Motion in Electromagnetic Fields
Introduction
Charged particles reveal fields through their motion. The guide focuses on paths in uniform electric fields, magnetic fields, crossed fields, and the forces on charges, conductors, and parallel wires.
Guide Focus
- Analyse charged-particle motion in uniform electric and magnetic fields.
- Use F = qvB sin(theta) and F = BIL sin(theta).
- Apply force per unit length between parallel current-carrying wires.
Key Concepts
1. Uniform electric fields
A charge in a uniform electric field experiences F = qE. If released or projected into the field, it accelerates in a direction depending on the sign of the charge.
2. Uniform magnetic fields
A moving charge experiences magnetic force F = qvB sin(theta). The force is perpendicular to velocity, so the magnetic field changes direction of motion without changing kinetic energy.
3. Circular paths and charge-to-mass ratio
When velocity is perpendicular to a uniform magnetic field, magnetic force provides centripetal force. The radius of the path can be used to determine q/m.
4. Forces on conductors and wires
A current-carrying conductor in a magnetic field experiences F = BIL sin(theta). Parallel currents attract if they flow in the same direction and repel if they flow in opposite directions. Force per unit length is F/L = mu0 I1I2 / (2 pi r).
Common Mistakes
- Assuming magnetic force changes a particle’s speed.
- Using qvB when velocity is parallel to the magnetic field.
- Mixing the direction rule for conventional current with electron motion.
Exam Tips
- Check the angle first: sin(theta) decides the magnitude.
- Use circular motion equations when magnetic force is perpendicular to velocity.
- In crossed fields, straight-line motion usually means electric and magnetic forces balance.
Practice Questions
Question 1 (Multiple Choice)
A charged particle enters a uniform magnetic field with velocity parallel to the field. The magnetic force is:
A. Zero. B. Maximum. C. qvB. D. Directed along the field.
Solution Architecture
F = qvB sin(theta), and theta = 0 degrees for parallel motion, so the force is zero.
Question 2 (Structured Paper 2 Style)
An electron of charge magnitude 1.60 x 10^-19 C moves at 2.0 x 10^6 m s-1 perpendicular to a 0.30 T magnetic field.
(a) Calculate the magnetic force magnitude. [2 marks]
(b) State what happens to the electron’s kinetic energy. [1 mark]
Markscheme Breakdown
Part (a) Solution:
F = qvB = 1.60 x 10^-19 x 2.0 x 10^6 x 0.30 = 9.6 x 10^-14 N.
Part (b) Solution:
The kinetic energy remains constant because the magnetic force is perpendicular to velocity.