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 0.115 T magnetic field.a. What electric field strength, in volts per mater, is needed to select a speed of 3.7 x 10^6 m/s?b. What is the voltage, in kilovolts, between the plates if they are separeted by 0.75 cm?

Answer 1

(a) $4.26\cdot 10^5 V/m$

In a velocity selector, the speed of the beam is related to the magnitude of the electric field and of the magnetic field by the formula:

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

where

E is the magnitude of the electric field

B is the magnitude of the magnetic field

In this problem, we have

$B=0.115 T$ (magnetic field)

$v=3.7\cdot 10^6 m/s$ (speed of the particles)

Solving the equation for E, we find the electric field:

$E=vB=(3.7\cdot 10^6 m/s)(0.115 T)=4.26\cdot 10^5 V/m$

(b) 3.2 kV

The relationship between electric field and potential difference between the two plates is:

$V=Ed$

where, in this problem:

$E=4.26\cdot 10^5 V/m$ is the magnitude of the electric field

$d=0.75 cm=0.0075 m$ is the separation between the plates

Substituting into the equation, we find the potential difference:

$V=(4.26\cdot 10^5 V/m)(0.0075 m)=3195 V=3.2 kV$