PHY C22: Charged Particles in Electric & Magnetic Fields

A tough one (but an interesting one!):

  • Velocity selection
  • Hall voltage

What happens when a charged particle enters an electric field AND a magnetic field?
In my fields post, I mentioned that when multiple fields act on an object, the object will experience the SUM of those forces. The same applies here:

All you have to do is find the net force, which will dictate how the particle will accelerate & deflect.


One application of this is:

Velocity Selection

What is a velocity selector?
A device used to produce a beam of particles all travelling at the same specific velocity.

It contains perpendicular magnetic & electric fields, with a beam of particles entering perpendicular to both:

How does it work?

Consider a positively-charged particle entering the velocity selector:



The force due to the electric field:


The force due to the magnetic field:
The net force is given by FE – FB:
We can vary the strength of E & B all we want, producing different net forces & different deflections.
IF the force from the electric field is equal to the force from the magnetic field, the particle has NO net force, so it will have NO deflection.

This allows it to pass through the slit.
For there to be no deflection,
FE = FB
qE = Bqv

v = E/B
Only particles travelling at velocity v = E/B will pass through the slit.
Particles travelling at any other velocity will be deflected!

In general,

Fnet = FE – FB
Fnet = qV – Bvq

Fnet = q(V – Bv)

For particles which pass through the selector undeviated:

v = E/B

This has a few implications for velocity selectors:

  • mass does not affect the net force acting on a particle:
    • particles of the selected velocity WILL pass through regardless of their mass
  • charge only affects the net force acting on a particle when it is non-zero:
    • particles of different charge will have different paths IF they are deviated, but all particles at the selected velocity WILL pass through undeviated regardless of their charge

*You must also be able to identify the direction of deviation for particles at velocities above or below the selected velocity. Just use Fnet = q(V – Bv) & identify the direction of the net force. Remember to check the direction each force effects the particle!


We have seen how charged particles react to the presence of magnetic & electric fields. But now, we will investigate how the deflection of the charged particles themselves creates an electric field:

The Hall Effect

Consider a thin slice of a conductor normal to a magnetic field.

It has width d & thickness t
When current (which we consider to be + in charge) is sent through the conductor, it is deflected according to Fleming’s left hand rule:
Now there are more positive charges on 1 side of the conductor, & less on the other.

The imbalance of charges creates a POTENTIAL DIFFERENCE across the conductor.
We call this the HALL VOLTAGE, VH

This works whether we consider (+) current flowing 1 direction:

…or (-) electrons flowing in the opposite direction:
As more particles get deflected, the electric field across the conductor increases.

For a uniform electric field,
E = V/d
EH = VH/d

BUT, remember that charged particles will be affected by any electric field (even ones they induce!).

Hall voltage reaches a constant value when:
magnetic force on the particles = electric force in the opposite direction

At this point, (+) charges entering the slice are no longer deflected – they experience equal & opposite forces from the magnetic & electric fields.

FE = FB
This relation allows us to derive an expression for the Hall Voltage!

The 2 forces are given by:
FE = qVH/d
FB = Bvq

When FE = FB,
qVH/d = Bvq
VH/d = Bv

VH = Bvd
From the electric fields chapter, remember this equation for current:
I = Anvq

Rearranging to find v,
v = I/Anq

For this slice of a conductor, cross sectional area is:
A = td

So, v = I/tdnq

Subbing this into our expression for VH:
VH = BId/tdnq
VH = BI/ntq

…& there’s our final equation!

Hall Voltage VH = BI/ntq

How can we apply this?
In a Hall probe.

  • It works by:
    • providing a thin slice of a conductor which can be placed perpendicular to a magnetic field
    • a Hall Voltage is created between the sides of the slice
    • the VH across these 2 terminals is measured
    • the value of B can be calculated by rearranging VH = BI/ntq (I, n, t, & q are specific to that Hall probe)
  • It can be optimised by increasing VH for a given B (to reduce % error). This is done by:
    • reducing the value of n: a semiconductor is used instead of a metal (less number density of charge carriers)
    • reducing the value of t: slice is made to be thin as possible

Extra Resources

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