Unit 6

Magnetic Fields & Forces

An electric field pushes on any charge. A magnetic field is stranger: it ignores charges at rest and pushes on moving ones — sideways, perpendicular to both the motion and the field. That sideways push bends electron beams into circles, sorts atoms by mass, produces a measurable voltage across a current-carrying chip (the Hall sensor in your phone and in every brushless motor), and turns a current loop into a rotating shaft (every electric motor). This unit is about what magnetic fields do; Unit 7 is about where they come from. The lab lets you fire charged particles through adjustable E and B fields and watch the circles, helices and straight-line “selected” paths appear.

1 · The magnetic force on a moving charge

Picture first. Hold a magnet near a CRT screen (an old TV) and the picture warps — the electron beam is pushed sideways. Experiments show the push is: zero if the charge is at rest; zero if it moves along the field; largest when it moves across the field; proportional to the charge and the speed; and always at right angles to both the velocity and the field. One formula captures all of it:

\[ \mathbf{F}_B = q\,\mathbf{v}\times\mathbf{B},\qquad |\mathbf{F}_B| = |q|vB\sin\phi \] \(q\) = charge (with its sign), \(\mathbf{v}\) = velocity, \(\mathbf{B}\) = magnetic field, \(\phi\) = angle between \(\mathbf{v}\) and \(\mathbf{B}\). The unit of \(B\) is the tesla: \(1\ \text{T} = 1\ \text{N}/(\text{C}\cdot\text{m/s}) = 1\ \text{N}/(\text{A}\cdot\text{m})\). The older gauss is \(10^{-4}\) T.

With an electric field present too, the total is the Lorentz force \(\mathbf{F} = q(\mathbf{E} + \mathbf{v}\times\mathbf{B})\) — the complete rule for how fields push charges, and the starting point for every device in this unit.

Source	Typical \(B\)
Earth's field at the surface	\(\approx 5\times10^{-5}\) T (50 µT)
Fridge magnet	\(\approx 5\times10^{-3}\) T
Neodymium magnet surface; loudspeaker gap	\(\approx 1\) T
Hospital MRI	1.5–3 T

Right-hand rule: point the fingers of your right hand along \(\mathbf{v}\), curl them toward \(\mathbf{B}\) (through the smaller angle \(\phi\)); the thumb gives \(\mathbf{v}\times\mathbf{B}\). For a positive charge that is the force; for a negative charge (an electron) the force is the opposite way. Convention: ⊗ = into the page (the tail feathers of an arrow flying away), ⊙ = out of the page (the arrow's point coming at you).

Common mistakes

Forgetting the sign of \(q\). Do the right-hand rule, then flip it for electrons. Doing it with your left hand to “handle” electrons is how signs get lost twice.
Thinking \(\mathbf{F}\) points along \(\mathbf{B}\). It never does. Magnetic field lines are not lines of force the way electric field lines are.
Using the cosine. The magnitude has \(\sin\phi\): motion along \(\mathbf{B}\) feels nothing.

The magnetic force does no work

Because \(\mathbf{F}_B\perp\mathbf{v}\), the power \(\mathbf{F}_B\cdot\mathbf{v} = 0\). A static magnetic field can change a particle's direction but never its speed or kinetic energy. To speed particles up you need an electric field — which is why accelerators use E to accelerate and B to steer.

Unit vectors make cross products mechanical: \(\hat{\imath}\times\hat{\jmath} = \hat{k}\), \(\hat{\jmath}\times\hat{k} = \hat{\imath}\), \(\hat{k}\times\hat{\imath} = \hat{\jmath}\), and reversing the order flips the sign. Example: an electron with \(\mathbf{v} = v\,\hat{\imath}\) in \(\mathbf{B} = B\,\hat{\jmath}\) feels \(\mathbf{F} = (-e)vB(\hat{\imath}\times\hat{\jmath}) = -evB\,\hat{k}\).

