Question

A uniform magnetic field points North; its magnitude is 1.5 T. A proton with kinetic energy 8.0x10–13 J (5.0 MeV) is moving vertically downward in this field. What is the magnetic force acting on it?

I have no idea where to start with this question, can I just have the formula's I would need to work it out?

Thanks:)

Answer 1

F=qvB(sinθ) but sinθ=1 so we can ignore it.

We have B =1.5T
q=1.6x10^-19
We just need v from our Ke.

Ke=1/2mv^2
v^2=2Ke/m
Punch in 8.0x10^13J and look up the mass of a proton (1.672x10^27) and solve for v, which turns out to be 3.09x10^7.

F=(1.6x10^-19)(3.09x10^7)(1.5)=
7.4x10^-12.
If you do the right hand rule you will find out it points east. (Look up right hand rule if you don't know it).