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Magnetic fields (AQA specification 3.7.5) describe the forces experienced by moving charges and current-carrying conductors. This topic introduces magnetic flux density, the motor effect, and the motion of charged particles in magnetic fields.
Spec mapping: AQA 7408 A-Level Physics, Section 3.7.5 (Magnetic fields). This lesson covers magnetic flux density as a vector field, the motor effect F = BIL sin θ on current-carrying conductors, the force F = BQv sin θ on moving charged particles, and the qualitative comparison of magnetic field behaviour with gravitational and electric fields. The mathematics of motion of charged particles inside the field (circular paths, cyclotron geometry) is developed in the next lesson; here the focus is on defining B and applying the two force laws. Refer to the official AQA 7408 specification document for the authoritative wording.
Synoptic links: (i) Newton's second law and circular motion (3.4.1, 3.6.1) — the magnetic force on a moving charge supplies the centripetal force in cyclotrons and mass spectrometers, so BQv = mv²/r is a direct fusion of magnetism with mechanics. (ii) Electric fields (3.7.3) — the velocity selector pairs an electric force QE with a magnetic force BQv acting in opposition, and the comparison of how each field exerts force on a charge is a frequent six-mark synoptic prompt. (iii) Particle physics (3.2.1) — the deflection of alpha, beta and proton beams in magnetic fields was historically how charge-to-mass ratio of the electron was measured (Thomson's experiment), so the lesson connects forward to mass spectrometry and backward to the discovery of the fundamental particles.
Key Definition: The magnetic flux density (B) at a point is the force per unit length per unit current on a conductor carrying current perpendicular to the field at that point.
The SI unit of magnetic flux density is the tesla (T), where 1 T = 1 kg s⁻² A⁻¹.
Alternatively, 1 T is the flux density that produces a force of 1 N on a 1 m length of wire carrying a current of 1 A perpendicular to the field.
Typical values of B:
| Source | Flux density (T) |
|---|---|
| Earth's magnetic field | 5 × 10⁻⁵ |
| Bar magnet (at pole) | 0.01–0.1 |
| Laboratory electromagnet | 0.1–2 |
| MRI scanner | 1.5–7 |
| Strong research magnet | 10–45 |
When a straight conductor of length L carrying current I is placed in a uniform magnetic field of flux density B, the force on the conductor is:
F = BIL sin θ
where θ is the angle between the current direction and the magnetic field direction.
When the current is perpendicular to the field (θ = 90°, sin θ = 1):
F = BIL
When the current is parallel to the field (θ = 0°), there is no force (F = 0).
Fleming's left-hand rule gives the direction of the force:
All three are mutually perpendicular.
Question: A straight wire of length 0.25 m carries a current of 4.0 A in a uniform magnetic field of flux density 0.30 T. The wire is perpendicular to the field. Calculate the force on the wire.
Solution:
F = BIL = 0.30 × 4.0 × 0.25 = 0.30 N
Question: The same wire is now oriented at 30° to the magnetic field. Calculate the new force.
Solution:
F = BIL sin θ = 0.30 × 4.0 × 0.25 × sin 30° = 0.30 × 0.50 = 0.15 N
A charged particle moving through a magnetic field experiences a force given by:
F = BQv sin θ
where B is the magnetic flux density (T), Q is the charge (C), v is the speed (m s⁻¹), and θ is the angle between the velocity and the field.
When the velocity is perpendicular to the field (θ = 90°):
F = BQv
Key properties of this force:
When a charged particle enters a uniform magnetic field perpendicular to its velocity, the magnetic force provides centripetal acceleration, causing the particle to move in a circle.
Setting the magnetic force equal to the centripetal force:
BQv = mv²/r
Solving for the radius:
r = mv/(BQ)
This shows:
Question: A proton (mass = 1.67 × 10⁻²⁷ kg, charge = 1.60 × 10⁻¹⁹ C) enters a uniform magnetic field of flux density 0.50 T at right angles with a speed of 2.0 × 10⁶ m s⁻¹. Calculate: (a) the force on the proton; (b) the radius of its circular path; (c) the period of one complete revolution.