2 · Circular motion in a uniform field

If \(\mathbf{v}\perp\mathbf{B}\), the force has constant size \(|q|vB\) and is always perpendicular to the velocity — exactly the recipe for uniform circular motion. The magnetic force supplies the centripetal force:

\[ |q|vB = \frac{mv^2}{r}\quad\Longrightarrow\quad r = \frac{mv}{|q|B} = \frac{p}{|q|B} \] \(m\) = mass, \(p = mv\) = momentum. Faster or heavier → bigger circle; stronger field or more charge → tighter circle. Measuring \(r\) in a known \(B\) measures momentum — that is how particle detectors work.

\[ \omega = \frac{v}{r} = \frac{|q|B}{m},\qquad f = \frac{|q|B}{2\pi m},\qquad T = \frac{2\pi m}{|q|B} \] The cyclotron frequency. The speed has cancelled: fast particles run bigger circles in the same time. Proton: \(f = 15.2\) MHz per tesla. Electron: \(28.0\) GHz per tesla. (Valid while \(v\ll c\); near light speed relativity lowers the frequency to \(f = |q|B/(2\pi\gamma m)\).)

A positive charge in a field into the page circles counterclockwise; an electron in the same field circles clockwise. The force always points to the centre, the speed never changes, and the period \(2\pi m/(|q|B)\) does not depend on the speed.

EE connection

The cavity magnetron in every microwave oven is electrons circling in a magnetic field near their cyclotron frequency; for 2.45 GHz that needs \(B = 2\pi m_ef/e = 2\pi(9.11\times10^{-31})(2.45\times10^{9})/(1.60\times10^{-19}) \approx 0.088\) T. That sets the scale: real magnetrons use fields of roughly a tenth of a tesla from two ring magnets. Old CRT televisions and oscilloscopes steered their electron beam with magnetic deflection coils. And the cyclotron accelerator works because \(f\) is independent of speed: a fixed-frequency AC voltage kicks the particle each half-turn as it spirals outward.

Interactive lab Charged particles in magnetic fields Open it in Interactive Laboratories →

3 · Helical motion

If the velocity has a component along \(\mathbf{B}\), split it: \(v_\parallel\) (along \(\mathbf{B}\)) and \(v_\perp\) (across it). The field ignores \(v_\parallel\) — that component stays constant — while \(v_\perp\) goes round in a circle. Circle + steady drift = a helix, like a stretched spring.

\[ v_\perp = v\sin\phi,\quad v_\parallel = v\cos\phi,\qquad r = \frac{mv_\perp}{|q|B},\qquad \text{pitch } p = v_\parallel T = v_\parallel\frac{2\pi m}{|q|B} \] The pitch \(p\) is the distance along \(\mathbf{B}\) per turn. Only \(v_\perp\) goes in the radius; only \(v_\parallel\) in the pitch. Using the full \(v\) in either is the standard mistake.

Nature's version

Charged particles from the Sun spiral along Earth's field lines toward the poles, where they hit the atmosphere — the aurora. Where field lines converge, the spiralling particle can be reflected back (a magnetic mirror), which is how the Van Allen belts trap particles and how some fusion experiments confine plasma.

4 · Velocity selector and mass spectrometer

Cross an electric field and a magnetic field at right angles, both perpendicular to a beam. For a positive charge moving right, with \(\mathbf{E}\) pointing down and \(\mathbf{B}\) into the page, the electric force \(qE\) is down and the magnetic force \(qvB\) is up. Only one speed balances them:

\[ qE = qvB\quad\Longrightarrow\quad v = \frac{E}{B} \] The velocity selector (Wien filter). The charge and mass cancel, so it passes particles of speed \(E/B\) whatever their charge sign or mass. Faster ones are bent toward the magnetic force, slower ones toward the electric force, and a slit blocks both.

Send the selected beam into a region of field \(B_0\) alone. Every ion now has the same speed, so its radius depends only on mass-to-charge ratio:

\[ r = \frac{mv}{qB_0} = \frac{mE}{qBB_0}\quad\Longrightarrow\quad m = \frac{qBB_0\,r}{E} \] The mass spectrometer. Ions land a distance \(2r\) from the entrance slit, so heavier isotopes land farther away. Measure the landing spot, get the mass.

