Velocity Selection

A velocity selector is defined as:

A device consisting of perpendicular electric and magnetic fields where charged particles with a specific velocity can be filtered

Velocity selectors are used in devices, such as mass spectrometers, to produce a beam of charged particles all travelling at the same velocity
A velocity selector consists of two oppositely charged parallel plates situated in a vacuum chamber
- The plates provide a uniform electric field with strength E between them

There is also a uniform magnetic field with flux density B applied perpendicular to the electric field
- If a beam of charged particles enters between the plates, they may all have the same charge Q but travel at different speeds

Velocity selector

Velocity selection diagram, downloadable AS & A Level Physics revision notes

The particles travelling at the desired speed v will travel through undeflected due to the equal and opposite electric and magnetic forces on them

The electric force does not depend on the velocity: $F_{E} = E Q$
However, the magnetic force does depend on the velocity:
- The magnetic force will be greater for particles which are travelling faster

To select particles travelling at exactly the desired speed v, the electric and magnetic force must therefore be equal, but in opposite directions

$F_{E} = F_{B}$

The resultant force on the particles at speed v will be zero, so they will remain undeflected and pass straight through between the plates
By equating the electric and magnetic force equations:

$E Q = B Q v$

The charge Q will cancel out on both sides to give the selected velocity v equation:

$v = \frac{E}{B}$

Therefore, the speed v in which a particle will remain undeflected is found by the ratio of the electric and magnetic field strength
- If a particle has a speed greater or less than v, the magnetic force will deflect it and collide with one of the charged plates
- This would remove the particles in the beam that are not exactly at speed v
Note: the gravitational force on the charged particles will be negligible compared to the electric and magnetic forces and therefore can be ignored in these calculations

Worked example

A positive ion travels between two charged plates towards a slit S

(a) State the direction of the electric and magnetic fields on the ion

(b) Calculate the speed of the ion emerging from slit S when the magnetic flux density is 0.50 T and the electric field strength is 2.8 kV m^-1

Answer:

Part (a)

Step 1: Determine the direction of E field

Electric field lines point from the positive to negative to charge
Therefore, it must be directed vertically upwards

Step 2: Determine the direction of B field

Using Fleming’s left-hand rule:
- The charge, or current I, is directed to the right
- B is out of the page
- Therefore, the force F is vertically downwards

Part (b)

Electric field strength, E = 2.8 kV m^-1 = 2.8 × 10³ V m^-1
Magnetic flux density, B = 0.50 T

$v = \frac{E}{B} = \frac{2.8 \times 10^{3}}{0.50} = 5600 m s^{- 1}$

Part (c)

If the speed increases, the magnetic force must be greater because F_B ∝ v
Since the magnetic force would direct the ion downwards in the direction of the field, the ion will be deflected towards the positive plate

Velocity Selection (CIE A Level Physics)

Revision Note

Velocity selection

Velocity selector

Worked example

You've read 0 of your 10 free revision notes

Unlock more, it's free!

Join the 100,000+ Students that ❤️ Save My Exams

Author: Ashika