Mass Spectrometer

Question

 

What is the speed of a proton that travels in a circular path of radius 0.045 cm, perpendicular to a
3.0 T magnetic field?

 

 

 

 

An electron traveling at speed v enters a magnetic field of strength B, and is forced into a circular path of radius R. If the strength of the field is doubled, and the speed of the electron is also doubled, what will the radius of the circular path become?

 

 

Answer 1

Hello Jamie,

 

The magnetic force on a charged, moving particle is:

 

     F_mag = qvBsinθ

 

where q is the particle's charge in Coulombs, v is the particle's velocity in meters per second, B is the strength of the magnetic field in Teslas, and θ is the angle between the directions of the particle's velocity and the magnetic field.

 

Also, if you recall from several chapters ago, the centripetal force is:

 

     F_cent = mv²/r

 

where m is the object's mass in kilograms, v is the object's velocity in meters per second, and r is the radius of the circular path in meters.

 

For a proton in circular motion inside of a uniform magnetic field, the magnetic force on the proton is the centripetal force keeping that proton in a circular path.

 

With these three bits of information, you should be able to solve this problem, Jamie.

 

Try your best to solve the entire problem, but if you end up needing more help, reach back out to us.

 

Best of luck! :)

Answer 2

Jamie,

 

I just want to add to Amos' fine explanation that when you put all this together, you get

 

mv²/r = qvB sinθ

 

In these problems, θ = 90° (velocity perpendicular to field), and a v cancels out.

 

mv/r = qB

 

Look up the charge and mass in your book.  The other quantities are given to you.  At this point, the physics part of the problem is done, and you just need to do the algebra.