A Bainbridge mass spectrometer. Only ions with \(v = E/B\) pass the selector straight. In the field \(B_0\) they make half-circles of radius \(r\propto m\) and land at \(2r\) from the slit. (Radii drawn in the true 12 : 14 ratio.)

EE connection

Mass spectrometers check the purity of semiconductor materials and gases, and the same physics drives ion implantation — the step in chip fabrication that dopes silicon. An implanter uses an analysing magnet to pick out exactly the boron or phosphorus isotope wanted, then fires it into the wafer. The doping profile of every transistor you will ever design was set by \(r = mv/(qB)\).

5 · The Hall effect

Now put the moving charges inside a conductor. A current \(I\) flows along a flat strip of width \(w\) and thickness \(t\); a field \(\mathbf{B}\) points through the strip's face. The carriers are pushed sideways by \(q\mathbf{v}_d\times\mathbf{B}\) and pile up on one edge, leaving the opposite charge on the other edge. That separated charge makes a transverse electric field \(E_H\) that grows until it cancels the magnetic push — a velocity selector built into the wire:

\[ qE_H = qv_dB,\qquad V_H = E_Hw = v_dBw,\qquad v_d = \frac{I}{nqwt}\quad\Longrightarrow\quad V_H = \frac{IB}{nqt} \] \(V_H\) = Hall voltage across the width; \(n\) = carrier density; \(t\) = thickness measured along \(\mathbf{B}\). The width cancels. \(V_H\) is proportional to \(B\) — so it measures \(B\) — and inversely proportional to \(n\), so a thin semiconductor (small \(n\), small \(t\)) gives a far bigger signal than a metal.

Drawn for positive carriers: moving right in a field into the page, they are pushed up, so the top edge is +. In a metal the carriers are electrons moving left; \((-e)(-v\,\hat{\imath})\times(-B\,\hat{k})\) also points up, so electrons pile up on top and the top edge is −. The sign of \(V_H\) reveals the sign of the carriers — this is how “holes” in p-type semiconductors were shown to act as positive charges.

EE connection: Hall sensors

A Hall sensor is a tiny semiconductor plate with a bias current and an amplifier. It is everywhere: brushless DC motors (three Hall sensors tell the controller where the rotor magnets are so it can switch the coils at the right moment), clamp meters and current sensors that measure DC current without breaking the circuit (they sense the wire's magnetic field, Unit 7), wheel-speed and crankshaft sensors in cars, lid-closed switches in laptops, contactless joysticks and keyboard switches. Hall sensitivity is quoted in mV/mT, and it equals \(I/(nqt)\).

6 · Force on a current-carrying wire

A current is moving charge, so a wire in a field feels the sum of the forces on its carriers. A straight segment of length \(L\) and area \(A\) contains \(nAL\) carriers, each feeling \(q\mathbf{v}_d\times\mathbf{B}\):

\[ \mathbf{F} = (nAL)\,q\mathbf{v}_d\times\mathbf{B} = (nqv_dA)\,\mathbf{L}\times\mathbf{B} = I\,\mathbf{L}\times\mathbf{B},\qquad |\mathbf{F}| = ILB\sin\phi \] \(\mathbf{L}\) = vector of length \(L\) pointing along the conventional current. Electrons drift backwards, but \(q\) is negative too, so the result depends only on \(I\) — you never need to think about electrons here.

\[ d\mathbf{F} = I\,d\mathbf{l}\times\mathbf{B},\qquad \mathbf{F} = I\int_a^b d\mathbf{l}\times\mathbf{B} \] For a curved wire, add up tiny straight pieces \(d\mathbf{l}\). In a uniform field \(\mathbf{B}\) comes out of the integral and \(\int_a^b d\mathbf{l}\) is just the straight vector from \(a\) to \(b\): any wire from \(a\) to \(b\) feels the same force as a straight one. For a closed loop in a uniform field, \(a = b\) and the net force is zero — but, as the next section shows, the torque is not.