Solution:
(a) F = BQv = 0.50 × 1.60 × 10⁻¹⁹ × 2.0 × 10⁶ = 1.60 × 10⁻¹³ N
(b) r = mv/(BQ) = (1.67 × 10⁻²⁷ × 2.0 × 10⁶) / (0.50 × 1.60 × 10⁻¹⁹) r = 3.34 × 10⁻²¹ / 8.0 × 10⁻²⁰ r = 0.042 m = 4.2 cm
(c) Period T = circumference / speed = 2πr / v T = 2π × 0.042 / (2.0 × 10⁶) = 0.264 / (2.0 × 10⁶) = 1.31 × 10⁻⁷ s ≈ 131 ns
Alternatively: T = 2πm/(BQ) = 2π × 1.67 × 10⁻²⁷ / (0.50 × 1.60 × 10⁻¹⁹) = 1.05 × 10⁻²⁶ / 8.0 × 10⁻²⁰ = 1.31 × 10⁻⁷ s ✓
Note: The period T = 2πm/(BQ) is independent of the speed v. This is the principle behind the cyclotron — particles at different speeds all complete a half-circle in the same time, allowing them to be accelerated in sync with an alternating voltage.
A velocity selector uses crossed electric and magnetic fields to select charged particles with a specific speed.
Described diagram — Velocity selector: A beam of charged particles enters from the left and passes between two parallel plates (creating a uniform electric field E pointing upward) and through a region of uniform magnetic field B (pointing into the page). The electric force on a positive charge is upward (F_E = QE) and the magnetic force is downward (F_B = BQv). Only particles for which these forces balance pass through undeflected.
For undeflected particles:
QE = BQv
v = E/B
Particles with v > E/B are deflected by the magnetic force; those with v < E/B are deflected by the electric force. This device is used in mass spectrometers to ensure all particles entering the deflecting region have the same speed.
Question: A velocity selector has an electric field of 3.0 × 10⁴ V m⁻¹ and a magnetic field of 0.20 T. Calculate the speed of particles that pass through undeflected.
Solution:
v = E/B = 3.0 × 10⁴ / 0.20 = 1.5 × 10⁵ m s⁻¹
After selection by a velocity selector, ions of known speed v enter a region of uniform magnetic field B. They follow a semicircular path of radius r = mv/(BQ). By measuring r, the mass m can be determined:
m = BQr/v
This allows isotopes with different masses to be separated and identified.
A cyclotron accelerates charged particles using an alternating electric field between two D-shaped electrodes (dees) placed in a uniform magnetic field. The magnetic field bends the particles in semicircles, and the electric field accelerates them each time they cross the gap. Because T = 2πm/(BQ) is independent of speed, the alternating field frequency remains constant as the particles spiral outward at increasing speed.
When a current-carrying conductor is placed in a magnetic field perpendicular to the current, the magnetic force on the charge carriers deflects them to one side, creating a potential difference (the Hall voltage) across the conductor. At equilibrium:
QE_H = BQv_d → E_H = Bv_d → V_H = Bv_d d
where v_d is the drift velocity and d is the width of the conductor. The Hall voltage is used to measure magnetic flux density.
Exam Tip: When a question asks for the direction of the force on a negative charge, be very careful with Fleming's left-hand rule. Conventional current flows opposite to the motion of negative charges. Point the second finger opposite to the electron's velocity, then the thumb gives the force direction. Alternatively, find the force on a positive charge moving in the same direction and reverse it.
A rectangular coil of N turns carrying current I sits with its plane parallel to a uniform field B. The two long sides of length L (perpendicular to B) each experience a force F = BIL pushing in opposite directions; the two short sides (parallel to B) carry no force. These two equal-and-opposite forces form a couple with arm equal to the width w of the coil, producing a torque:
τ = NBILw = NBIA
where A = Lw is the coil area. This torque rotates the coil, and the result underlies the d.c. motor, the moving-coil ammeter, and the simple electric generator. Notice the structure of the formula: torque is proportional to N (more turns multiply the force), to B (stronger field), to I (more current), and to A (larger coil). At the moment the coil's plane lies perpendicular to B (i.e., the coil normal is parallel to B), the forces on the two sides become collinear and the torque collapses to zero — this is why d.c. motors need a split-ring commutator to reverse current at exactly this instant and maintain rotation.
