — on a moving charged particle, or on a magnetic object such as a magnet. You already know a little of this if you have ever pinned a note to a refrigerator with a small magnet, or accidentally wiped a credit card by waving it past one. The magnet acts on the door, or on the card's magnetic stripe, through its magnetic field.
The applications are everywhere and multiplying. Magnets spin the motors in a car's window lifts, wipers, and ignition; they drive the speaker cones in your headphones and phone; they run the read heads of hard drives and the latches of doorbells and alarms. The science of magnetic fields is physics; their application is engineering. Both begin with one question: what produces a magnetic field?
What Produces a Magnetic Field?
Since an electric field is produced by an electric charge, we might expect a magnetic field to be produced by a magnetic charge — a magnetic monopole. Such monopoles are predicted by some theories, but none has ever been found. So how are magnetic fields actually made? In two ways.
The first is by moving charge — a current in a wire — which makes an electromagnet; this is the subject of the next chapter. The second is more subtle: certain elementary particles, the electron above all, carry an intrinsic magnetic field as a basic property, like their mass and charge. In some materials the electrons' tiny fields add up to give a permanent magnet; in most materials they cancel, which is why you do not snap onto every refrigerator you pass. Either way, our first job is to define the magnetic field, and we do so through the force it exerts on a moving charge.
The Definition of B
We defined the electric field by holding a test charge at rest and measuring the force on it, \(\vec{E} = \vec{F}_{E}/q\). We cannot do the same for \(\vec{B}\), because a magnet test-charge (a monopole) does not exist. Instead we fire a charged particle through the point in many directions and watch the force. We find a special axis along which the force vanishes; for any other direction the force grows as \(\sin\phi\), where \(\phi\) is the angle from that zero-force axis, and the force is always perpendicular to the velocity. These clues point straight at a cross product.
Because \(\vec{F}_{B}\perp\vec{v}\), the force has no component along the motion and does zero work. It cannot change the kinetic energy — only steer. This single fact explains every circular and helical path in this chapter.
The SI unit of \(\vec{B}\) follows from the definition: a newton per coulomb-metre-per-second, named the tesla.
Magnetic Field Lines
As with electric fields, we picture \(\vec{B}\) with field lines: the tangent gives the field's direction, and closer spacing means a stronger field. But magnetic field lines differ in one decisive way — they form closed loops, with no beginning and no end. They emerge from the north pole of a magnet and re-enter at its south pole, then continue through the magnet back to the north. Every magnet therefore has two poles; it is a magnetic dipole, and a single isolated pole has never been found.
Crossed Fields and the Discovery of the Electron
When an electric and a magnetic field act on the same charge at right angles to each other, they are called crossed fields. In 1897 J. J. Thomson used exactly this arrangement to discover the electron. A beam of charged particles is accelerated through a potential difference and sent between charged plates (an \(\vec{E}\) field) that also sit in a magnetic field, with the electric and magnetic deflections opposing each other.
Turn on \(\vec{E}\) alone and the beam deflects; now add \(\vec{B}\) and tune it until the deflections exactly cancel and the beam goes straight. At that balance the electric and magnetic forces are equal, \(|q|E = |q|vB\), which fixes the particle's speed.
Crossed Fields: The Hall Effect
Can a magnetic field deflect the drifting conduction electrons inside a wire, not just a beam in vacuum? In 1879 Edwin Hall, then a graduate student, showed it can. Send a current through a flat strip and apply \(\vec{B}\) across its thickness: the magnetic force pushes the moving carriers to one edge, which charges up until the resulting transverse electric field exactly balances the magnetic push. At equilibrium \(eE = ev_{d}B\), and a voltmeter reads a Hall potential difference across the width.
A Circulating Charged Particle
Send a charged particle into a uniform field perpendicular to \(\vec{B}\). The magnetic force has constant magnitude \(|q|vB\) and always points at right angles to the motion — exactly the recipe for uniform circular motion. Setting the magnetic force equal to the centripetal requirement \(mv^{2}/r\) gives the radius.
Helical Paths, Cyclotrons, and Synchrotrons
If the velocity is not perpendicular to \(\vec{B}\) but tilted at an angle \(\phi\), split it into a part along the field, \(v_{\parallel} = v\cos\phi\), and a part across it, \(v_{\perp} = v\sin\phi\). The perpendicular part drives the circling; the parallel part drifts steadily along the field. Together they trace a helix, whose pitch (the advance per turn) is \(p = v_{\parallel}T\). In a field that gets stronger at both ends, a particle can be reflected back and forth — trapped in a magnetic bottle.
