Question

Find the value of the magnetic field necessary to cause a proton moving at a speed of 2.50 ×10^3 m/s to go into a circular orbit of 15.5-cm radius.

An electron has an energy of 100 eV as it enters a magnetic field of 3.50 × 10−2 T. Find the radius of the orbit.

A velocity selector has a magnetic field of 0.300 T. If a perpendicular electric field of 10,000 V/m is applied, what will be the speed of the particles that will pass through the selector?

Answer 1

All of these three questions can be solved by applying the rule F = q V B
which applies when V is perpendicular to V.

I. e V B = Mp * V^2/R

You have been given R and V; solve for B.
e is the proton charge and Mp is the proton mass.

Try the others yourself