Question: A rectangular coil of 50 turns measuring 5.0 cm by 3.0 cm carries a current of 0.40 A. The coil sits with its plane parallel to a uniform magnetic field of 0.080 T. Calculate the torque on the coil.
Solution:
A = 0.050 × 0.030 = 1.5 × 10⁻³ m²
τ = NBIA = 50 × 0.080 × 0.40 × 1.5 × 10⁻³ = 2.4 × 10⁻³ N m
Specimen question modelled on the AQA A-Level Physics 7408 Paper 2 format. Not from any published paper.
Q. (a) State what is meant by the magnetic flux density at a point. (2 marks) (b) A horizontal wire of length 0.080 m and mass 1.4 g hangs between the poles of a U-shaped magnet, perpendicular to the magnetic field. When a current of 4.5 A flows through the wire in a direction such that the magnetic force is vertically upward, the wire just lifts. Calculate the magnetic flux density between the magnet poles. Take g = 9.81 N kg⁻¹. (3 marks) (c) A proton enters a region of uniform magnetic field of flux density 0.30 T with a velocity of 1.2 × 10⁶ m s⁻¹ perpendicular to the field. Calculate the magnitude of the magnetic force on the proton, and explain why the proton subsequently moves in a circular path. State the radius of that path. (4 marks) (d) Compare the way a magnetic field exerts a force on a charged particle with the way an electric field exerts a force on the same particle. Refer to dependence on velocity, the direction of the resulting force, and whether work is done on the particle. (9 marks)
(Total: 18 marks)
Grade C response (~200 words):
A magnetic field exerts a force on a charged particle only when it is moving, since the formula is F = BQv sin θ. If the particle is stationary, v = 0 and so the magnetic force is zero. An electric field, on the other hand, exerts a force F = QE on a charged particle even when it is stationary. The magnetic force is also perpendicular to the velocity, so the particle moves in a circle, whereas the electric force is in the direction of the field (for a positive charge), so the particle accelerates in a straight line. Because the magnetic force is perpendicular to the velocity, it does no work on the particle, and the speed is constant. The electric force is in the direction of motion (if the particle moves along the field), so it does work, and the kinetic energy of the particle changes. Both forces depend on the charge Q, and both can be used to deflect a charged particle. Both fields are vector fields and have associated field lines.
Examiner commentary: This response earns the basic comparison marks (M1: B force requires motion / E force does not; M1: B force perpendicular to v / E force parallel to E; M1: B does no work / E does do work; M1: both depend on Q). It picks up roughly 5/9. The candidate has not addressed the consequences for kinetic energy in detail, has not commented on the direction-reversal possible with magnetic deflection, and has not linked the no-work property of the magnetic force to the existence of circular motion at constant speed. Mark-scheme structure is present but the evaluative depth is missing.
Grade B response (~260 words):
The magnetic force F = BQv sin θ depends on the speed v of the particle, the angle between v and B, and the charge Q. It is zero for a stationary particle and zero for motion parallel to B. By contrast, the electric force F = QE depends only on Q and E, and acts regardless of whether the particle is moving or stationary. The directions differ fundamentally: the magnetic force is perpendicular to both v and B, given by Fleming's left-hand rule, while the electric force is along the field direction (or opposite for a negative charge). Because the magnetic force is always perpendicular to the velocity, it has no component along the direction of motion, and the work done is W = ∫F·ds = 0. The speed of the particle is therefore unchanged, and only the direction of the velocity changes — producing circular motion when the particle moves perpendicular to a uniform B field. The electric force, however, can do positive or negative work on the particle, transferring energy between the field and the kinetic energy of the particle. A charged particle accelerated through a potential difference V gains kinetic energy QV. Both fields can be combined in a velocity selector, where the electric and magnetic forces oppose for a particle moving at v = E/B. In atomic physics, electric attraction binds the electron to the nucleus, while magnetic forces are responsible for orbital and spin contributions to the magnetic moment.
