Magnetic Force

A force that only notices things that move

Put a proton at rest in the gap of a strong magnet and nothing happens. It just sits there. Now flick it sideways at a hundred kilometres a second and it swings around in a tight circle, comes back, and keeps circling forever. Nothing about the magnet changed. The only thing that changed was the proton's velocity.

That is the strange part of magnetism, and it is worth sitting with before you touch a formula. In the circuits you have just come from, current was pushed by a potential difference and the force on a charge pointed along the field. Here the force depends on how fast the charge is going, and it does not point along the field, and it does not point along the velocity either. It points off to the side, perpendicular to both.

The force law, and the direction

Experiment says the force on a charge $q$ moving with velocity $\vec{v}$ through a magnetic field $\vec{B}$ has three properties. It is proportional to $q$, proportional to $v$, and it vanishes when $\vec{v}$ lies along $\vec{B}$. It is also always perpendicular to both $\vec{v}$ and $\vec{B}$. There is exactly one product of two vectors that behaves that way, the cross product, so

$\vec{F} = q\,\vec{v} \times \vec{B}$

The magnitude follows from the definition of the cross product:

$F = |q|\,v\,B\sin\theta$

where $\theta$ is the angle between $\vec{v}$ and $\vec{B}$. If the charge moves straight across the field, $\theta = 90^\circ$ and you get the maximum, $F = qvB$. If it moves along the field, $\sin 0 = 0$ and there is no force at all. $B$ is measured in tesla: one tesla pushes a one-coulomb charge moving at one metre per second across it with one newton of force. A tesla is a big field. The Earth's is about $5\times10^{-5}$ T, a fridge magnet a hundredth of a tesla, an MRI magnet a few tesla.

The direction comes from the right-hand rule. Point your fingers along $\vec{v}$, curl them toward $\vec{B}$, and your thumb points along $\vec{v}\times\vec{B}$. That is the force if $q$ is positive. For an electron the force is the opposite way, because $q$ is negative.

Try the easy end first. A proton at $2.0\times10^{5}$ m/s crossing a $0.50$ T field perpendicular feels

$F = (1.6\times10^{-19})(2.0\times10^{5})(0.50) = 1.6\times10^{-14}\ \text{N}$

Tiny in absolute terms, but the proton's mass is $1.67\times10^{-27}$ kg, so the acceleration is about $10^{13}$ m/s$^2$. Small forces do dramatic things to small masses.

Magnetic forces do no work

Because $\vec{F}$ is always perpendicular to $\vec{v}$, the power it delivers is

$P = \vec{F}\cdot\vec{v} = q(\vec{v}\times\vec{B})\cdot\vec{v} = 0$

A magnetic field can never change a particle's speed or its kinetic energy. It can only bend the path. That is one of the most useful constraints in this whole unit. If a solution has a magnetic field speeding a particle up, the solution is wrong. If a particle enters a field region at $1.0\times10^{6}$ m/s, it leaves at $1.0\times10^{6}$ m/s, however long it spent inside and however much it turned.

Circles, periods, and the radius formula

Fire a charge into a uniform field perpendicular to the field lines. The force has constant magnitude $qvB$ (the speed never changes, so neither does the force) and it always points at right angles to the motion. Constant-magnitude, always-perpendicular force is exactly the recipe for uniform circular motion. So set the magnetic force equal to what circular motion demands:

$qvB = \frac{mv^2}{r}$

One power of $v$ cancels off each side, and

$r = \frac{mv}{qB} = \frac{p}{qB}$

Read that. Radius is proportional to momentum and inversely proportional to field. A heavier or faster particle is harder to bend, a stronger field bends it more. Bending a beam through a known field and measuring the curvature is how particle detectors weigh what went through them.

The period is stranger. Going once around covers $2\pi r$ at speed $v$, so

$T = \frac{2\pi r}{v} = \frac{2\pi m}{qB}$

The speed has vanished. A fast particle traces a big circle, a slow one traces a small circle, and they take exactly the same time. That fixed frequency, $f = qB/2\pi m$, is called the cyclotron frequency, and it is what lets a cyclotron drive every orbit, from the first tiny one to the last huge one, with a single fixed-frequency voltage.

Worked example: a mass spectrometer

A singly charged ion starts from rest, falls through a potential difference of $2000$ V, then enters a $0.50$ T field perpendicular to its velocity and bends on a circle of radius $0.15$ m. What is its mass?

