Motion in Electromagnetic Fields: notes and practice questions
- This topic describes the motion of charged particles and current-carrying conductors in uniform electric and magnetic fields.
- The magnetic force on a charge moving with velocity in a magnetic field is given by .
- The magnetic force on a current-carrying conductor of length with current in a magnetic field is .
- The force per unit length between two parallel wires carrying currents and separated by distance is .
- The kinetic energy of a charged particle remains constant in a uniform magnetic field.
- The direction of these forces can be determined.
How it is examined
Both papers, both levels, and a reliable multi-part Paper 2 question. May 2025 SL Paper 2 TZ1 question 5 and HL question 6 were the same electron-between-plates problem: state the direction of acceleration (1), show the acceleration is about 10¹⁴ m s⁻² (2), determine the horizontal distance travelled before the electron hits a plate (3), then calculate a magnetic field strength that balances the deflection (2). The HL version added a preceding calculate of the accelerating potential difference. The mark scheme allowed error carried forward across all four parts, and told examiners to ignore negative signs in the force answer.
, , the parallel-wire force per unit length, and the electron charge and mass in the constants table. The circular-path radius is not printed as such: it comes from equating with the centripetal force from A.2. Left-hand and right-hand rules are not in the booklet at all and are pure recall.
- the motion of a charged particle in a uniform electric field
- the motion of a charged particle in a uniform magnetic field
- the motion of a charged particle in perpendicularly orientated uniform electric and magnetic fields
- the magnitude and direction of the force on a charge moving in a magnetic field, as given by , where B is the magnetic field strength
Guiding questions
- How do charged particles move in magnetic fields?
- What can be deduced about the nature of a charged particle from observations of it moving in electric and magnetic fields?
Linking questions
- What causes circular motion of charged particles in a field?
- How can the orbital radius of a charged particle moving in a field be used to determine the nature of the particle?
- How can conservation of energy be applied to motion in electromagnetic fields?
- How are the concepts of energy, forces and fields used to determine the size of an atom?
- How are the properties of electric and magnetic fields represented? (NOS)
Practice questions
5 questions · 2 easy · 2 medium · 1 hardQuestion 1
EasyPaper 1A · calculator1 markThree statements about alpha particles are:
I. They have a greater mass than beta-minus particles.
II. They have a shorter range in air than beta-minus particles.
III. They are deflected in the opposite direction to beta-minus particles in a uniform magnetic field.
Which statements are correct?
A. I and II only
B. I and III only
C. II and III only
D. I, II and III
Recall the fundamental properties of alpha and beta particles. Consider their composition (protons, neutrons, electrons), their charge, and how these properties affect their interaction with matter and with magnetic fields.
Question 2
MediumPaper 2 · calculator4 marksAn alpha particle () is emitted from a radioactive source with a kinetic energy of J. It then enters a uniform magnetic field of strength T, moving perpendicular to the field lines in a vacuum.
(a) Calculate the radius of the circular path followed by the alpha particle.
First, use the kinetic energy to determine the speed of the alpha particle. Remember the mass and charge of an alpha particle. Then, apply the principle that the magnetic force provides the centripetal force for circular motion.
Question 3
HardPaper 2 · calculator8 marksTwo parallel overhead transmission cables are separated by a distance of mm. They carry currents of A and A respectively, flowing in the same direction.
(a) Explain whether or not one cable experiences a greater magnetic force than the other.
(b) Determine the magnitude of the magnetic force exerted on a m section of the cable carrying A.
(c) Determine the location between the two cables where the resultant magnetic field is zero.
Consider Newton's third law and the formula for the force between two parallel current-carrying wires. How do the currents of both wires contribute to the force on each wire?
Recall the formula for the force between two parallel current-carrying wires. Remember to use the permeability of free space constant, T m A.
For the resultant magnetic field to be zero, the magnetic fields produced by each wire must be equal in magnitude and opposite in direction. Since the currents are in the same direction, where would this condition be met?
Question 4
EasyPaper 1A · calculator1 markA velocity selector is a device used in mass spectrometry to ensure that ions entering the main spectrometer all have the same velocity. It consists of a region with a uniform electric field and a uniform magnetic field, arranged to be perpendicular to each other and to the direction of motion of the ions.
A beam of singly ionized potassium-39 ions () is found to pass through the selector undeflected at a speed of .
The ion source is then changed to produce doubly ionized calcium-40 ions (). The electric and magnetic fields in the selector are unchanged.
At what speed must the calcium-40 ions travel to pass through the velocity selector without deflection?
A.
B.
C.
D.
For an ion to pass undeflected, the net force on it must be zero. Consider the electric and magnetic forces acting on the ion. How does the condition for zero net force depend on the ion's charge and mass?
Question 5
MediumPaper 1A · calculator1 markTwo long parallel wires, P and Q, carry steady currents. The magnitude of the magnetic force per unit length acting on wire P is .
The current in wire P is doubled and the distance between the wires is halved. What is the magnitude of the new magnetic force per unit length acting on wire P and on wire Q?
| Force on P | Force on Q | |
|---|---|---|
| A. | ||
| B. | ||
| C. | ||
| D. |
Consider how the formula for the magnetic force per unit length depends on the currents and the separation. Also, recall Newton's third law regarding the mutual forces between two interacting objects.
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Where marks are lost
- Stopping one step short of the conclusion. Two numbers and no sentence is two marks out of three.
- Answering a procedure question with a platitude.
- Losing precision in Paper 1B. Uniquely to this paper, quoting the right number badly loses marks.