Examiner commentary: This response is solid and earns roughly 7/9. The candidate correctly identifies the velocity-dependence, the perpendicular nature of the magnetic force, the no-work consequence, the connection to circular motion at constant speed, and the contrasting energy transfer for the electric force. The velocity-selector application is a good synoptic move. To reach Grade A* the answer needs explicit treatment of how each force modifies the trajectory in kinematic terms — magnetic force changes direction without changing |v|, electric force changes speed along the field direction — and a precise statement of when each does and does not change kinetic energy, ideally with a one-line energy balance.
Grade A response (~360 words):*
A charged particle in a magnetic field experiences F = BQv sin θ, a force that vanishes for a stationary particle and for motion parallel to B. The same particle in an electric field experiences F = QE, which is independent of velocity. This single distinction underlies most of the comparison: the magnetic force is a velocity-dependent interaction, while the electric force is not. Direction is the next key difference. The magnetic force is perpendicular to both v and B, lying in the plane perpendicular to B and rotating with the particle's velocity. The electric force is along E (or anti-parallel for a negative charge), independent of v. Because Fmag is always perpendicular to v, the work done by the magnetic force is W = ∫F·v dt = 0 over any path. The kinetic energy ½mv² is therefore conserved, and the speed |v| is constant; only the direction of v can change. This is the formal reason that a charged particle in a uniform B field perpendicular to its initial velocity executes uniform circular motion with r = mv/(BQ) and angular frequency ω = BQ/m. The electric force, by contrast, does work W = QE·d, with sign depending on whether motion is along or against E. The kinetic-energy change is ΔKE = QV for an electron accelerated through a potential difference V — the basis of the electron gun in a cathode-ray tube. A practical consequence: a particle can be turned in a magnetic field without changing its energy, but it cannot be accelerated by a magnetic field. Cyclotrons therefore use crossed B and E fields — B to bend the path into a circle of fixed period T = 2πm/(BQ), and an alternating E across the gap between the dees to do the work of acceleration. Mass spectrometers use a velocity selector (E ⊥ B with v = E/B) to filter particles by speed, then a pure B-field region to separate them by m/Q. Both fields are indispensable in modern physics; one supplies the geometry, the other supplies the energy.
Examiner commentary: This response would secure full marks. The candidate has gone beyond the basic comparison to integrate (i) the kinematic consequence — circular motion at constant speed — with the no-work property of the magnetic force, (ii) the energy bookkeeping for the electric case ΔKE = QV, and (iii) the cyclotron and mass-spectrometer applications that pair the two fields. The mathematics is precise (r = mv/(BQ), ω = BQ/m, v = E/B), and the language is the scholarly register expected at A*. The synoptic move to particle accelerators links the topic forward to particle physics. Mark-scheme bracket-by-bracket: M1 velocity dependence; M1 direction perpendicular vs parallel; M1 no work / circular motion; M1 energy transfer in E field; M1 r = mv/(BQ); M1 T independent of v; M1 work-energy theorem; M1 velocity selector v = E/B; M1 synoptic synthesis (cyclotron / mass spec).
Many candidates lose marks by using F = BIL when the wire is not perpendicular to the field, instead of F = BIL sin θ. A clue is that the question states the wire and field directions are at an angle other than 90° — always check before substituting.
A second common error is reversing Fleming's left-hand rule when the moving charge is negative. Conventional current flows in the opposite direction to electron motion. The safe technique is to use the rule for the positive-charge equivalent (treat the electron beam as if it were a positive-charge beam moving the opposite way) and confirm the direction matches.
Candidates frequently invent a magnetic potential energy formula by analogy with the electric case. The magnetic force on a moving charge does no work, so the concept of a scalar magnetic potential for charged-particle motion is not part of the A-Level treatment. Do not write expressions like "V_mag = BQr" — these are not standard physics.