EE connection

A loudspeaker is \(F = ILB\) wound into a coil: the voice coil sits in a radial field in the gap of a ring magnet, so every piece of wire feels an axial force in the same direction. The audio current \(I(t)\) becomes a force \(F(t)\) that drives the cone. Relays, solenoid valves and the rail gun all start from this formula.

7 · Torque on a current loop

Take a rectangular loop (sides \(a\) and \(b\)) carrying current \(I\) in a uniform field. The net force is zero, but the forces on opposite sides are pushed in opposite directions along different lines — a couple — so the loop twists.

Counterclockwise current; \(\mathbf{B}\) to the right, in the plane of the loop. The right side (current up) is pushed into the page, the left side (current down) out of it: the loop starts to rotate about the vertical axis so that its moment \(\mathbf{\mu}\) (out of the page) swings toward \(\mathbf{B}\). Here \(\theta = 90^\circ\) and the torque is maximal, \(\tau = IabB\).

Each vertical side has force \(IaB\), with lever arm \(\tfrac{b}{2}\sin\theta\) about the axis, where \(\theta\) is the angle between the loop's normal and \(\mathbf{B}\). Two sides: \(\tau = 2(IaB)(\tfrac b2\sin\theta) = I(ab)B\sin\theta\). With \(N\) turns, and area \(A = ab\):

\[ \boldsymbol{\mu} = NIA\,\hat{\mathbf{n}},\qquad \boldsymbol{\tau} = \boldsymbol{\mu}\times\mathbf{B},\qquad |\boldsymbol{\tau}| = NIAB\sin\theta \] \(\boldsymbol{\mu}\) = magnetic dipole moment (A·m²). Its direction \(\hat{\mathbf{n}}\) comes from a second right-hand rule: curl your fingers along the current, and your thumb points along \(\boldsymbol{\mu}\). The result holds for any flat loop shape — only the area matters.

\[ U = -\boldsymbol{\mu}\cdot\mathbf{B} = -\mu B\cos\theta \] Potential energy of the dipole: lowest (\(-\mu B\)) when \(\boldsymbol{\mu}\) is aligned with \(\mathbf{B}\), highest (\(+\mu B\)) when anti-aligned. A dipole is a compass needle: it turns to line up with the field. This is exactly the electric dipole of Unit 1 (\(\boldsymbol{\tau} = \mathbf{p}\times\mathbf{E}\), \(U = -\mathbf{p}\cdot\mathbf{E}\)) with \(\mathbf{p}\to\boldsymbol{\mu}\), \(\mathbf{E}\to\mathbf{B}\).

Common mistake

\(\theta\) is the angle between \(\boldsymbol{\mu}\) (the normal to the loop) and \(\mathbf{B}\) — not the angle between the plane of the loop and \(\mathbf{B}\). When the field lies in the plane of the loop (as in the figure), \(\theta = 90^\circ\) and the torque is maximum; when the field passes straight through the loop, the torque is zero.

8 · The DC motor

Put a coil on a shaft between magnet poles. The torque \(NIAB\sin\theta\) turns it toward alignment — and then, as it overshoots, the torque reverses and pulls it back. Left alone, the coil would just oscillate about \(\theta = 0\) like a compass needle.

The fix is the commutator: a split ring on the shaft, touched by two brushes, that reverses the current in the coil every half-turn — exactly when \(\boldsymbol{\mu}\) passes through alignment. Reversing \(I\) flips \(\boldsymbol{\mu}\), so the torque keeps the same sense of rotation all the way round. A single coil gives a torque \(NIAB|\sin\theta|\), averaging \(\tfrac{2}{\pi}NIAB\); real motors use many coils at different angles so the torque is nearly constant.

Coming in Unit 8

A spinning coil in a magnetic field is also a generator. A running motor generates a back EMF that opposes the supply and grows with speed, which is why a motor draws a huge current when stalled or starting and much less at full speed.

