Question

This situation in question is that shown in Figure 17.33 of Etkina, which is posted in a PDF in BB. Please refer to that figure when answering questions. A) When a charged particle enters a region of uniform magnetic field, where the magnetic field is perpendicular to the velocity of the particle, the particle will execute uniform circular motion. There is a relationship between the radius, charge, mass, speed and magnetic field – you’ll find it at the top of the PDF from Etkina that includes Figure 17.33. Derive this relationship using: Newton’s Second Law, the Lorentz force law and expression for centripetal acceleration in uniform circular motion. B) In the mass spectrometer shown in Figure 17.33, we accelerated singly ionized He atoms through a 5500 Volt potential difference. They then enter a velocity selector region where a magnetic field of 0.75 Tesla is applied, perpendicular to the plane of the figure. Then the ions enter a region of uniform magnetic field with a strength of 0.35 Tesla. i) Look up the mass of singly ionized He. What is the speed of the ionized He particles after being accelerated through 5500 Volts? Use conservation of energy to determine this speed – like dropping a marble of some mass in a gravitational field. ii) The speed found in i) is the one we want to select for in the velocity selector. What is the electric field we should apply in the selector to do this? iii) What’s the radius of motion for these He ions in the mass spectrometer?

Answer 1

A)

The centripetal force must be equal to the Lorentz force.

Lorentz force is given by: where v and B are the velocity of the particle and the magnetic field respectively. q is the charge of the particle.

For, circular motion, the force must always be perpendicular to the velocity. Also, the magnetic field is perpendicular to the direction of velocity.

The centripetal force is given by: where m is the mass of the charged particle and r is the radius of rotation.

Dividing both sides by v

rearranging the equation:

[This the relation between radius of the circular path and the parameters m,v,q and B]

B)

i) The energy due to applied electric field is

Since the particle in question is an He+ ion, the charge is equal to the electronic charge and it is positive in polarity.

Therefore,

This energy is responsible for the kinetic energy of the ion,

Rearranging the equation,

Using

ii) In the velocity selector, the force due to magnetic field must be equal to the force due to electric field.

Equating the two,

Dividing both sides by e,

B=0.75 T

iii) Using the relation for r we derived in part A,

Here, q=e