Magnetic Fields and Forces

Magnetic fields are produced by moving charges — whether that is a current flowing through a wire, electrons orbiting a nucleus, or the aligned magnetic domains in a permanent magnet. This lesson covers the forces that magnetic fields exert on current-carrying conductors and moving charges, which form the basis of electric motors, loudspeakers, and particle accelerators.

Magnetic Flux Density

Magnetic flux density (B) — often loosely called “magnetic field strength” — is defined by the force it exerts on a current-carrying conductor:

B = F / (IL sinθ)

or equivalently, the force on a current-carrying conductor in a magnetic field is:

F = BIL sinθ

where:

F is the force on the conductor (N)
B is the magnetic flux density (T — tesla)
I is the current (A)
L is the length of conductor in the field (m)
θ is the angle between the conductor and the field direction

When the conductor is perpendicular to the field (θ = 90°), sinθ = 1 and the force is maximum:

F = BIL

When the conductor is parallel to the field (θ = 0°), sinθ = 0 and the force is zero.

The tesla (T) is defined as: 1 T = 1 N A⁻¹ m⁻¹ — the magnetic flux density that produces a force of 1 N on a 1 m conductor carrying 1 A perpendicular to the field.

Typical Magnetic Flux Density Values

Source	Approximate B
Earth’s magnetic field	50 μT (5 × 10⁻⁵ T)
Fridge magnet	5 mT
School bar magnet	10–50 mT
Strong permanent magnet (neodymium)	0.5–1.5 T
MRI scanner	1.5–3 T
Particle accelerator	5–10 T

Fleming’s Left-Hand Rule

The direction of the force on a current-carrying conductor is given by Fleming’s left-hand rule:

First finger — Field direction (north to south)
seCond finger — Current direction (conventional, positive to negative)
thuMb — Motion (force) direction

Hold the left hand so that the first finger, second finger, and thumb are all at right angles to each other. The thumb gives the direction of the force (and therefore the motion of the conductor).

Important Notes

This rule uses conventional current (the direction positive charges would flow). If the question involves electron flow, reverse the current direction before applying the rule.
The force is always perpendicular to both the field and the current — this is a consequence of the cross product nature of the magnetic force.

Force on a Moving Charge

A single charge Q moving with velocity v through a magnetic field B experiences a force:

F = BQv sinθ

where θ is the angle between the velocity and the field.

When the charge moves perpendicular to the field:

F = BQv

This force is always perpendicular to the velocity of the charge. This means it cannot do work on the charge (the force and displacement are always at right angles), so the speed remains constant. However, the direction changes continuously — the charge follows a circular path.

Circular Motion of a Charged Particle

For a charge moving perpendicular to a uniform magnetic field, the magnetic force provides the centripetal force:

BQv = mv²/r

Solving for the radius:

r = mv / (BQ)

This relationship is the basis of mass spectrometry and cyclotron particle accelerators.

Key Insight: The Period is Independent of Speed

The time for one complete revolution is:

T = 2πr / v = 2πm / (BQ)

The period T depends only on the mass, charge, and field strength — not on the speed. Faster particles move in larger circles but complete each orbit in the same time. This is the principle that makes cyclotrons work: particles are accelerated in a constant-frequency oscillating field and spiral outward.

Worked Example 1

A proton (mass = 1.67 × 10⁻²⁷ kg, charge = 1.6 × 10⁻¹⁹ C) moves at 3.0 × 10⁶ m s⁻¹ perpendicular to a magnetic field of 0.50 T. What is the radius of its circular path?

r = mv / (BQ) = (1.67 × 10⁻²⁷ × 3.0 × 10⁶) / (0.50 × 1.6 × 10⁻¹⁹)

r = 5.01 × 10⁻²¹ / 8.0 × 10⁻²⁰ = 0.063 m ≈ 6.3 cm

Worked Example 2: Comparing Particles in the Same Field

A proton and an alpha particle (charge = 2e, mass = 4mₚ) are both accelerated from rest through the same potential difference V and then enter the same magnetic field. Compare their circular path radii.

After acceleration: ½mv² = QV, so v = √(2QV/m)

For the proton: vₚ = √(2eV/mₚ)

For the alpha: vα = √(2 × 2e × V / 4mₚ) = √(eV/mₚ) = vₚ/√2

Radius: r = mv / (BQ)

rₚ = mₚvₚ / (Be)

rα = 4mₚvα / (B × 2e) = 4mₚ × (vₚ/√2) / (2Be) = 2mₚvₚ / (√2 × Be) = √2 × mₚvₚ / (Be)

rα / rₚ = √2 ≈ 1.41

The alpha particle has a radius √2 times larger than the proton.

Applications: The Electric Motor

The DC electric motor exploits the force on a current-carrying conductor in a magnetic field:

A rectangular coil of wire is placed in a magnetic field between the poles of a permanent magnet
When current flows through the coil, the forces on the two sides of the coil (which carry current in opposite directions) are in opposite directions — creating a turning effect (torque)
A commutator (split ring) reverses the current direction every half turn, ensuring the coil continues to rotate in the same direction

The torque depends on B, I, the number of turns N, and the area of the coil A.

