Edexcel IAL Physics revision · A2 — Fields
Magnetic fields
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.
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What the syllabus demands
- —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 velocity selectors (crossed fields) and the Hall effect
Definitions that earn marks
Clear definitions to practise — check your course mark scheme
- 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.
The equations
More equations to practise: the Edexcel IAL formula sheet.
Where the marks die
Common mistakes to check
- 01
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.
- 02
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.
- 03
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.
- 04
Deriving r = mv/Bq without stating that the magnetic force provides the centripetal force: Bqv = mv²/r. The equating line is the mark.
One worked example, done properly
Question
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)
Method
- 1.The magnetic force provides the centripetal force: Bqv = mv²/r.
- 2.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)