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:
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!
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