In circular-motion problems, a frequent slip is forgetting that r = mv/(BQ) gives the radius only when v is perpendicular to B. If the particle enters the field at an angle, the velocity component parallel to B passes through unaffected (giving helical motion) and only the perpendicular component v sin θ contributes to the circular radius.
Finally, candidates sometimes treat the tesla as the SI unit of magnetic flux. The tesla is the unit of flux density (T = Wb m⁻²); the weber is the unit of flux. The distinction matters in induction problems where Φ = BA is asked for separately from B.
University-level electromagnetism replaces F = BQv with the full Lorentz force F = Q(E + v × B), which packages the electric and magnetic forces into a single vector equation. The cross-product notation makes the perpendicularity of the magnetic force automatic. In special relativity, electric and magnetic fields are recognised as components of a single rank-2 antisymmetric tensor F^μν — what looks like a pure magnetic field in one inertial frame can appear partly electric in another. This is the content of the unification of electricity and magnetism attributed to Maxwell's equations, and it is the reason Einstein's 1905 paper "On the Electrodynamics of Moving Bodies" became the foundation of special relativity.
For Oxbridge / Imperial admissions interviews, classic prompts include: "Why does the magnetic force on a moving charge do no work?", "Can a magnetic field ever change the kinetic energy of a particle?" (the careful answer involves time-varying B and induced E), and "Why is the cyclotron frequency independent of the particle's speed?". Recommended undergraduate reading: Griffiths, Introduction to Electrodynamics, chapters 5 and 6.
Top-grade candidates know that "circular motion in a magnetic field" only holds when v ⊥ B. If the velocity has a component parallel to B, the parallel component is unaffected (no force on it) and the perpendicular component rotates — producing a helix rather than a circle. This is a frequent A* discriminator on extended questions.
A second subtlety: the period T = 2πm/(BQ) is independent of v only for non-relativistic speeds. At v approaching c, the relativistic mass m_rel = γm₀ grows with speed, and a fixed-frequency cyclotron loses synchronisation. The synchrocyclotron and synchrotron were invented precisely to compensate (by varying B or the AC frequency). Mentioning this distinction in the right context is an A* move.
Third, the tesla is a very large unit. The 7 T field of a clinical MRI magnet is one of the strongest sustained fields encountered in everyday life. Don't confuse milli-teslas (laboratory electromagnets) with teslas (research magnets) when interpreting question values.
Exam Tip 1 — Always check the angle. F = BIL sin θ and F = BQv sin θ both include sin θ. If the question gives a non-perpendicular geometry, students who substitute as if sin θ = 1 lose method marks. Draw the field and current/velocity vectors before substituting.
Exam Tip 2 — Use the F = BIL setup for the RP11 balance. When a current-carrying wire sits on a top-pan balance between magnet poles, the change in reading multiplied by g gives the magnetic force. Then B = F/(IL). This recurs in 6-mark practical-skills questions.
Exam Tip 3 — Distinguish flux density from flux. Magnetic flux density B (in teslas) is the local field strength; magnetic flux Φ = BA cos θ (in webers) is the field integrated over an area. Examiners frequently catch candidates using "flux" when they mean "flux density".
Exam Tip 4 — For charged-particle circular motion, always derive r from BQv = mv²/r. Showing the working step — equating magnetic force to centripetal force — is worth a mark in itself. Do not just write down r = mv/(BQ) from memory.
Exam Tip 5 — On 9-mark comparison questions, structure your answer. Use sub-headings or numbered comparison points (velocity dependence, direction, work done, kinematic consequence). Examiners reward visible structure on extended-prose physics questions.
| Concept | Formula | Units |
|---|---|---|
| Force on a current-carrying wire | F = BIL sin θ | N |
| Force on a moving charge | F = BQv sin θ | N |
| Radius of circular path | r = mv/(BQ) | m |
| Period of circular orbit | T = 2πm/(BQ) | s |
| Velocity selector | v = E/B | m s⁻¹ |
| Torque on a current-carrying coil | τ = NBIA | N m |
This content is aligned with the AQA A-Level Physics (7408) specification.