Posted by J on Monday, February 25, 2008 at 5:27pm.
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?

Physics  Damon, Monday, February 25, 2008 at 6:16pm
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.
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