A proton moves through a uniform magnetic field given by =(6.97-9.54+35.0) mT. At time t1, the proton has a velocity given by =vx+vy+(1.58 km/s) and the magnetic force on the proton is FB=(4.72 X 10-17N)+(3.45 X 10-17N).
(a) At that instant, what is vx?
(b) At that instant, what is vy?
F=qV x B
The Attempt at a Solution
So, in order to find the velocity in the x direction I am assuming you do this:
That is really all I have because I cant figure out the cross product. I end up with vx terms in the j direction and I don't know what to do with them. I hope I am making sense. Please let me know if youcan help. THANK YOU!!!