# Proton moving through magnetic field

## 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?

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 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!!!

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berkeman
Mentor
Could you please clarify your vector velocity and B-field components? And then show how you derived the vector force components? Please label the x, y, and z components explicitly in each equation.

they are given in the problem

b= (6.97i-9.54j+35.0k)

and the vector force was not derives as is also given in the equation as: (vx)i+(vy)j+(1.58 km/s)k

berkeman
Mentor
You have the field and the charge and the initial veloity. You have what you need. What are the relevant equations?

You did leave off the i, j, and k unit vectors when describing the problem which left a little to the imagination. But it looks like your equation:

(4.27E-17)i +(3.45E-17)j=e[(vx+vy+1.58k)x(6.97i-9.54j+35k)]
looks correct so far. Could you show us the cross product you ended up getting so we could see if you did something wrong? And getting a vx term in the j direction is alright. Once you have the cross product, you just equate the i-terms on both sides of the equation, and do the same for the j- and k-terms and solve.

for the cross product i got:
(35vy+15.07)i+(35vx-11.07)j+(-9.54vx-6.97vy)k

if I do this:
(4.72E-17)=(35vy+15.07) and I solve for vy that is what the velocity in the x directions is? .431? Why am I solving for vy if i am looking for vx?

hmm I bet I need to convert the mT to T huh?