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Homework Help: Magnetic force and electric field, analytic problem about Lorentz Force.

  1. May 3, 2010 #1
    1. The problem statement, all variables and given/known data

    Question is E=(x,y,z) B=(x,y,z) v=(x,y,z) (in vector form)
    For all E,B,v find v(t) and r(t) v and r of course have their vector arrows.

    2. Relevant equations
    F=ma= qE + qv x B

    There is a hint to take second derivatives across and some terms will clear up. I was thinking more about the integrating it back to velocity and position functions.

    3. The attempt at a solution

    dv/dt - (q/m)v x B= qE

    V= Vxi+Vyj+Vzk

    so VxB = (VyBz-VzBy)i-(VxBz-VzBx)j+(VxBy-VyBx)k

    dvx/dt - q/m(VyBz-VzBy) = q Ex

    So I got the cross product, and put the Lorentz equation in to its differentiate form, but I am stuck here since I don't understand what I am supposed to do when the question is asking "for all v(t) and r(t)". I guess I am more stuck on the format of the answer, like how is it supposed to look like, or how many vectors/equations there needs to be by the time I am done. At the end am I supposed to get just two equations that are the derivative/integral of each other that describe the position and velocity of any given charge if data plugged in right? I would assume so since it is all analytic.

    Appreciate the help.
  2. jcsd
  3. May 3, 2010 #2


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    Welcome to PF!

    Hi AdamP ! Welcome to PF! :smile:

    (on this forum, you can use bold letters for vectors, and try using the X2 tag just above the Reply box :wink:)

    Yes, you need to find equations for r (and v) in terms of r0 and v0.

    There's two ways you could go about this …

    i] transform to a frame of reference in which either E or B = 0, solve that much easier equation, then transform back again :wink:; or

    ii] try various simple algebraic manipulations, such as "dotting" or "crossing" the Lorentz force equation with r0 or v0. :smile:

    (and remember eg a2 = a.a)
  4. May 4, 2010 #3
    Hello and thank you for your help.

    As far as transferring to a frame of reference where E or B are 0, could you elaborate on that a little? I am in over my head in this course and lack the necessary background to truly succeed, which is why I am struggling with this question so much. However I am willing to try as much as possible to get it done.

    Also, why would you cross or dot r and v? I think I don't understand why you would dot and cross the position function with the velocity function and what would you gain out of that.

    let me try a2 I did it yay :)
  5. May 4, 2010 #4


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    Hello AdamP! :wink:
    oh, if you haven't covered that, just forget about it …
    Well, for example, if you "dot" the Lorentz equation with v, the v x B goes to zero, and the dv/dt becomes v.dv/dt, which is d/dt of … ? :smile:
  6. May 4, 2010 #5
    Ok ok I am slowly getting it. I think i struggled because I didint know about the (AxB).C = (CxA).B. So ok (VxB).V =0 from VxV=0 . If I did that I am getting vdv/dt=qEv .
    Then I played around a little v=dx/dt and dv/dt=d2x/dt=a . Probably vdv/dt equals something else..hmmm

    I mean can I now divide by V and just get dv/dt=qE=a . It doesnt make any sense. Would you seperate the variables and integrate dv to v and qEdt, to qEt? I dont know :(
  7. May 5, 2010 #6


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    No, you can't divide v.dv/dt by v

    you can't divide a scalar by a vector!! :wink:
    Yes … v.dv/dt = 1/2 d(v.v)/dt :smile:
  8. May 16, 2010 #7
    I was away due to other exams and did not do any additional work on this problem, but I still need to solve it.

    It looks like we are making progress all though you are doing most of the work. :) Anyhow, I flat out don't get that last part.

    I don't see how you got 1/2 d(v.v)/dt . after that v dot v is v squared magnitude, but how does that help me. dv/dt is a...

    Sorry man I am not trying to make you solve this, its just I don't get it :(
  9. May 16, 2010 #8


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    If you have an equation with a (like F = ma), you can dot it with v

    then v.a = v.dv/dt = 1/2 dv2/dt, which is easy to integrate.
  10. May 16, 2010 #9
    Oh excellent tnx tim.
    So dv/dt is a, and we dotted that with v, which gave us vdv/dt and got rid of the -(q/m)vXB.

    Then vdv/dt integrated to 1/2dv^2 /dt and so then it is equal to intg(qEv).

    1) Now by doing all this are we closing in on the E? Like single out E on the right hand side.

    2) How is the integral on the right hand side of the equation is going to work? ( I am assuming this since you integrated v(dv/dt) to 1/2 dv^2 /dt, so both sides must get the same operator.
  11. May 16, 2010 #10


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    (Not Ev, but E.v.)

    So you have a energy equation … 1/2 mv2 = q∫E.v dt + constant …

    that's progress, isn't it? :smile:

    (and you can similarly get an angular momentum equation by differentiating r x v)

    btw, in this question, are E and B constants, or do they depend on position and time?
  12. May 16, 2010 #11
    Well, they are given as vectors with (x,y,z) coordinateness but no function of (t) is implied. However velocity and position r, are supposed to be found as functions of time v(t), r (t). I am not sure if that forces the E B and v to be as functions of time also.

    How did 1/2dv^2/dt become 1/2 mv^2?

    RxV ( u mean position crossed with velocity?) Is this the position function r(t) that we are looking for?
    When are we going to use the VXB, that I have all the way on top? Do we even need it to go forward?
  13. May 16, 2010 #12


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    The fact that they have (x,y,z) coordinates doesn't mean that they depend on x y and z, it only means that they are vectors.

    I'm hoping they're constants, since that makes everything much easier. :smile:
    By integrating.
    Yes, and no.
    We dotted with v to get rid of it, but we'll need it back later, or the result won't depend on B at all, will it? :smile:
  14. May 16, 2010 #13
    Ok the assignment states for partial credit you can assume Bi=B(i) for all i, Bx=B(x) By=(y)
    I guess that is what you are talking about since supposedly this makes the problem much easier, however does not provide full credit.
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