Question

A proton moving at 5.00 106 m/s through a magnetic field of magnitude 1.78 t experiences a magnetic force of magnitude 7.40 10-13 n. what is the angle between the proton's velocity and the field? (enter both possible answers from smallest to largest. enter only positive values between 0 and 360.)

Answer 1

The magnetic part using the Lorentz force is: F = q v x B, 
where v and B are vectors and v x B is the vector cross product. 

Magnitude of the force: F = q v B sin(α) 

So, sin(α) = F/( e v B), with e the proton charge. 

This will give you a value for sin(α), and two potentials for its opposite.

You will now look for: 

sin(α) = 7.40 10^-13/( 1.60 10^-19 * 5 10^6 * 1.78) 
= 0.520

So either sin(α) = 0.502 or sin(α) = -0.502 
The 1st α = 30.1 degrees or α = 150 degrees. 
The 2nd α = 210 degrees or α = 330 degrees. 
So we can say that 30.1 degrees and 330 degrees would be minimum and biggest on [0,360]