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

This guide teaches D.3, Motion in electromagnetic fields, from Theme D (Fields) of the IB Physics guide, first assessment 2025. D.3 is taught at both standard level (SL) and higher level (HL), and the guide lists no additional HL content for it, so everything here applies to both. It follows on from the field ideas in D.2. For the rest of the course, see the IB Physics hub and the printable checklist.

The other two pages for this unit are the revision notes and the practice questions. Circular motion itself is covered under forces and momentum, so see IB Physics: forces and momentum if centripetal force is rusty. If you are unsure where you stand, take a free diagnostic first.

What D.3 covers

Guide reference	What you must be able to do	Level
D.3	Describe and analyse how a charge moves when a uniform electric field acts on it	SL and HL
D.3	Describe and analyse how it moves when a uniform magnetic field acts on it	SL and HL
D.3	Describe how it moves when uniform electric and magnetic fields are set perpendicular to each other	SL and HL
D.3	Find the size and sense of the magnetic force on a moving charge using F = qvB sin θ	SL and HL
D.3	Find the size and sense of the magnetic force on a wire carrying current using F = BIL sin θ	SL and HL
D.3	Use F/L = μ0 I1 I2 / 2πr for the force per unit length between two parallel wires	SL and HL
D.3 guidance	Recognise that kinetic energy is constant in a magnetic field	SL and HL
D.3 guidance	Determine a charge-to-mass ratio from the path in a uniform magnetic field	SL and HL
D.3 guidance	Find the field direction from the current direction around a straight wire	SL and HL
D.3 guidance	Know that parallel currents in the same direction attract	SL and HL

The symbols: q is charge, v is speed, B is magnetic field strength (tesla, T), I is current, L is how much of the conductor lies in the field, and θ is the angle between the velocity (or current) and the field.

Charge in a uniform electric field

A charge q in an electric field of strength E feels a force F = qE. The force is along the field for a positive charge and against it for a negative charge. In a uniform field this force is constant, so the acceleration a = qE/m is constant, and you can use the equations of uniform acceleration from kinematics.

If the charge starts at rest, or moves parallel to the field, the motion is a straight line with constant acceleration. If it enters at right angles to the field, the motion splits into two independent parts:

• along the plates the velocity component is constant, because no force acts that way
• across the plates the particle accelerates uniformly from zero

This is the same maths as a projectile, with the electric force replacing weight. The path is a parabola.

Worked example 1. An electron enters the gap between two horizontal plates at 4.0 × 10^6 m/s, parallel to the plates. The plates are 6.0 cm long and the field between them is 5.0 × 10^2 N C^-1 (taken as uniform, upward). Find the vertical deflection on leaving and the direction of travel on leaving. Use e = 1.60 × 10^-19 C and electron mass 9.11 × 10^-31 kg.

1. The field points up, the electron is negative, so the force and acceleration point down.
2. a = eE/m = (1.60 × 10^-19 × 5.0 × 10^2) / (9.11 × 10^-31) = 8.78 × 10^13 m s^-2.
3. Time between the plates: t = 0.060 / (4.0 × 10^6) = 1.5 × 10^-8 s.
4. Deflection: y = ½at² = ½ × 8.78 × 10^13 × (1.5 × 10^-8)² = 9.9 × 10^-3 m, about 9.9 mm downward.
5. Vertical speed on leaving: v_y = at = 1.32 × 10^6 m s^-1.
6. Angle below the original direction: tan φ = 1.32 × 10^6 / 4.0 × 10^6, so φ = 18°.

Gravity is ignored here. The weight of an electron is tiny compared with eE, and you should say so if you are asked to justify it.

Charge in a uniform magnetic field

Magnitude and direction of the force

For a charge q moving at speed v at angle θ to a uniform magnetic field B:

F = qvB sin θ

• If θ = 90°, sin θ = 1 and the force is at its maximum, qvB.
• If the charge moves parallel (or antiparallel) to the field, θ = 0° or 180°, sin θ = 0 and there is no force.
• A charge at rest feels no magnetic force.

The force is perpendicular to both the velocity and the field. A right-hand rule gives the direction for a positive charge: point the fingers along the velocity, curl them towards the field direction, and the thumb gives the force. For a negative charge, reverse the result. Equivalently, treat the moving negative charge as a current in the opposite direction (conventional current flows the way positive charge would).

For example, a proton moving to the right in a field directed into the page: fingers right, curl into the page, thumb up. The force is up the page. An electron on the same path is pushed down the page.