The speed-independent period is exploited in the cyclotron. Two hollow "dees" sit in a uniform field, and an oscillator flips the voltage across the gap between them. A proton spirals outward, getting a kick each time it crosses the gap. The trick works only if the oscillator's frequency matches the circulation frequency.
Magnetic Force on a Current-Carrying Wire
A current is just moving charge, so a magnetic field pushes on a current-carrying wire too — the sideways shove on each drifting electron is handed off to the wire itself. Add up the force on all the carriers in a straight length \(L\) and the drift speed cancels out, leaving a clean result.
Torque on a Current Loop and the Magnetic Dipole Moment
Put a current loop in a field and the forces on its opposite sides, though they cancel as a net force, do not share a line of action — so they twist the loop. This torque is what turns every electric motor. For a flat coil of \(N\) turns, each enclosing area \(A\) and carrying current \(i\), the torque depends on the angle \(\theta\) between the field and the coil's normal.
It is tidier to bundle the coil's properties into one vector, the magnetic dipole moment \(\vec{\mu}\), pointing along the coil's normal by the right-hand rule. Then the torque and the orientation energy take exactly the same elegant form as for an electric dipole in Chapter 22.
Bar magnets, rotating charged spheres, planet Earth, and even electrons and protons can all be treated as magnetic dipoles. Each tends to swing so that its \(\vec{\mu}\) lines up with the applied field — the principle behind the compass and the motor alike.
Putting It to Work
Problem. A uniform field of 1.2 mT points vertically up in a chamber. A proton (mass \(1.67\times10^{-27}\,\mathrm{kg}\)) with kinetic energy 5.3 MeV enters moving horizontally, south to north. What magnetic force acts on it?
Solution. First get the speed from \(K = \tfrac{1}{2}mv^{2}\), then apply \(F_{B} = |q|vB\sin\phi\) with \(\phi = 90^\circ\).
By the right-hand rule (positive charge, \(\vec{v}\) north into \(\vec{B}\) up) the force points west to east. Tiny as it looks, on so light a particle it gives an acceleration of nearly \(4\times10^{12}\,\mathrm{m/s^{2}}\).
Problem. A metal cube of edge \(d = 1.5\,\mathrm{cm}\) moves at \(v = 4.0\,\mathrm{m/s}\) through a field \(B = 0.050\,\mathrm{T}\) perpendicular to the motion. Find the potential difference set up across the cube at equilibrium.
Solution. At equilibrium the electric and magnetic forces on a conduction electron balance, \(|q|E = |q|vB\), so \(E = vB\) and \(V = Ed\).
This motional voltage is the same physics as the Hall effect — moving the conductor through the field, instead of pushing carriers through a stationary strip.
Problem. A proton moves perpendicular to a uniform 0.50 T field at \(v = 3.0\times10^{6}\,\mathrm{m/s}\). Find the radius of its circular path and the period.
Solution. Use \(r = mv/|q|B\) and \(T = 2\pi m/|q|B\).
Double the speed and the circle doubles in size — but the period is unchanged, because it depends only on \(m\), \(|q|\), and \(B\).
Problem. A horizontal copper wire of linear density 46.6 g/m carries a current of 28 A. What is the minimum magnetic field that would suspend it — balancing gravity?
Solution. Set the upward magnetic force equal to weight, \(iLB\sin\phi = mg\). The field is smallest when \(\phi = 90^\circ\), giving \(B = (m/L)g/i\).
That is about 160 times Earth's surface field — a reminder that magnetic levitation of everyday objects takes a serious field.
Problem. A circular coil of 250 turns, area \(2.52\times10^{-4}\,\mathrm{m^{2}}\), carries 100 µA, sitting in a 0.85 T field with \(\vec{\mu}\) aligned with \(\vec{B}\). How much work must an external agent do to rotate it 90° (to \(\vec{\mu}\perp\vec{B}\))?
Solution. The work equals the change in orientation energy, \(W_{a} = U_{f} - U_{i} = -\mu B\cos 90^\circ - (-\mu B\cos 0^\circ) = \mu B\), with \(\mu = NiA\).
Lifting the dipole out of alignment costs energy; left free, it would swing back to line up with the field, just like a compass needle.
Chapter Summary
Made by moving charge (electromagnets) and by the intrinsic fields of particles; no monopoles exist.
