Key Idea: This topic is about the forces that electric and magnetic fields put on charges — and the motion those forces cause. Three set-ups recur: a current-carrying wire pushed by a magnetic field (the motor effect), a charge in an electric field accelerated like a projectile, and a moving charge in a magnetic field bent into a circle, used in the velocity selector. It is examined on both papers. Paper 1A is quick multiple-choice — the direction of a force from a left-hand rule, when the force is zero, what path a particle follows, whether the selected speed depends on the charge. Paper 2 is longer structured work — find a field strength from F = BIL, a two-step F = qE then a = F/m calculation, a 'show that' on a huge acceleration, a projectile-style deflection with s = ½at², or a crossed-fields balance qE = qvB.
📋 Key formulas
Most of these carry the data-booklet badge (look for it). The selector speed v = E/B and the circular-path radius r = mv/(qB) are not printed separately — they come straight from balancing or equating the given forces, so you remember those.
- force on the wire (N)
- magnetic field strength / flux density (T, tesla)
- current in the wire (A)
- length of wire in the field (m)
- angle between the current and the magnetic field
- electric force on the charge (N)
- the charge in the field (C, coulombs)
- electric field strength (N C⁻¹, or V m⁻¹)
- net (electric) force on the charge (N)
- mass of the charged particle (kg)
- acceleration of the particle (m s⁻²)
- electric field strength between the plates (V m⁻¹, or N C⁻¹)
- potential difference (voltage) between the plates (V)
- separation (gap) between the plates (m)
- magnetic force on the moving charge (N)
- size of the moving charge (C)
- speed of the charge (m s⁻¹)
- magnetic field strength (T, tesla)
- radius of the circular path (m)
- mass of the charged particle (kg)
- speed of the particle (m s⁻¹)
- size of the charge (C)
- magnetic field strength (T)
- selected speed — the speed that passes straight through (m s⁻¹)
- electric field strength between the plates (N C⁻¹ or V m⁻¹)
- magnetic field strength (T)
⚖️ The three force set-ups side by side
🧲 Electric force vs magnetic force on a charge
Force on a wire → Fleming's left-hand rule: First finger Field, seCond finger Current, thuMb Force (Motion). Reverse the current OR the field and the force flips. Force on a charge → for a positive charge it follows the field directions above; for a negative charge (an electron) every force is the opposite way.
✏️ Worked exam-style questions
IB-style question — force on a wire, then the field strength
A straight wire of length 0.20 m lies at right angles to a uniform magnetic field and carries a current of 5.0 A. (a) The field strength is 0.30 T — find the force on the wire. (b) In a second experiment the same wire (same length, same 5.0 A current) feels a force of 0.45 N. Find the new field strength B.
Solution:
(a) Start with the given formula; the wire is perpendicular, so θ = 90° and sinθ = 1:
(a) Work it out — keep the unit:
(b) Rearrange F = BIL to make B the subject:
(b) Evaluate:
(a) F = 0.30 N. (b) B = 0.45 T. Show the step where sinθ = 1 so a marker sees you handled the angle; L is the length IN the field, not the whole wire.
IB-style question — accelerating an electron in a field
An electron (charge 1.6 × 10⁻¹⁹ C, mass 9.1 × 10⁻³¹ kg) sits in the uniform field between two parallel plates 0.025 m apart with 250 V across them. (a) Find the field strength E. (b) Find the electric force on the electron. (c) Show that its acceleration is of order 10¹⁵ m s⁻², and state its direction relative to the field.
Solution:
(a) Uniform field between plates — use the given E = V/d:
(b) The force on the charge is F = qE:
(c) Newton's second law gives the acceleration:
(c) Evaluate:
(a) E = 1.0 × 10⁴ N C⁻¹. (b) F = 1.6 × 10⁻¹⁵ N. (c) a ≈ 1.8 × 10¹⁵ m s⁻², of order 10¹⁵ as required; the acceleration is OPPOSITE to the field, because the electron is negative. Always two steps: F = qE, then a = F ÷ m — don't read the field strength as the acceleration.
IB-style question — a charge fired across the plates (projectile)
An electron enters the gap between two parallel plates moving parallel to them at 5.0 × 10⁷ m s⁻¹. The plates are 0.060 m long and the field gives the electron a sideways acceleration of 3.2 × 10¹⁴ m s⁻². (a) Find the time the electron spends between the plates. (b) Find how far it is deflected sideways as it crosses, and explain why s = ½at² is used rather than s = vt.
Solution:
(a) Along the plates there is no force, so the speed is constant — time = distance ÷ speed:
(b) Across the plates the electron starts from rest sideways (u = 0), so the suvat equation is:
(b) Substitute a = 3.2 × 10¹⁴ and t = 1.2 × 10⁻⁹:
(b) Evaluate — keep the unit:
(a) t = 1.2 × 10⁻⁹ s. (b) sideways deflection ≈ 2.3 × 10⁻⁴ m (about 0.23 mm). Use s = ½at², NOT s = vt, because the sideways motion is accelerated (a constant force qE acts across the field) — there is no single constant sideways speed. The path is a parabola, exactly like a projectile.
IB-style question — velocity selector, then the circular path
An ion of charge 1.6 × 10⁻¹⁹ C and mass 2.5 × 10⁻²⁶ kg passes undeflected through a velocity selector whose crossed fields are E = 3.6 × 10⁴ N C⁻¹ and B₁ = 0.18 T. After the selector the ion enters a region of magnetic field B₂ = 0.40 T alone, at right angles to its motion. (a) Find the selected speed. (b) Find the radius of the circular path the ion then follows.
Solution:
(a) Undeflected ⇒ the forces balance (qE = qvB₁); the charge cancels:
(a) Evaluate:
(b) In the second region the magnetic force is the centripetal force, giving the radius:
(b) Evaluate top and bottom, then divide:
(a) v = 2.0 × 10⁵ m s⁻¹ — the same for any charge in those fields, since q cancels. (b) r ≈ 7.8 × 10⁻² m (about 7.8 cm). Two-step crossed-fields problem: get v = E/B from the selector first, then feed that v into r = mv/(qB). Keep the powers of ten lined up.
🧠 Quick self-check
Tap each card to reveal the answer.
🎯 Exam tips
Exam Tips
- Force on a wire: use F = BIL sinθ; set sinθ = 1 when the wire is at right angles to the field, and remember the force is ZERO when the current runs along the field. F is proportional to both B and I, so doubling either doubles the force.
- Direction of the force on a wire = Fleming's LEFT-hand rule: First finger Field, seCond finger Current, thuMb Force. Reverse the current or the field and the force flips.
- Charge in an electric field is ALWAYS two steps: F = qE first, then a = F ÷ m. Don't read the field strength E as the acceleration. A positive charge accelerates along the field; an electron accelerates opposite to it.
- Use E = V/d to turn a plate voltage into a field, then F = qE for the force. Watch the units — convert a plate gap in cm or mm to metres before dividing.
- Fired across the field = a projectile: constant velocity along the plates (gives the time), s = ½at² across them (gives the deflection). Never use s = vt for the accelerated sideways direction.
- A magnetic force F = qvB acts only on a MOVING charge and is perpendicular to v, so it does no work — it bends the path into a circle of radius r = mv/(qB) but never changes the speed.
- Velocity selector: balance qE = qvB → v = E ÷ B. The charge cancels, so the selected speed is the same for any particle. Too slow → the electric force wins; too fast → the magnetic force wins.