Question

/ 7, Consider the mass spectrometer shown schematically in the figure. The electric field between the plates of the velocity selector is 2500V/m, and the magnetic field in both the velocity selector and the deflection chamber has a magnitude of 0.035T I IA a) Show that only particles with a speed of 6.25*10mls enter the mass spectrometer. b) Calculate the radius of the circular part of the path for a singly charged ion with a mass of 2.18*102kg. Design the solenoid which would produce the magnetic field for the instrument with a 1.0A current. c)
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Answer #1

A velocity selector is a region in which there is a uniform electric field and a uniform magnetic field. The fields are perpendicular to one another and perpendicular to the initial velocity of the charged particles that are passing through the region.

The force excreted on a charged particle by the electric field is given by:

F = qE

The magnitude of the force exerted by the magnetic field is

F = qvB

as long as the velocity is perpendicular to the field.

The idea is that, if the two forces are equal and opposite, the net force is zero and the particle passes through the region without changing direction.

ie, qE=qvBE=vBv=rac{E}{B}

(a) Given that,

Electric field, E = 2500 V/m

Magnetic field, B = 0.035 T = 0.04 T

Therefore, the velocity of the particles that can pass through the region is,

  2500 0.04 6.25 x 104 m/s

(b)  In the chamber, the magnetic force provides a centripetal force, mu2

Given, mass = 2.18 × 10-26 and the charge of an electron is 1.6 × 10-19 C

ie,   qvB=rac{mv^2}{r}

Tr Bq

2.18 × 10-26 × 6.25 × 104 0.04 × 1.60 × 10-19 0.213 m

(c) The magnetic field associated with a solenoid is given by,

B=mu_0 nI

  rl 1401 0.04 4π × 10-7 × 1 = 31831

A solenoid with 31831 wounds would produce a magnetic field of 0.04 T for the instrument with a current of 1A.

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