Charged Particles in Magnetic Fields

When a charged particle enters a uniform magnetic field, it travels in a circular path
This is because the direction of the magnetic force F will always be
- perpendicular to the particle's velocity v
- directed towards the centre of the path, resulting in circular motion

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

In a magnetic field, a charged particle travels in a circular path as the force, velocity and field are all perpendicular

The magnetic force F provides the centripetal force on the particle
The equation for centripetal force is:

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

Equating this to the magnetic force on a moving charged particle gives the expression:

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

Rearranging for the radius r gives an expression for the radius of the path of a charged particle in a perpendicular magnetic field:

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

Where:
- r = radius of the path (m)
- m = mass of the particle (kg)
- v = linear velocity of the particle (m s⁻¹)
- B = magnetic field strength (T)
- Q = charge of the particle (C)

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

The centripetal acceleration is in the same direction as the magnetic (centripetal) force
This can be found using Newton's second law:

$F = m a$

Worked example

An electron travels 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 circular path of the electron.

Answer:

Step 1: List the known quantities

Electron charge-to-mass ratio = $\frac{e}{m_{e}}$ = 1.76 × 10¹¹ C kg⁻¹ (from formula sheet)
Magnetic flux density, B = 6.2 mT = 6.2 × 10⁻³ T
Speed of the electron, v = 3.0 × 10⁶ m s⁻¹

Step 2: Write an expression for the radius of an electron in a magnetic field

centripetal force = magnetic force

$\frac{m_{e} v^{2}}{r} = B e v$

$r = \frac{m_{e} v}{e B}$

Step 3: Substitute the known values into the expression

$\frac{m_{e}}{e} = \frac{1}{1.76 \times 10^{11}}$

$r = \frac{3.0 \times 10^{6}}{(1.76 \times 10^{11}) \times (6.2 \times 10^{- 3})} = 2.7 \times 10^{- 3} = 2.7$ mm

Exam Tip

Make sure you can derive the equation for the radius of the path of a particle travelling in a magnetic field.

As with orbits in a gravitational field, any object moving in circular motion will have a centripetal force and a centripetal acceleration. Make sure to refresh your knowledge of these equations.

Charged Particles in Magnetic Fields (HL IB Physics)

Revision Note