Worked Example 3: Force on a Motor Coil

A motor coil has 200 turns, each of length 0.08 m in a field of 0.35 T, carrying a current of 2.5 A. Calculate the force on one side of the coil.

F = NBIL = 200 × 0.35 × 2.5 × 0.08 = 14 N

If the coil has a width of 0.05 m, the torque on the coil is:

τ = F × d = 14 × 0.05 = 0.70 N m (when the coil plane is parallel to the field)

The Hall Effect

When a current-carrying conductor is placed in a magnetic field perpendicular to the current, the magnetic force pushes charge carriers to one side. This creates a charge imbalance — a potential difference across the conductor called the Hall voltage.

The Hall voltage is:

V_H = BI / (nqt)

where n is the number density of charge carriers, q is the charge on each carrier, and t is the thickness of the conductor in the direction of the magnetic force.

The Hall voltage is proportional to B and I, and inversely proportional to the number density of charge carriers and the thickness of the conductor. Hall probes use this effect to measure magnetic flux density.

Why Semiconductors Give Larger Hall Voltages

In semiconductors, n (the charge carrier density) is much smaller than in metals. Since V_H is inversely proportional to n, semiconductor Hall probes produce much larger (and easier to measure) Hall voltages than metal ones.

Real-World Applications

Loudspeakers: A coil attached to a speaker cone carries an alternating current in a permanent magnetic field. The varying force (F = BIL) causes the cone to vibrate at the frequency of the current, producing sound waves.

Mass spectrometers: Ions are accelerated through a known potential difference and then enter a magnetic field. The radius of their circular path (r = mv/BQ) depends on their mass-to-charge ratio, allowing different isotopes to be separated and identified.

Cyclotrons: Charged particles are accelerated in a spiral path by a combination of a magnetic field (which bends them in circles) and an oscillating electric field (which accelerates them each half-turn). Used to produce medical isotopes and in research.

Common Exam Mistakes

Using the wrong hand: Fleming’s LEFT-hand rule is for the motor effect (force on a conductor). The RIGHT-hand rule is for the generator effect (covered in the electromagnetic induction lesson).
Forgetting sinθ: The force is BIL sinθ. If the wire is at an angle to the field, you must include the sin term.
Confusing speed and velocity: The magnetic force changes the direction of a charged particle but not its speed. The kinetic energy is constant.
Wrong direction for electrons: Electrons move opposite to conventional current. Reverse the current direction before applying Fleming’s rule.

Key Points

Force on a current-carrying conductor: F = BIL sinθ
Fleming’s left-hand rule gives the force direction (First finger = Field, seCond finger = Current, thuMb = Motion)
Force on a moving charge: F = BQv sinθ
Charged particles in a uniform magnetic field follow circular paths: r = mv / (BQ)
The magnetic force on a moving charge is always perpendicular to the velocity, so it changes direction but not speed
The period of circular motion is T = 2πm/(BQ), independent of speed
Electric motors use the force F = BIL on current-carrying coils

A-Level Deep Dive: Magnetic Fields and Forces

Spec mapping

Edexcel 9PH0 specification Topic 7 — Electric and Magnetic Fields, magnetic-fields sub-strand, covers magnetic flux density $B$ , the force on a current-carrying conductor ( $F = BIL\sin\theta$ ), the force on a moving charge ( $F = BQv\sin\theta$ ) and the resulting circular motion of charged particles in uniform magnetic fields (refer to the official specification document for exact wording). The same topic threads forward into electromagnetic induction (Topic 7's flux/flux-linkage sub-strand), into Topic 12 (Particle Physics, where $r = mv/(BQ)$ underpins mass spectrometers and cyclotrons), and into Topic 11 (Nuclear and Particle Physics, where particle accelerators and detectors depend on these results). Magnetic-field questions appear most prominently in Paper 2 (Advanced Physics II) but synoptic-style questions on Paper 3 routinely revisit $F = BIL$ and $r = mv/(BQ)$ in unfamiliar contexts. The Edexcel formula booklet does list $F = BIL\sin\theta$ , $F = BQv\sin\theta$ and $r = p/(BQ)$ — but you must still know the geometry and the right-hand rule.

Worked example with full mark scheme

Question (8 marks):

A proton enters a region of uniform magnetic flux density $B = 0.40\,\text{T}$ at right angles to the field. The proton has been accelerated from rest through a potential difference of $2.0\,\text{kV}$ . (Mass of proton $m_p = 1.67 \times 10^{-27}\,\text{kg}$ ; charge $e = 1.60 \times 10^{-19}\,\text{C}$ .)

(a) Show that the speed of the proton on entering the field is approximately $6.2 \times 10^5\,\text{m s}^{-1}$ . (3)

(b) Calculate the radius of the circular path followed by the proton in the field. (3)

Solution with mark scheme:

(a) Step 1 — equate work done by the accelerating p.d. to the kinetic energy gained.

$eV = \tfrac{1}{2} m_p v^2$

M1 — correct energy equation. A common slip is to write $qV = mv^2$ (forgetting the $\tfrac{1}{2}$ ), which loses both M1 and the A1.