EE connection

Brushes spark and wear out, so modern drones, disk drives, fans and electric cars use brushless DC motors: the magnets spin, the coils stay still, and a transistor bridge does the commutation electronically — timed by the Hall sensors of Section 5 or by sensing the back EMF. Designing that controller is a classic power-electronics job.

9 · Worked examples

1. Bending an electron beam

Electrons are accelerated from rest through 2.0 kV and enter a uniform \(1.0\) mT field perpendicular to their velocity. Find their speed, the radius of their path, the force on each, and the cyclotron frequency.

\[ \tfrac12m_ev^2 = eV\;\Rightarrow\; v = \sqrt{\frac{2eV}{m_e}} = \sqrt{\frac{2(1.60\times10^{-19})(2000)}{9.11\times10^{-31}}} = \mathbf{2.65\times10^{7}\ m/s} \] \[ r = \frac{m_ev}{eB} = \frac{(9.11\times10^{-31})(2.65\times10^{7})}{(1.60\times10^{-19})(1.0\times10^{-3})} = \mathbf{0.15\ m} \] \[ F = evB = (1.60\times10^{-19})(2.65\times10^{7})(1.0\times10^{-3}) = \mathbf{4.2\times10^{-15}\ N},\qquad f = \frac{eB}{2\pi m_e} = \mathbf{28\ MHz} \]

\(v\approx 0.09c\), so the non-relativistic formulas are fine (error well under 1 %). A field of only 1 mT — twenty times Earth's — bends the beam on a 15 cm radius: that is why CRTs needed magnetic shielding.

2. Separating carbon isotopes

Singly charged carbon ions pass a selector with \(E = 1.0\times10^{5}\) V/m and \(B = 0.50\) T, then enter \(B_0 = 0.50\) T. Find where C-12 and C-14 land (\(m = 12u\), \(14u\), \(u = 1.66\times10^{-27}\) kg).

\[ v = \frac{E}{B} = \frac{1.0\times10^{5}}{0.50} = 2.0\times10^{5}\ \text{m/s} \] \[ r_{12} = \frac{(12)(1.66\times10^{-27})(2.0\times10^{5})}{(1.60\times10^{-19})(0.50)} = 4.98\ \text{cm},\qquad r_{14} = \tfrac{14}{12}r_{12} = 5.81\ \text{cm} \] \[ \text{landing points: } 2r_{12} = \mathbf{9.96\ cm},\quad 2r_{14} = \mathbf{11.6\ cm};\qquad \text{separation } 2(r_{14} - r_{12}) = \mathbf{1.66\ cm} \]

Easily resolved on a detector. Note \(r\propto m\) at fixed \(v\): the 14 : 12 mass ratio is exactly the radius ratio. The same principle, at much higher energies, is used in accelerator mass spectrometry for radiocarbon dating.

3. A helix

An electron moves at \(2.0\times10^{6}\) m/s at \(30^\circ\) to a uniform \(1.0\) mT field. Find the radius and pitch of its helix.

\[ v_\perp = v\sin30^\circ = 1.0\times10^{6}\ \text{m/s},\qquad v_\parallel = v\cos30^\circ = 1.73\times10^{6}\ \text{m/s} \] \[ r = \frac{m_ev_\perp}{eB} = \frac{(9.11\times10^{-31})(1.0\times10^{6})}{(1.60\times10^{-19})(1.0\times10^{-3})} = \mathbf{5.7\ mm} \] \[ T = \frac{2\pi m_e}{eB} = 3.58\times10^{-8}\ \text{s},\qquad p = v_\parallel T = (1.73\times10^{6})(3.58\times10^{-8}) = \mathbf{6.2\ cm} \]

A tight helix, stretched out along the field: each turn advances 6.2 cm but is only 1.1 cm across.

