The key idea
Two force formulas — one for currents, one for moving charges — and one glorious consequence: a charge moving perpendicular to a magnetic field goes in a circle. Half of A2 particle physics apparatus is built on that circle.
Understand it · 01
Include the angle and the charge sign
A charge moving through a magnetic field experiences force magnitude |q|vB sin θ, where θ is the angle between velocity and field. The force is zero for motion parallel to the field and greatest for perpendicular motion. Its direction is perpendicular to both vectors and reverses if the sign of charge reverses. For a wire carrying conventional current, the corresponding force is BIL sin θ for the portion of wire within a uniform field.
Field diagrams use crosses for vectors into the page and dots for vectors out of the page. Interpret these before applying a hand rule. A force direction obtained for conventional current or a positive charge must be reversed for electrons moving in the stated direction. Magnetic field strength B is measured in teslas. The current-carrying wire feels a force because the moving charge carriers interact with the field and transfer that effect to the conductor.
Understand it · 02
Use magnetic force as the inward resultant
For a particle moving perpendicular to a uniform magnetic field, the force stays perpendicular to the velocity. It changes the velocity direction without transferring energy to the particle, so speed stays constant when other forces are negligible. Equating |q|vB with mv²/r gives r = mv/(|q|B), or r = p/(|q|B) using momentum. A stronger field produces a smaller radius for the same particle momentum.
If the particle also has a velocity component parallel to the field, that component is not changed by the magnetic force. The combination can produce a helical path. The simple circular-radius expression uses the perpendicular component of momentum in that case. Do not conclude that a magnetic field accelerates a particle to a greater speed merely because it bends the path; acceleration can describe a change in direction alone.
Understand it · 03
Combine electric and magnetic forces carefully
A velocity selector uses electric and magnetic forces in opposite directions. For a particle that passes undeflected, |q|E = |q|vB and therefore v = E/B. This selects speed rather than mass when the geometry is arranged correctly. Particles with different speeds are deflected because the magnetic force depends on speed while the electric force does not. The directions must be checked for the actual signs of the charges and fields.
In a force-on-a-wire experiment, vary current while keeping the wire length within the field and its orientation fixed. A straight-line force–current relationship supports the model, with gradient BL for a perpendicular wire. Include only the effective length in the approximately uniform field and subtract any zero reading from the balance. Reversing current reverses the magnetic force and can help distinguish it from a constant offset such as the weight of the apparatus.
Key definitions
- Magnetic flux density (B)
- The force per unit current per unit length on a conductor placed at right angles to the field: B = F ÷ IL. Units: tesla (T).
- The tesla
- One tesla is the flux density producing a force of 1 N on each metre of a conductor carrying 1 A at right angles to the field.
- Hall voltage
- The p.d. that develops across a conductor in a magnetic field, when the magnetic force on the moving charge carriers pushes them to one side until the electric and magnetic forces balance: V_H = BI ÷ ntq.
Equations, units & conditions
Force on a conductor
F = B I L sin θ
Units: N
Force on a moving charge
F = B |q| v sin θ
Units: N
Magnitude; reverse the positive-charge force direction for a negative charge.
Radius of circular path
r = p ÷ (B |q|)
Units: m
Perpendicular motion; p = mv in the non-relativistic model.
Velocity selector
v = E ÷ B
Worked examples, step by step
Example 1
An electron moving at 4.0 × 10⁶ m/s enters a field of 2.0 mT at right angles. Find the radius of its circular path. (m = 9.11 × 10⁻³¹ kg, q = 1.6 × 10⁻¹⁹ C)
Follow the method
- The magnetic force provides the centripetal force: Bqv = mv²/r.
- r = mv ÷ Bq = (9.11 × 10⁻³¹ × 4.0 × 10⁶) ÷ (2.0 × 10⁻³ × 1.6 × 10⁻¹⁹).
r ≈ 1.1 × 10⁻² m (about 1 cm)
Practice questions with worked solutions
Try each question on paper before opening the solution. These are original academy questions; the suggested marks are a self-checking guide, not an official exam-board mark scheme. Difficulty labels describe this practice set, not predicted grades.
Build · 3 suggested marks
1. Force on a wire
A 0.080 m wire carries 3.0 A perpendicular to a uniform 0.25 T field. Find the magnetic force magnitude.
Show solution and self-check: question 1
- For perpendicular orientation, sin θ = 1.
- F = BIL = 0.25 × 3.0 × 0.080.
0.060 N.
Self-check · one mark per point
- Use the length actually inside the field.
- Apply F = BIL.
- Give 0.060 N.
Build · 3 suggested marks
2. Velocity selector
A selector has electric field strength 3.6 × 10⁴ V m⁻¹ and magnetic field strength 0.12 T, arranged so the forces oppose. Find the selected speed.
Show solution and self-check: question 2
- For no deflection, |q|E = |q|vB.
- v = E/B = 3.6 × 10⁴/0.12.
3.0 × 10⁵ m s⁻¹.
Self-check · one mark per point
- Equate opposing electric and magnetic force magnitudes.
- Cancel charge and calculate E/B.
- Obtain 3.0 × 10⁵ m s⁻¹.
Common mistakes & how to avoid them
- Forgetting the magnetic force does no work. It is always perpendicular to the velocity, so it changes direction, never speed — which is exactly why the path is circular at constant speed.
- Applying the left-hand rule to an electron with the current pointing the way it moves. Conventional current is opposite to electron motion — reverse the second finger or reverse the answer.
- sin θ dropped when the motion is not perpendicular to B. Parallel to the field, the force is zero; a charge launched at an angle spirals (helix), a favourite extension question.
- Deriving r = mv/Bq without stating that the magnetic force provides the centripetal force: Bqv = mv²/r. The equating line is the mark.