Circular motion and constant kinetic energy

The magnetic force is always at right angles to the velocity. A force perpendicular to the motion does no work, so the kinetic energy and speed stay constant. Only the direction of the velocity changes. When the velocity is at right angles to a uniform field, the force provides the centripetal force:

qvB = mv²/r, so r = mv/(qB)

The particle moves on a circle in a plane perpendicular to B. From r = mv/(qB):

• a faster particle moves on a larger circle
• a stronger field gives a smaller circle
• a heavier particle gives a larger circle
• the period T = 2πr/v = 2πm/(qB) does not depend on the speed

Worked example 2. A proton (mass 1.67 × 10^-27 kg) moves at 3.0 × 10^5 m s^-1, perpendicular to a uniform 0.30 T field. Find the force, the radius and the period.

1. F = qvB = 1.60 × 10^-19 × 3.0 × 10^5 × 0.30 = 1.44 × 10^-14 N.
2. r = mv/(qB) = (1.67 × 10^-27 × 3.0 × 10^5) / (1.60 × 10^-19 × 0.30) = 1.04 × 10^-2 m.
3. T = 2πm/(qB) = 2π × 1.67 × 10^-27 / (1.60 × 10^-19 × 0.30) = 2.19 × 10^-7 s.

If the same proton instead entered at 30° to the field, the force would be 1.44 × 10^-14 × sin 30° = 7.2 × 10^-15 N.

Crossed electric and magnetic fields

Now set the electric and magnetic fields at right angles to each other, with the particle moving at right angles to both. Arrange them so the electric force and the magnetic force on the particle point in opposite directions. The particle passes straight through when the two forces are equal:

qE = qvB, so v = E/B

The charge and mass cancel. Only particles with speed E/B go straight; this is why the arrangement is used as a velocity selector. A faster particle has the larger magnetic force and bends in the direction of that force. A slower one has the larger electric force and bends the other way.

Charge-to-mass ratio from a circular path

The guide requires you to find q/m for a particle from the curved path it follows in a uniform B-field. Rearrange r = mv/(qB):

q/m = v/(Br)

You need the speed v, the field B and the measured radius r. The speed may be given, or found with crossed fields (v = E/B), or from the accelerating arrangement used to launch the particle. Comparing q/m with known values helps identify the particle. Different particles with the same speed and field move on circles of different radii, so they can be separated.

Worked example 3. Particles pass undeflected through crossed fields with E = 6.0 × 10^3 V m^-1 and B = 0.040 T. They then enter a second, uniform field of 0.25 T at right angles and follow a circle of radius 6.3 mm. Find q/m.

1. v = E/B = 6.0 × 10^3 / 0.040 = 1.5 × 10^5 m s^-1.
2. q/m = v/(Br) = 1.5 × 10^5 / (0.25 × 6.3 × 10^-3) = 9.5 × 10^7 C kg^-1.

This is close to the value for a proton, 1.60 × 10^-19 / 1.67 × 10^-27 = 9.58 × 10^7 C kg^-1, so the particles are probably protons. Ratio matching does not prove it, because other particles can share a ratio.

Force on a current-carrying conductor

A current is moving charge, so a wire in a magnetic field feels a force. For a straight conductor of length L in the field, carrying current I, at angle θ to the field:

F = BIL sin θ

The direction uses the same right-hand rule, with the fingers along the conventional current. The force is zero when the wire is parallel to the field.

Worked example 4. A straight wire of length 0.35 m carries 4.5 A in a region where B = 0.12 T. The wire makes 30° with the field. Find the force.

F = BIL sin θ = 0.12 × 4.5 × 0.35 × sin 30° = 9.5 × 10^-2 N (0.0945 N before rounding).

Only the part of the wire inside the field counts.

Parallel current-carrying wires

A current in a straight wire produces circular field lines around it. To find the direction, point the right thumb along the conventional current: the fingers curl the way the field circulates. Each wire sits in the field of the other, so each feels a force. For two long parallel wires a distance r apart, the force per unit length on each is

F/L = μ0 I1 I2 / (2πr)

where μ0 is the permeability of free space, 4π × 10^-7 T m A^-1. The two forces are equal in magnitude and opposite in direction (Newton’s third law). Currents in the same direction attract. Currents in opposite directions repel.

Worked example 5. Two long parallel wires 4.0 cm apart carry 12 A and 8.0 A in the same direction. Find the force per unit length, and the force on a 1.5 m length.

1. F/L = (4π × 10^-7 × 12 × 8.0) / (2π × 0.040) = 4.8 × 10^-4 N m^-1, attractive.
2. F = 4.8 × 10^-4 × 1.5 = 7.2 × 10^-4 N.

Doubling the separation halves the force per unit length. Doubling one current doubles it.

Common errors

• Applying sin θ with θ measured to the wrong line. θ is the angle between velocity (or current) and the field, not the angle to the field’s normal.
• Treating the magnetic force as doing work on the particle, or saying the speed increases.
• Forgetting that the right-hand rule gives the force on a positive charge. Reverse it for electrons.
• Using electron motion direction as the current direction for a wire.
• Mixing up the two field-line rules: one gives the force on a charge or wire, the other gives the field around a wire.
• Writing that parallel wires with currents in the same direction repel.

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.