Motion of a Charged Particle in a Magnetic Field | CIE A Level Physics Revision Notes 2025

Motion of a Charged Particle in a Uniform Magnetic Field

A charged particle in uniform magnetic field which is perpendicular to its direction of motion travels in a circular path
This is because the magnetic force F_B will always be perpendicular to its velocity v
- F_B will always be directed towards the centre of the path

Circular motion in a magnetic field

Circular motion of charged particle, downloadable AS & A Level Physics revision notes

A charged particle travels in a circular path in a magnetic field

The magnetic force F_B provides the centripetal force on the particle
Recall the equation for centripetal force:

$F = \frac{m v^{2}}{r}$

Where:
- m = mass of the particle (kg)
- v = linear velocity of the particle (m s^-1)
- r = radius of the orbit (m)
Equating this to the force on a moving charged particle gives the equation:

$\frac{m v^{2}}{r} = B q v$

Rearranging for the radius r obtains the equation for the radius of the orbit of a charged particle in a perpendicular magnetic field:

$r = \frac{m v}{B q}$

This equation shows that:
- Faster moving particles with speed v move in larger circles (larger r): r ∝ v
- Particles with greater mass m move in larger circles: r ∝ m
- Particles with greater charge q move in smaller circles: r ∝ $\frac{1}{q}$
- Particles moving in a strong magnetic field B move in smaller circles: r ∝ $\frac{1}{B}$

Worked example

An electron with charge-to-mass ratio of 1.8 × 10¹¹ C kg^-1 is travelling at right angles to a uniform magnetic field of flux density 6.2 mT. The speed of the electron is 3.0 × 10⁶ m s^-1.

Calculate the radius of the circle path of the electron.

Answer:

Step 1: Write down the known quantities

Charge-to-mass ratio: $\frac{q}{m} = 1.8 \times 10^{11} C {kg}^{- 1}$
Magnetic flux density, B = 6.2 mT
Electron speed, v = 3.0 × 10⁶ m s^-1

Step 2: Write down the equation for the radius of a charged particle in a perpendicular magnetic field

$r = \frac{m v}{B q}$

Step 3: Substitute in values

$\frac{m}{q} = \frac{1}{1.8 \times 10^{11}}$

$r = \frac{(3.0 \times 10^{6})}{(1.8 \times 10^{11}) (6.2 \times 10^{- 3})} = 2.688 \times 10^{- 3} m = 2.7 mm (2 s . f)$

Motion of a Charged Particle in a Magnetic Field (CIE A Level Physics)

Revision Note

Motion of a Charged Particle in a Uniform Magnetic Field

Circular motion in a magnetic field

Worked example

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Author: Ashika