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Proton moving through magnetic field

  1. May 1, 2009 #1
    1. The problem statement, all variables and given/known data

    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?

    2. Relevant equations
    F=qV x B

    3. 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!!!
     
  2. jcsd
  3. May 1, 2009 #2

    berkeman

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    Staff: 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.
     
  4. May 1, 2009 #3
    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
     
  5. May 1, 2009 #4

    berkeman

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    Staff: Mentor

    You have the field and the charge and the initial veloity. You have what you need. What are the relevant equations?
     
  6. May 1, 2009 #5
    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:

    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.
     
  7. May 3, 2009 #6
    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?
     
  8. May 3, 2009 #7
    hmm I bet I need to convert the mT to T huh?
     
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