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Magnetic force per unit charge and work done on a proton

  1. May 28, 2013 #1
    1. The problem statement, all variables and given/known data

    a proton moves with a velocity <6*105,3*105> m/s through a magnetic field given by <.2,.2> T

    a. Find the magnetic force per unit charge on the proton

    b. Find the work done by this force


    2. Relevant equations

    Fm=qv x B


    3. The attempt at a solution

    For part a, I used a property of the cross product to simply divide the force by the charge

    Fm/q=v x B

    So by finding the cross product, I believe that I reached the force per unit charge which gives (3/5)*105 N/C

    For part b, I know that the charge is moving with constant velocity (no net force). Therefore the work done by the magnetic force must be negative. (some outside force must do work against it?)

    I have no idea how I would go about finding the "work done by this force", I almost feel as if there's not enough information provided. Wouldn't I at least need the time period over which to consider?

    Thank you.
     
    Last edited: May 29, 2013
  2. jcsd
  3. May 28, 2013 #2

    gneill

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    Staff: Mentor

    Yes, that looks okay. Your value can be written as 6.0 x 104 N/C
    If there were no force, then the work done by that force would be zero, right? But you calculated the force (well, the force per unit charge), and it wasn't zero...

    What is the expression that gives the work done by a force? (hint: the vector version). What's the direction of the force with regards the motion of the proton?
    No, there's enough information. It has to do with a characteristic of the force that results from how it is generated. Take a close look at how you calculated it.
     
  4. May 28, 2013 #3
    Thanks so much for your response

    1. There is no NET force, if I understand the question statement correctly, because the velocity is constant. No acceleration means no force. I know that the magnetic force is not zero.

    2. I know that work=force*distance, I did a search for a vector formula but couldn't find anything I think you are talking about. The direction of the force is perpendicular to the two vectors I crossed, making it in the z direction.

    but the proton is not moving in the z direction at all, so how is there any work being done?!?

    Is it zero then?

    If I'm missing something fundamental I could really use some guidance as to what it is.
    Thank you
     
    Last edited: May 28, 2013
  5. May 28, 2013 #4

    gneill

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    Staff: Mentor

    Consider and compare with circular motion. Consider also that there the velocity is NOT constant, but speed is...they are not the same thing. In uniform circular motion there is a constant force, a continuous acceleration, yet the speed is constant. Why?
    What distance is involved in the work formula? The vector version tells you something about how the work is related to the direction of the force and the direction of the displacement...
    It's not initially moving in the z-direction...
     
  6. May 29, 2013 #5
    I just read in my book the following statement

    "the magnetic force acting on a charged particle can never do work because at every instant the force is perpendicular to the velocity"

    So the work is zero like I said.
     
    Last edited: May 29, 2013
  7. May 29, 2013 #6

    gneill

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    Staff: Mentor

    That is correct. [itex]W = \vec{F} \cdot \vec{d} = Fd\;cos(\theta)[/itex]
    The work is zero, but not like you said before. This time your reason is correct :smile:
     
    Last edited: May 29, 2013
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