# Time Period of a Proton in a Magnetic Field

• January 25th 2011, 01:59 PM
alexgeek
Time Period of a Proton in a Magnetic Field
Spent half an hour looking at this problem, but I can't see anyway to get any of the multiple choice answers:
Quote:

Originally Posted by AQA Specimen Paper
Protons, each of mass mand charge e, follow a circular path when travelling perpendicular to a magnetic field of uniform flux density B. What is the time period for on complete orbit?
• A $\frac{2 \pi e B}{m}$
• B $\frac{m}{2 \pi e B}$
• C $\frac{eB}{2 \pi m}$
• D $\frac{2 \pi m}{eB}$

All I know is that it should have something to do with:
$F = m \omega^2 r = m \frac{v^2}{r} = BQv \: , \: \omega = \frac{2 \pi}{T}$

Thanks

Edit: Oh and the answer is D
• January 25th 2011, 05:16 PM
Aryth
You're missing an equation.

Remember that the tangential velocity of an object in circular motion is: $v = \omega r$ so that:

$F = QvB = m\frac{v^2}{r}$

$QB = m\frac{v}{r} = m\omega$

$\omega = \frac{QB}{m}$

We know the formula for period, which is:

$T = \frac{2\pi}{\omega}$

$T = \frac{2\pi m}{QB}$

Since we know the charge, we can fill that in to get:

$T = \frac{2\pi m}{eB}$

I'm not sure why D doesn't match up to this, but I'm almost 95% sure this is right.
• January 25th 2011, 11:47 PM
alexgeek
Thanks very much for the reply, will look again when I got some caffeine.
You were right also, I just had a typo in my LaTeX (\pim instead of \pi m).