# Proton moving through magnetic field

• mariahkraft
In summary: So if I do that..I get something like .00000697In summary, the problem involves a proton moving through a uniform magnetic field with given components and velocity. The question is to find the velocity in the x and y directions at a specific time. The relevant equation is F=qV x B. With the given information, the cross product is calculated and equated to determine the velocity components. It is important to convert the magnetic field from mT to T in the calculations.

## 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 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!

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

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?