Two stages, so two equations. First, the electrical stage gives it energy:

$qV = \tfrac{1}{2}mv^2 \quad\Longrightarrow\quad v = \sqrt{\frac{2qV}{m}}$

Second, the magnetic stage sets the radius, $r = mv/qB$. You do not know $v$ and you do not know $m$, so eliminate $v$. Substituting,

$r = \frac{m}{qB}\sqrt{\frac{2qV}{m}} = \frac{1}{B}\sqrt{\frac{2Vm}{q}}$

Square it and solve for the mass:

$m = \frac{qB^2r^2}{2V}$

Now the arithmetic. $B^2 = 0.25$ T$^2$, $r^2 = 0.0225$ m$^2$, so $B^2r^2 = 5.625\times10^{-3}$. Then

$m = \frac{(1.6\times10^{-19})(5.625\times10^{-3})}{2(2000)} = \frac{9.0\times10^{-22}}{4000} = 2.25\times10^{-25}\ \text{kg}$

About 135 atomic mass units. That $m \propto B^2r^2/V$ relation is the whole instrument: two isotopes accelerated through the same voltage arrive at radii in the ratio $\sqrt{m}$, and if you bend them a half circle onto a plate, the landing spots are $2(r_2 - r_1)$ apart.

If the velocity is not perpendicular to the field, split it. The component $v_\parallel = v\cos\theta$ along $\vec{B}$ feels no force at all and just coasts. The component $v_\perp = v\sin\theta$ across the field circles as before. Together they trace a helix: a circle of radius $mv_\perp/qB$ drifting steadily along the field. The distance it advances in one full turn is the pitch,

$\text{pitch} = v_\parallel T = v\cos\theta \cdot \frac{2\pi m}{qB}$

Crossed fields

Put $\vec{E}$ and $\vec{B}$ perpendicular to each other and send a charge through perpendicular to both. The electric force $qE$ pushes one way, the magnetic force $qvB$ the other. They cancel when

$qE = qvB \quad\Longrightarrow\quad v = \frac{E}{B}$

The charge and the mass both drop out. Every particle at exactly that speed sails straight through whatever it is; everything slower or faster gets deflected into a wall. That is a velocity selector, and it is the front end of most mass spectrometers, because $r = mv/qB$ is only useful if you know $v$.

A charge released from rest in crossed fields does something more interesting. It has no magnetic force at first, so $E$ accelerates it; as it picks up speed the magnetic force grows and curves it around, eventually turning it back and slowing it to rest again. The path is a cycloid, the curve a point on a rolling wheel traces. Since magnetism does no work, all the kinetic energy came from $E$, and the particle is momentarily at rest again after drifting a distance $d$ sideways. The fastest point is halfway, and working through the geometry gives a maximum speed of $v_{\max} = 2E/B$: for $E = 1.0\times10^{4}$ V/m and $B = 0.20$ T, that is $1.0\times10^{5}$ m/s.

Wires, because a wire is just charges moving

A current is a stream of charges with a drift velocity. Take a straight segment of length $L$ carrying current $I$. In a time $\Delta t$ the charge that passes is $q = I\Delta t$, and it moves a distance $v\Delta t = L$. So summing $q\vec{v}\times\vec{B}$ over all the carriers gives $qvB\sin\theta = (I\Delta t)(L/\Delta t)B\sin\theta$, and the $\Delta t$ cancels:

$\vec{F} = I\vec{L}\times\vec{B}, \qquad F = ILB\sin\theta$

$\vec{L}$ points along the wire in the direction of conventional current. A $0.40$ m wire carrying $5.0$ A perpendicular to a $0.30$ T field feels $F = (5.0)(0.40)(0.30) = 0.60$ N. That is a force you can feel with your hand, which is why motors are built out of current-carrying wire and not out of individual protons.

For a bent wire you integrate $I\,d\vec{\ell}\times\vec{B}$, but in a uniform field $\vec{B}$ comes outside the integral and $\int d\vec{\ell}$ depends only on the endpoints. So a bent wire in a uniform field feels the same force as the straight wire joining its ends. A semicircle of radius $0.10$ m carrying $4.0$ A perpendicular to a $0.25$ T field feels the force of its $0.20$ m diameter: $F = (4.0)(0.20)(0.25) = 0.20$ N. And a closed loop in a uniform field feels zero net force, since its endpoints coincide.

Two parallel wires each sit in the other's field. Using the field of a long wire that you will derive properly in the next lesson, $B = \mu_0 I/2\pi d$, the force per unit length is

$\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi d}$

Same direction, they attract; opposite, they repel. Two bus bars $5.0$ cm apart carrying $10$ A each pull on one another with $(4\pi\times10^{-7})(100)/(2\pi \times 0.05) = 4.0\times10^{-4}$ N per metre.

Worked example: starting a rod against friction

A $0.20$ kg rod lies across horizontal rails $0.40$ m apart in a vertical $0.50$ T field. The static friction coefficient is $0.30$. What current just starts it moving?

Start with what is on the paper: the field is vertical, the rod is horizontal and perpendicular to it, so $\vec{I}\vec{L}\times\vec{B}$ is horizontal, along the rails. Good, that is the direction the rod would slide. Its magnitude is $ILB$ with no $\sin\theta$ correction.

