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## Homework Statement

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?

## Homework Equations

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:

(4.27E-17)i +(3.45E-17)j=e[(vx+vy+1.58k)x(6.97i-9.54j+35k)]

That is really all I have because I can't 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!