IB DP Physics: Charged particles and conductors in electric and magnetic fields (DP Physics) – Revision Notes

These are condensed notes for D.3, Motion in electromagnetic fields, from the IB Physics guide, first assessment 2025. The content is the same at SL and HL, because the guide gives D.3 no additional HL content. For full explanations, use the study guide. Then test yourself with the practice questions. Course pages: IB Physics hub, checklist and free diagnostics.

Formulas

Situation	Result	Notes
Force on charge in electric field	F = qE	Along E for positive q, against E for negative q
Acceleration in electric field	a = qE/m	Constant in a uniform field
Force on moving charge in magnetic field	F = qvB sin θ	θ is the angle between v and B
Force on conductor in magnetic field	F = BIL sin θ	L is the length inside the field
Circular path (v at 90° to B)	r = mv/(qB)	From qvB = mv²/r
Period of circular path	T = 2πm/(qB)	Independent of speed
Velocity selector (crossed fields)	v = E/B	From qE = qvB
Charge-to-mass ratio	q/m = v/(Br)	Rearranged from the radius
Parallel wires	F/L = μ0 I1 I2 / (2πr)	μ0 = 4π × 10^-7 T m A^-1

Key facts

• Kinetic energy stays constant in a magnetic field. The force is at right angles to the velocity, so it does no work. The speed does not change; the direction does.
• No magnetic force acts on a charge at rest, or on one moving parallel (or antiparallel) to the field.
• The magnetic force is at its largest when velocity and field are at 90°.
• Same-direction currents attract. Opposite-direction currents repel. The two forces form a Newton’s third law pair, so they are equal and opposite.
• Charge and mass cancel in qE = qvB, so a velocity selector passes any particle of speed E/B.

Direction rules

1. Force on a positive charge. Fingers along v, curl towards B, thumb gives F.
2. Negative charge. Do the same, then reverse the result.
3. Force on a wire. Fingers along the conventional current, curl towards B, thumb gives F.
4. Field around a straight wire. Thumb along the conventional current, fingers curl in the direction of the field lines.
5. Electric field. Force on a positive charge is along E. Force on a negative charge is opposite to E.

A worked reminder for rule 1: a proton moves to the right in a field into the page. Fingers right, curl into the page, thumb up. The force is up, so the proton curves upward. An electron curves downward.

Method boxes

Charge crossing an electric field at right angles

1. Find a = qE/m. Work out the direction of the force from the sign of the charge.
2. Time between the plates from the speed along them: t = length/v.
3. Sideways displacement: y = ½at². Sideways speed: v_y = at.
4. Final angle from tan φ = v_y/v, and speed from Pythagoras.

Circular path in a magnetic field

1. Check the velocity is at right angles to B. If not, use only the component that is, or ask what the question intends.
2. Set qvB = mv²/r and solve for the unknown (r, v, B, m or q/m).
3. Direction of circulation from the right-hand rule, remembering the sign of the charge.
4. State that speed and kinetic energy are unchanged.

Charge-to-mass ratio

1. Find the speed: given, or from crossed fields using v = E/B.
2. Measure the radius r in a known B.
3. q/m = v/(Br). Compare with the proton value 9.58 × 10^7 C kg^-1 or the electron value 1.76 × 10^11 C kg^-1.

Parallel wires

1. Use the currents and separation in F/L = μ0 I1 I2 / (2πr).
2. Multiply by the length to get the total force.
3. Decide attraction or repulsion from the current directions.
4. For three or more wires, find each force separately, then combine them as vectors along the line.

Small worked reminders

Electron in a magnetic field. v = 5.0 × 10^6 m s^-1 at right angles to B = 0.20 T. F = qvB = 1.60 × 10^-19 × 5.0 × 10^6 × 0.20 = 1.6 × 10^-13 N. Radius r = mv/(qB) = 9.11 × 10^-31 × 5.0 × 10^6 / (1.60 × 10^-19 × 0.20) = 1.4 × 10^-4 m.

Wire. 0.80 m carrying 3.0 A in a 0.15 T field at 90°: F = 0.15 × 3.0 × 0.80 = 0.36 N.