\(\vec{F}_{B} = q\vec{v}\times\vec{B}\), magnitude \(F_{B} = |q|vB\sin\phi\); force \(\perp\) to \(\vec{v}\), so it does no work.
The tesla: \(1\,\mathrm{T} = 1\,\mathrm{N/(A\cdot m)} = 10^{4}\) gauss. Field lines are closed loops.
Speed selector \(v = E/B\); the Hall effect gives the carrier sign and density \(n = Bi/Vle\).
\(r = \dfrac{mv}{|q|B}\), \(T = \dfrac{2\pi m}{|q|B}\); period independent of speed.
Pitch \(p = v_{\parallel}T\); resonance \(|q|B = 2\pi m f_{\text{osc}}\).
\(\vec{F}_{B} = i\vec{L}\times\vec{B}\); differential form \(d\vec{F}_{B} = i\,d\vec{L}\times\vec{B}\).
\(\mu = NiA\), \(\vec{\tau} = \vec{\mu}\times\vec{B}\), \(U = -\vec{\mu}\cdot\vec{B}\).
Problems
Take \(e = 1.60\times10^{-19}\,\mathrm{C}\), proton mass \(1.67\times10^{-27}\,\mathrm{kg}\), electron mass \(9.11\times10^{-31}\,\mathrm{kg}\), and \(1\,\mathrm{u} = 1.66\times10^{-27}\,\mathrm{kg}\). Throughout, decide first whether \(\vec{v}\) is along, across, or tilted to \(\vec{B}\) — that angle controls everything — and use the right-hand rule for every direction.
- A proton moves at \(23^\circ\) to a 2.60 mT field and feels a force of \(6.50\times10^{-17}\,\mathrm{N}\). Find (a) its speed and (b) its kinetic energy in electron-volts.
- An electron has velocity \(\vec{v} = (2.0\times10^{6}\,\hat{\imath} + 3.0\times10^{6}\,\hat{\jmath})\,\mathrm{m/s}\) in a field \(\vec{B} = (0.030\,\hat{\imath} - 0.15\,\hat{\jmath})\,\mathrm{T}\). Find the force on it, then repeat for a proton of the same velocity.
- An alpha particle (\(q = 3.2\times10^{-19}\,\mathrm{C}\), \(m = 6.6\times10^{-27}\,\mathrm{kg}\)) travels at 550 m/s through a 0.045 T field at \(52^\circ\). Find (a) the force and (b) the acceleration; (c) does its speed change?
- An electric field of 1.50 kV/m and a perpendicular magnetic field of 0.400 T produce no net force on a moving electron. What is the electron's speed?
- A copper strip 150 µm thick and 4.5 mm wide carries 23 A in a 0.65 T field perpendicular to the strip. With \(n = 8.47\times10^{28}\,\mathrm{m^{-3}}\), find the Hall potential difference.
- An alpha particle (\(q = 2e\), \(m = 4.00\,\mathrm{u}\)) circles at radius 4.50 cm in a 1.20 T field. Find (a) its speed, (b) its period, and (c) its kinetic energy.
- An electron of kinetic energy 1.20 keV circles in a plane perpendicular to a uniform field at orbit radius 25.0 cm. Find (a) the speed, (b) the field magnitude, (c) the frequency, and (d) the period.
- What field, perpendicular to a beam of electrons moving at \(1.30\times10^{6}\,\mathrm{m/s}\), makes them travel in a circular arc of radius 0.350 m?
- A positron of kinetic energy 2.00 keV is projected into a 0.100 T field with its velocity at \(89.0^\circ\) to the field. Find (a) the period, (b) the pitch, and (c) the radius of its helical path.
- A cyclotron of dee radius 53.0 cm runs at an oscillator frequency of 12.0 MHz to accelerate protons. (a) What field gives resonance? (b) What is the kinetic energy of an emerging proton?
- A wire 1.80 m long carries 13.0 A at \(35.0^\circ\) to a uniform 1.50 T field. Calculate the magnetic force on the wire.
- A 13.0 g wire of length 62.0 cm hangs from flexible leads in a 0.440 T field. What current (magnitude and direction) removes the tension in the leads?
- A 20-turn rectangular coil, 10 cm by 5.0 cm, carries 0.10 A and lies at \(30^\circ\) to a 0.50 T field. Find the magnitude of the torque on the coil.
- A circular coil of 160 turns has radius 1.90 cm. (a) What current gives a dipole moment of magnitude \(2.30\,\mathrm{A\cdot m^{2}}\)? (b) Find the maximum torque it can feel in a 35.0 mT field.