4. Why Hall sensors are semiconductors

(a) A copper strip \(0.50\) mm thick carries 5.0 A in a 1.2 T field. (b) A semiconductor Hall element \(0.20\) mm thick, \(n = 1.0\times10^{21}\ \text{m}^{-3}\), carries 1.0 mA in a 0.10 T field. Find both Hall voltages.

\[ \text{(a)}\ V_H = \frac{IB}{net} = \frac{(5.0)(1.2)}{(8.49\times10^{28})(1.60\times10^{-19})(5.0\times10^{-4})} = \mathbf{8.8\times10^{-7}\ V}\ (0.88\ \mu\text{V}) \] \[ \text{(b)}\ V_H = \frac{(1.0\times10^{-3})(0.10)}{(1.0\times10^{21})(1.60\times10^{-19})(2.0\times10^{-4})} = \mathbf{3.1\times10^{-3}\ V}\ (3.1\ \text{mV}) \]

With 5000 times less current and 12 times less field, the semiconductor gives 3500 times more voltage — because its carrier density is \(10^{8}\) times smaller, so the carriers drift much faster for a given current. Copper's sub-microvolt signal would drown in noise.

5. The force on a loudspeaker voice coil

A voice coil has 50 turns of radius 1.25 cm in a radial field of 1.0 T. Find the force when the current is 1.0 A.

The field is radial and the wire runs around the circle, so \(d\mathbf{l}\perp\mathbf{B}\) everywhere and every piece is pushed along the axis in the same direction. Total wire length \(L = N(2\pi r)\):

\[ L = 50(2\pi)(0.0125) = 3.93\ \text{m},\qquad F = ILB = (1.0)(3.93)(1.0) = \mathbf{3.9\ N} \]

Reverse the current and the force reverses — an AC audio signal pushes the cone in and out. The ratio \(F/I = NB(2\pi r) = 3.9\) N/A is the speaker's \(Bl\) “force factor” printed on its data sheet.

6. Torque and energy of a motor coil

A 100-turn rectangular coil, 4.0 cm × 5.0 cm, carries 2.0 A in a 0.50 T field. Find \(\mu\), the maximum torque, the torque at \(\theta = 30^\circ\), the work to flip it from aligned to anti-aligned, and the average torque with a commutator.

\[ \mu = NIA = (100)(2.0)(0.040\times0.050) = \mathbf{0.40\ A\cdot m^2},\qquad \tau_{\max} = \mu B = \mathbf{0.20\ N\cdot m} \] \[ \tau(30^\circ) = \mu B\sin30^\circ = \mathbf{0.10\ N\cdot m} \] \[ W = U(180^\circ) - U(0^\circ) = (+\mu B) - (-\mu B) = 2\mu B = \mathbf{0.40\ J},\qquad \bar\tau = \tfrac{2}{\pi}\mu B = \mathbf{0.13\ N\cdot m} \]

Note \(\theta = 30^\circ\) means the normal is 30° from \(\mathbf{B}\) — the coil is nearly aligned, so the torque is only half its maximum.

Checklist

I can find the direction of \(q\mathbf{v}\times\mathbf{B}\) with the right-hand rule and with unit vectors, and I flip it for negative charges.
I know the magnetic force does no work and cannot change a particle's speed.
I can derive \(r = mv/(|q|B)\) and \(f = |q|B/(2\pi m)\), and explain why \(f\) is independent of speed.
I split \(\mathbf{v}\) into \(v_\parallel\) and \(v_\perp\) for helical motion and compute radius and pitch.
I can explain the velocity selector (\(v = E/B\)) and the mass spectrometer.
I can derive \(V_H = IB/(nqt)\) and say what its sign and size tell you.
I use \(\mathbf{F} = I\mathbf{L}\times\mathbf{B}\), and know a closed loop in a uniform field feels zero net force.
I use \(\boldsymbol{\mu} = NIA\hat{\mathbf{n}}\), \(\boldsymbol{\tau} = \boldsymbol{\mu}\times\mathbf{B}\), \(U = -\boldsymbol{\mu}\cdot\mathbf{B}\), with \(\theta\) measured from the normal.
I can explain how a commutator turns an oscillating torque into continuous rotation.