The rod is on the verge of moving when that force equals the largest static friction the rails can supply. Nothing pushes the rod down or up, so the normal force is just the weight, $N = mg$, and

$ILB = \mu mg \quad\Longrightarrow\quad I = \frac{\mu m g}{LB}$

Numbers: $\mu m g = (0.30)(0.20)(9.8) = 0.588$ N, and $LB = (0.40)(0.50) = 0.20$ T$\cdot$m. So $I = 0.588/0.20 = 2.94$ A.

Loops: moment, torque, and small oscillations

A loop feels no net force in a uniform field, but it does feel a twist. Picture a rectangular loop of area $A$ carrying current $I$, with its plane containing the field. The two sides perpendicular to $\vec{B}$ feel forces $ILB$ in opposite directions, offset by the width of the loop. Equal and opposite forces on different lines of action are a couple: no net force, pure torque. Working out the lever arms gives $\tau = IAB$ for that orientation, and $N$ turns multiply it by $N$.

Define the magnetic dipole moment $\mu = NIA$, a vector pointing along the normal to the loop (right hand curled the way the current flows, thumb along $\vec{\mu}$). Then for any orientation

$\vec{\tau} = \vec{\mu}\times\vec{B}, \qquad \tau = \mu B \sin\alpha$

where $\alpha$ is the angle between $\vec{\mu}$ and $\vec{B}$. Maximum torque when the loop's plane contains the field ($\alpha = 90^\circ$); zero torque when the normal lines up with the field. The associated potential energy is $U = -\vec{\mu}\cdot\vec{B} = -\mu B\cos\alpha$, lowest when aligned. Rotating a $0.50$ A$\cdot$m$^2$ moment from aligned to perpendicular in a $0.20$ T field costs $\Delta U = 0 - (-0.10) = 0.10$ J of external work.

Worked example: a coil that oscillates

A coil of $200$ turns, area $4.0\times10^{-4}$ m$^2$, carries $0.50$ A and hangs so it can rotate about a vertical axis in a horizontal $0.020$ T field. Its moment of inertia is $1.2\times10^{-6}$ kg$\cdot$m$^2$. Find the period of small oscillations about the aligned position.

The aligned position is the energy minimum, so nudge it and it swings back. The restoring torque at a small angle $\alpha$ is

$\tau = -\mu B\sin\alpha \approx -\mu B \alpha$

using $\sin\alpha \approx \alpha$ for small angles. That is a restoring torque proportional to displacement, which is angular simple harmonic motion: $I_{\text{rot}}\ddot\alpha = -\mu B\alpha$, so the effective torsion constant is $\kappa = \mu B$ and

$T = 2\pi\sqrt{\frac{I_{\text{rot}}}{\mu B}}$

Compute the moment first. $\mu = NIA = (200)(0.50)(4.0\times10^{-4}) = 4.0\times10^{-2}$ A$\cdot$m$^2$. Then $\mu B = (4.0\times10^{-2})(0.020) = 8.0\times10^{-4}$ N$\cdot$m per radian. So

$T = 2\pi\sqrt{\frac{1.2\times10^{-6}}{8.0\times10^{-4}}} = 2\pi\sqrt{1.5\times10^{-3}} = 2\pi(0.0387) = 0.243\ \text{s}$

That is the pendulum formula with $\mu B$ playing the role of $mg\ell$. Whenever you see a dipole nudged from alignment, expect this.

Recognising the technique

Most magnetic force problems announce themselves in one of five ways.

A charge and a speed and a field. If the question gives you a moving charge and asks for a force, it is $qvB\sin\theta$ and the only work is finding $\theta$. If it asks for a radius, a period, a frequency, or a kinetic energy, you want $r = mv/qB$ and $T = 2\pi m/qB$, and you pick whichever one contains the quantity you were given. Energy questions go through $K = \frac{1}{2}mv^2$ with $v = qBr/m$.

Two stages, one electric and one magnetic. A potential difference accelerating an ion and then a field bending it means eliminate $v$ between $qV = \frac{1}{2}mv^2$ and $r = mv/qB$. Any problem with a voltage and a radius in the same sentence is this problem.

A velocity that is not perpendicular to the field. Split it into parallel and perpendicular pieces immediately. The parallel piece coasts, the perpendicular piece circles, and the words "helix" or "pitch" confirm it.

A wire or a rod. Then it is $ILB\sin\theta$, and the rest is a free body diagram of the sort you have drawn many times: balance it against gravity for a floating wire, against $\mu N$ for friction, against $ma$ for an accelerating rod. Two wires means $\mu_0 I_1 I_2/2\pi d$ per unit length.

A loop, a coil, or the word "turns". Compute $\mu = NIA$ first, always, then reach for $\tau = \mu B\sin\alpha$, $U = -\mu B\cos\alpha$, or the oscillation period. Do not try to add up forces on the individual sides; the dipole moment already did that for you.

Everything here has taken the field as given. The next lesson asks where the field came from in the first place, and gives you the tool for computing it from the currents that make it: the Biot-Savart law.