Step 2 — rearrange for $v$ .

$v = \sqrt{\dfrac{2eV}{m_p}} = \sqrt{\dfrac{2 \times 1.60 \times 10^{-19} \times 2000}{1.67 \times 10^{-27}}}$

M1 — correct substitution with consistent SI units ( $V$ in volts, charge in coulombs, mass in kg).

Step 3 — evaluate.

$v = \sqrt{3.83 \times 10^{11}} \approx 6.19 \times 10^5\,\text{m s}^{-1}$

A1 — answer to at least 2 s.f., consistent with the printed value $6.2 \times 10^5\,\text{m s}^{-1}$ . Because the answer is given ("show that"), the working must include at least one intermediate numerical step before the final value.

(b) Step 1 — equate magnetic force to centripetal force.

$BQv = \dfrac{m_p v^2}{r}$

M1 — recognising that the magnetic force provides the centripetal force.

Step 2 — rearrange for $r$ .

$r = \dfrac{m_p v}{BQ}$

M1 — correct rearrangement (this is also the formula $r = p/(BQ)$ from the data booklet).

Step 3 — substitute and evaluate.

$r = \dfrac{1.67 \times 10^{-27} \times 6.19 \times 10^5}{0.40 \times 1.60 \times 10^{-19}} \approx 1.6 \times 10^{-2}\,\text{m}$

A1 — $r \approx 1.6\,\text{cm}$ (accept $1.6 \times 10^{-2}\,\text{m}$ to 2 s.f.).

B1 — because the magnetic force is always perpendicular to the velocity, it does no work on the proton, so the speed (and therefore the KE) stays constant; only the direction of motion changes.

Total: 8 marks (M3 A2 B2, plus 1 M earned in (b) for the centripetal-force equation).

Specimen question modelled on the Edexcel 9PH0 Paper 2 format

Question (6 marks): A horizontal copper rod of length $0.25\,\text{m}$ and mass $15\,\text{g}$ is suspended by two flexible vertical wires connected to a power supply. The rod sits in a uniform horizontal magnetic field of flux density $B$ , perpendicular to the rod. When a current of $3.5\,\text{A}$ is switched on, the rod is observed to remain stationary in mid-air, with the wires going slack.

(a) Draw a free-body diagram and state the condition on the magnetic force for the rod to remain stationary with the wires slack. (2)

(b) Calculate the magnetic flux density $B$ required. (3)

Mark scheme decomposition by AO:

(a)

B1 (AO1.2) — free-body diagram showing weight $W = mg$ acting downward and magnetic force $F = BIL$ acting upward, with no tension in the wires.
B1 (AO2.1) — statement that the magnetic force must equal the weight in magnitude and act vertically upward ( $BIL = mg$ ).

(b)

M1 (AO1.1) — correct equation $BIL = mg$ .
M1 (AO2.1) — substitution $B = mg/(IL) = (0.015 \times 9.81)/(3.5 \times 0.25)$ .
A1 (AO1.2) — $B \approx 0.17\,\text{T}$ (accept $0.168\,\text{T}$ ).

(c)

B1 (AO3.2) — the rod is pushed downward (force reverses by Fleming's left-hand rule); the wires take up tension equal to $mg + BIL$ and the rod is pulled to its original suspended position (or equivalent — accept "rod accelerates downward until wires become taut").

Total: 6 marks split AO1 = 3, AO2 = 2, AO3 = 1. This is a standard "rod-in-field equilibrium" item — the AO2/AO3 marks reward candidates who interpret what "wires go slack" implies physically rather than just computing $B$ blindly.

Synoptic links

Connects to:

Topic 7 — Electromagnetic induction (Faraday's and Lenz's laws): the same field that exerts $F = BIL$ on a current-carrying wire will, when the wire is moved through the field, induce an EMF $\varepsilon = BLv$ . The motor effect and the generator effect are mirror images of one another, related by energy conservation: a motor converts electrical energy to mechanical, a generator the reverse, and both rely on $F = BIL$ at the conductor level.
Topic 8 — Motors and generators: the DC motor torque $\tau = NBIA\cos\phi$ is just $F = BIL$ summed around $N$ turns of a coil of area $A$ . Brushless DC motors, stepper motors and AC induction motors all reduce, in their underlying physics, to the force on current-carrying conductors derived in this lesson.
Topic 12 — Particle accelerators (cyclotrons, synchrotrons): the period $T = 2\pi m/(BQ)$ being independent of speed is the principle that lets a cyclotron use a single fixed-frequency oscillating electric field. At relativistic speeds ( $v \to c$ ), $m$ becomes $\gamma m_0$ and the period is no longer constant — this is why high-energy particle physics requires synchrotrons (which ramp $B$ as energy increases) rather than cyclotrons.
Earth's magnetic field and space physics: charged particles from the solar wind spiral along Earth's geomagnetic field lines (period $T = 2\pi m/(BQ)$ ) and concentrate near the magnetic poles, producing aurorae. The Van Allen radiation belts are particles trapped in the magnetic-bottle geometry of Earth's dipole field — exactly the physics of $F = BQv$ .