Question

A beam of protons is accelerated through a potential difference of 0.745 kV and then enters a uniform magnetic field traveling perpendicular to the field. (a) What magnitude of field is needed to bend these protons in a circular arc of diameter 1.75 m? (b) What magnetic field would be needed to produce a path with the same diameter if the particles were electrons haveing the same speed as the protons?

ok for part (a) I confused on what equation to incorporate all the givens! I am lost! and for part (b) you would just need to change the mass of the the protons to the mass of electrons or is there more to that?

Answer 1

The kinetic energy of the proton is (1/2) m v^2

Therefore (1/2) m v^2 = 745 volts * charge of proton

That gives you v, the speed of the proton.

Now use your force on a particle in a magnetic field to get the radius of path (if you do not have the equation in your book) If you do, just skip to the end of the next paragraph.

F = q v B if B perpendicular to v
= mass * centripetal acceleration = m v^2 /R.
So R = m v /(q B)
That is all you need.

In part B m changed for the proton. That changes the velocity from (1/2) m v^2
and also makes a change in the radius equation.
(1/2) m v^2 = q V
v = sqrt (2 q V/m)
then for th radius
R = m v/(qB) = sqrt(2 q V m)/(q B)
so the radius goes up with sqrt(m)
of course the sign of q changed, but that only changes the direction, not the radius.