Wires. 10 A and 5.0 A, same direction, 2.0 cm apart: F/L = 4π × 10^-7 × 10 × 5.0 / (2π × 0.020) = 5.0 × 10^-4 N m^-1, attractive.

Must-know distinctions

Idea A	Idea B
Electric force: acts on any charge, even at rest, does work, changes speed	Magnetic force: acts only on moving charge, perpendicular to v, does no work
Path in uniform E at right angles to v: parabola	Path in uniform B at right angles to v: circle
Conventional current direction (used in wire rules)	Electron drift direction (opposite)
Force on a wire in a field	Field produced by a wire
Same-direction currents: attract	Opposite-direction currents: repel

Worked check: crossed fields then a circle

Particles go straight through crossed fields with E = 4.0 × 10^3 V m^-1 and B = 0.020 T. So v = E/B = 2.0 × 10^5 m s^-1. They then enter a second field of 0.50 T at right angles and move on a circle of radius 2.1 mm. Then q/m = v/(Br) = 2.0 × 10^5 / (0.50 × 2.1 × 10^-3) = 1.9 × 10^8 C kg^-1.

Quick self-test

Constants: e = 1.60 × 10^-19 C, proton mass 1.67 × 10^-27 kg, μ0 = 4π × 10^-7 T m A^-1.

1. Explain why kinetic energy is unchanged for a charge moving in a magnetic field.
2. A proton moves at 2.0 × 10^6 m s^-1 at right angles to a 0.50 T field. Find the force.
3. The same proton now moves at 30° to the field. Find the force.
4. Find the radius of the circular path in question 2.
5. A wire of length 0.50 m carries 6.0 A at 90° to a 0.040 T field. Find the force.
6. Particles go straight through crossed fields with E = 3.0 × 10^3 V m^-1 and B = 0.015 T. Find their speed.
7. Do two parallel wires with currents in opposite directions attract or repel?
8. Find the force per unit length between wires 5.0 cm apart carrying 20 A and 30 A in the same direction.
9. State what happens to the force per unit length in question 8 if the separation is halved.
10. A particle of speed 4.0 × 10^5 m s^-1 moves on a circle of radius 0.020 m in a 0.30 T field. Find q/m.

Answers

1. The force is perpendicular to the velocity, so it does no work; only the direction changes.
2. F = qvB = 1.60 × 10^-19 × 2.0 × 10^6 × 0.50 = 1.6 × 10^-13 N.
3. F = 1.6 × 10^-13 × sin 30° = 8.0 × 10^-14 N.
4. r = mv/(qB) = 1.67 × 10^-27 × 2.0 × 10^6 / (1.60 × 10^-19 × 0.50) = 4.2 × 10^-2 m.
5. F = BIL = 0.040 × 6.0 × 0.50 = 0.12 N.
6. v = E/B = 3.0 × 10^3 / 0.015 = 2.0 × 10^5 m s^-1.
7. They repel.
8. F/L = 4π × 10^-7 × 20 × 30 / (2π × 0.050) = 2.4 × 10^-3 N m^-1, attractive.
9. It doubles, to 4.8 × 10^-3 N m^-1.
10. q/m = v/(Br) = 4.0 × 10^5 / (0.30 × 0.020) = 6.7 × 10^7 C kg^-1.

Where marks are usually lost

• Giving a force direction without saying which rule or which charge sign was used.
• Using the electron’s direction of travel as the current direction.
• Stating that the magnetic force speeds the particle up or does work.
• Leaving out sin θ when the angle is not 90°.
• Taking L as the full wire when only part of it sits in the field.
• Mixing up r in r = mv/(qB) with the wire separation r in the parallel-wire formula.
• Forgetting to convert centimetres or millimetres to metres.
• Writing that parallel wires with same-direction currents repel.
• Dropping the powers of ten in q/m, or giving it without units.

Official syllabus

IB Diploma Programme Physics guide, first assessment 2025, published by the International Baccalaureate Organization. This page covers D.3, Motion in electromagnetic fields.