Question

In a mass spectrometer, a specific velocity can be selected from a distribution by injecting charged particles between a set of plates with a constant electric field between them and a magnetic field across them (perpendicular to the direction of particle travel). If the fields are tuned exactly right, only particles of a specific velocity will pass through this region undeflected. Consider such a velocity selector in a mass spectrometer with a 0.105 T magnetic field.

Part (a) What electric field strength, in volts per meter, is needed to select a speed of 3.8 × 106 m/s?
Part (b) What is the voltage, in kilovolts, between the plates if they are separated by 0.75 cm?

Answer 1

a).
`$3.99 * 10^5 \ v/m$`

b). 2.9925 kV

Step-by-step explanation:

Given :

For mass spectrometer

The magnetic field = B

B = 0.105 T

a). Given speed, v =
`$3.8 * 10^6 \ m/s$`

We known

`$(E)/(B)=v$`

∴
`$E= 3.8 * 10^6 * 0.105$`

`$=3.99 * 10^5 \ v/m$`

b). Now spectrometer, d = 0.75 cm

`$d=0.75 * 10^(-2) \ m$`

We known

`$E=(V)/(d)$`

`$V = E* d$`

`$V = 3.99 * 10^5 * 0.75 * 10^(-2)$`

`$V = 2.9925 * 10^3 \ V$`

= 2.9925 kV