Understand the why
Questions, explained
Start with the short answer, then follow the reasoning. Each explanation has a permanent link you can share with a classmate or return to when revising.
Why does a charge move in a circle in a magnetic field?
A charged particle entering perpendicular to a uniform magnetic field experiences a force perpendicular to its velocity. With no other significant force, this inward force changes direction without changing speed, producing a circular path.
Equating |q|vB to mv²/r gives r = mv/(|q|B), or r = p/(|q|B). The absolute charge ensures a positive radius; the sign of charge determines which way the trajectory curves. If there is also a velocity component parallel to the field, that component continues unchanged and the resulting ideal path is a helix. A uniform field is essential to the simple circular description. Energy loss through collisions or another force can alter momentum and therefore the radius, so a spiralling detector track may carry information about more than the magnetic force alone.
How does a velocity selector work?
A velocity selector arranges electric and magnetic forces to oppose. Particles pass undeflected when |q|E = |q|vB, giving v = E/B. It selects speed under the stated geometry, independently of particle mass.
The electric force has no speed dependence in this model, while the magnetic force is proportional to speed for perpendicular motion. A faster particle therefore has a different resultant from a slower one. Check both force directions for the actual charge sign; reversing the charge reverses both forces, so the balance condition remains the same if the path geometry is unchanged. The selector does not make every incoming particle acquire the selected speed. It rejects or deflects those with unsuitable speeds, which is why apertures are used to define the transmitted beam.
Selected speed
A selector has E = 2.4 × 10⁴ V m⁻¹ and B = 0.080 T with opposing forces. Find the transmitted speed.
- For no deflection, |q|E = |q|vB.
- v = E/B = 2.4 × 10⁴/0.080.
3.0 × 10⁵ m s⁻¹.
What does the Hall effect measure?
The Hall effect produces a transverse voltage when a magnetic field deflects charge carriers sideways in a current-carrying material. A calibrated Hall probe uses this voltage to measure magnetic flux density; this is an extension beyond the listed core IAL equations.
Charge accumulates at opposite sides until the resulting electric force balances the sideways magnetic force on the carriers. The voltage depends on field strength and also on current, specimen dimensions and carrier properties, so those conditions must be controlled or calibrated. The polarity can provide information about the sign of the dominant charge carriers. A Hall probe is not simply a voltmeter placed anywhere near a magnet; it relies on a current through a suitable element and an appropriate orientation. Turning the probe changes the component of field that contributes to its response.
How can a magnetic field accelerate a particle but do no work?
The magnetic force on a moving point charge is perpendicular to its instantaneous velocity, so its mechanical power F·v is zero. It changes momentum direction but cannot by itself increase or decrease that particle’s kinetic energy.
This statement concerns the magnetic part of the force, not every device containing magnets. An electric motor can transfer energy because an electrical source supplies energy and forces act on moving conductors within the complete circuit. An accelerating particle can gain energy from an electric field even while a magnetic field bends its path. In a pure uniform magnetic field with no losses, constant speed follows from the perpendicular force. If speed changes in an observed track, consider collisions, electric fields or radiation losses rather than assigning tangential work to the magnetic force.
What determines the radius of a charged particle’s path in a magnetic field?
For perpendicular motion in a uniform magnetic field, radius is r = p/(|q|B). Larger momentum produces a wider curve, while larger field strength or charge magnitude produces a tighter curve, with other quantities fixed.
The momentum form remains more general than substituting mv, which is the non-relativistic expression. Compare particles using both momentum and charge: equal speeds alone do not imply equal radii if their masses differ, and equal momentum does not imply equal radii if charge magnitudes differ. Charge sign reverses the bending direction rather than making radius negative. In track interpretation, also check that both particles experience the same field component and that the visible projection corresponds to the perpendicular circular motion. A helical trajectory introduces a parallel component not represented by the radius alone.
Momentum from curvature
A singly charged particle curves with radius 0.050 m in a perpendicular 0.20 T field. Find momentum using |q| = 1.60 × 10⁻¹⁹ C.
- p = r|q|B.
- p = 0.050 × 1.60 × 10⁻¹⁹ × 0.20.
1.60 × 10⁻²¹ kg m s⁻¹.
How do I find the magnetic force direction on an electron?
Find the magnetic-force direction for a positive charge or conventional current, then reverse it for an electron travelling in the stated direction. Conventional current is opposite to electron motion, so using the electron’s velocity as current gives the wrong sign.
Read dots and crosses before using a hand rule: dots represent a field out of the page and crosses a field into it. If velocity is parallel to the field, no hand-rule direction is needed because the magnetic force is zero. For oblique entry, only the perpendicular velocity component contributes to the force magnitude. Reversing the field or reversing the velocity reverses the force for the same charge. Reversing both leaves the force direction unchanged. A small labelled vector sketch is usually more reliable than rotating the whole diagram mentally.
Your learning checklist
Use these goals to check your understanding. Requirements vary by specification, tier and exam year; the official references below define your full course.
- Define magnetic flux density B from F = BIL; the tesla
- Use F = BIL sin θ for a current-carrying conductor
- Use F = Bqv sin θ for a moving charge
- Show that a charge moving perpendicular to B follows a circular path: r = mv/Bq
- Describe crossed-field applications; the Hall-effect explanation below is an extension
Sources & how to cite these notes
Original explanations, illustrations and practice by Dr Desouky Physics Academy. Official specifications guide the course scope; questions and suggested marks on this page are our own revision material.
Reference this page
Dr Desouky Physics Academy. Edexcel IAL Physics: Magnetic fields. Updated 2026-09-25. https://www.drdesouky.com/revision/a-level-physics/magnetic-fields
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