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Topic D.3 · SL and HL

Motion in Electromagnetic Fields: notes and practice questions

Summary
  • This topic describes the motion of charged particles and current-carrying conductors in uniform electric and magnetic fields.
  • The magnetic force FF on a charge qq moving with velocity vv in a magnetic field BB is given by F=qvBsin⁡θF = qvB \sin \theta.
  • The magnetic force FF on a current-carrying conductor of length LL with current II in a magnetic field BB is F=BILsin⁡θF = BIL \sin \theta.
  • The force per unit length between two parallel wires carrying currents I1I_1 and I2I_2 separated by distance rr is FL=μ0I1I22πr\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi r}.
  • 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.

Given in the booklet

F=qvBsin⁡θF = qvB\sin\theta, F=BILsin⁡θF = BIL\sin\theta, the parallel-wire force per unit length, μ0\mu_0 and the electron charge and mass in the constants table. The circular-path radius r=mv/qBr = mv/qB is not printed as such: it comes from equating qvBqvB with the centripetal force from A.2. Left-hand and right-hand rules are not in the booklet at all and are pure recall.

Key ideas
  • 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 F=qvBsin⁡θF = qvB\sin\theta, 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 hard
Showing 5 of 5

Question 1

EasyPaper 1A · calculator1 mark

Three 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

Question 2

MediumPaper 2 · calculator4 marks

An alpha particle (24He2+^{4}_{2}\text{He}^{2+}) is emitted from a radioactive source with a kinetic energy of 6.0×10−136.0 \times 10^{-13} J. It then enters a uniform magnetic field of strength 0.500.50 T, moving perpendicular to the field lines in a vacuum.

(a) Calculate the radius of the circular path followed by the alpha particle.

Question 3

HardPaper 2 · calculator8 marks
(a)

Two parallel overhead transmission cables are separated by a distance of 120120 mm. They carry currents of 3.03.0 A and 5.05.0 A respectively, flowing in the same direction.

(a) Explain whether or not one cable experiences a greater magnetic force than the other.

[2]
(b)

(b) Determine the magnitude of the magnetic force exerted on a 0.500.50 m section of the cable carrying 3.03.0 A.

[3]
(c)

(c) Determine the location between the two cables where the resultant magnetic field is zero.

[3]

Question 4

EasyPaper 1A · calculator1 mark

A 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 (39K+^{39}K^+) is found to pass through the selector undeflected at a speed of v0v_0.

The ion source is then changed to produce doubly ionized calcium-40 ions (40Ca2+^{40}Ca^{2+}). 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. v0v_0

B. 2v02v_0

C. v02\frac{v_0}{2}

D. 4039v0\sqrt{\frac{40}{39}} v_0

Question 5

MediumPaper 1A · calculator1 mark

Two long parallel wires, P and Q, carry steady currents. The magnitude of the magnetic force per unit length acting on wire P is FF.

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 PForce on Q
A.2F2F2F2F
B.4F4FFF
C.4F4F2F2F
D.4F4F4F4F

Every Motion in Electromagnetic Fields question, marked for you

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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.
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What does Motion in Electromagnetic Fields cover in IB Physics?

This topic describes the motion of charged particles and current-carrying conductors in uniform electric and magnetic fields. The magnetic force F on a charge q moving with velocity v in a magnetic field B is given by F = qvB sin θ. The magnetic force F on a current-carrying conductor of length L with current I in a magnetic field B is F = BIL sin θ.

Is Motion in Electromagnetic Fields SL or HL?

Both. SL and HL students study Motion in Electromagnetic Fields to the same depth.

How do I revise Motion in Electromagnetic Fields for IB Physics?

Start from the core idea: this topic describes the motion of charged particles and current-carrying conductors in uniform electric and magnetic fields. In the exam: 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). Then practise exam-style questions, easiest first, writing out every step of your working before you check it.

How does FourtyFive help me practise Motion in Electromagnetic Fields?

FourtyFive has 5 Motion in Electromagnetic Fields questions. Every answer you write is marked mark by mark, IB-style, and you see where each mark was won or lost. Every part has a hint, the AI tutor helps you through the step you are stuck on, and your Study Profile picks what to practise next.

Is FourtyFive free for Motion in Electromagnetic Fields practice?

Yes. A free account gives you 50 marked answers a month, and you do not need a card to sign up.

Can I handwrite Motion in Electromagnetic Fields answers on an iPad?

Yes. In the FourtyFive iPad app you write your working by hand with Apple Pencil, the way you would on paper, and it is marked the same way.

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