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Homework Help: Electromagnetism of particle problem

  1. Aug 6, 2010 #1


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    1. The problem statement, all variables and given/known data
    We are given surface current density, K which is constant (K is negative) along the z-axis which is placed in x=a plane (the board which contains this density is neutral), in x=0 plane there's an infinite plane which is charged by a surface density [tex] \sigma[/tex] which is constant.
    we have a particle with charge q and mass m which starts at rest at r=0.
    1. given sigma find out what is the minimal K s.t the particles doesn't cross the boards.
    2. given the K you found assume now K is twice the K in 1, and show that:
    the minimal distance from the board with the current is
    [tex]d_{min} = a- \frac{m\sigma}{\pi q (3\sigma^2 +\frac{4m\sigma}{\pi q a})}[/tex]

    2. Relevant equations
    There's a file with a picture of the system attached to this post here:
    It's question number 1, you shouldn't mind that the text is in hebrew, I translated what it asks in English.

    3. The attempt at a solution
    For 1 I thought of looking at the force euqation:

    mv^2/a = qE+ qv x B where E is the electric field from the charged board and B is from the above board (situated at x=a), and then just need to find out when there's only one solution to this quadratic equation with regards to v, the veclotiy of the particle.

    The problem arises when I need to prove (2).
    Don't know how to do it.

    Last edited: Aug 6, 2010
  2. jcsd
  3. Aug 7, 2010 #2
    Re: electromagnetism.

    I guess things look like my picture? (see the picture attached) If so, then let me show you a way to obtain the general solution.

    It's easily shown that: [tex]\vec{B}=\frac{\mu _o|K|}{2}\hat{y}[/tex] and [tex]\vec{E}=\frac{\sigma}{2\epsilon _o}\hat{x}[/tex].

    The Newton's 2nd law equation for the particle: [tex]m\frac{d\vec{v}}{dt}=e\vec{E}+q\vec{v}\times\vec{B}[/tex]


    (1): [tex]\frac{dv_x}{dt}=A+Bv_z[/tex]

    (2): [tex]\frac{dv_z}{dt}=-Bv_x[/tex]

    [tex]v_y=0[/tex] (the particle only moves inside the y=0 plane)

    where: [tex]A=e\frac{\sigma}{2m\epsilon _o}[/tex] and [tex]B=\frac{\mu _o|K|}{2m}q[/tex]

    From (1) and (2), you can solve for a relation between vx and vz. Remember the initial condition: [tex]\vec{v}(\vec{r}=0)=0[/tex]. From (2), you can solve for the relation between x and vz. Then from the 2 relations, you should get an equation relating vx and x. Notice that when the particle reaches to the closest distance to the board with current, vx=0.

    For part 1, we have xmax=a when vx=0. Solve for K (or |K|). For part 2, we simply substitute K with 2 times value of K from part 1.

    By the way, I don't see [tex]\epsilon _o[/tex] and [tex]\mu _o[/tex] anywhere in dmin needed to show. Maybe the question assumes using another unit system rather than SI system?

    P.S: By "mv^2/a = qE+ qv x B", you assumed that at the time the particle reaches x=a, the particle's radius of curvature at that point is a, which hasn't been shown. It can be 2a, 3a, or 1000a; we don't know. You have to show it.

    Attached Files:

    Last edited: Aug 7, 2010
  4. Aug 7, 2010 #3


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    Re: electromagnetism.

    We're using cgs system of units and c=1.
    Thanks I think I understand what to do in 1.

    as of 2, I am not sure I understand what to do (I get in cgs units and c=1), that for part 1:

    K=m/(2qa), after solving the differential equation:

    [tex]d^2v_x/dt^2 = -(\frac{2\pi K q}{m})^2 v_x[/tex]
    which with v(0)=0 we get a sine function, and for x=a vx=0 we get what I wrote in the above.

    I think that in part 2, I need to write down xmax=d (where d is the distance from the lower plane situated at x=0), and then again vx=0, but I don't get the correct final solution.
  5. Aug 7, 2010 #4
    Re: electromagnetism.

    I've never worked with the CGS system before. Maybe that's how the term "pi" arises.
    You don't need to solve that differential equation. Just divide (1) by (2), you will get:
    Then do the integration: [tex]-\frac{B}{2}v_x^2=Av_z+\frac{B}{2}v_z^2[/tex]
    Integrate (2): [tex]v_z=-Bx[/tex]
    From the 2 equations above: [tex]-\frac{B}{2}v_x^2=-ABx+\frac{B^3}{2}x^2[/tex]
    I don't get the answer either. However, I think it's wrong. Consider the extreme case where [tex]\sigma = 0[/tex]. That means there is no electric field, no initial force to cause the particle to move, and the particle remains at rest at r=0. In that case, dmin=a. But from the answer:
    [tex]d_{min}=a-\frac{m\sigma}{3\pi q\sigma ^2+4m\sigma /a}=a-\frac{m}{3\pi q\sigma +4m/a}[/tex]
    when [tex]\sigma = 0[/tex] , we get [tex]d_{min}=3a/4[/tex] which is wrong.
  6. Aug 7, 2010 #5


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    Re: electromagnetism.

    Ok I understnad what you write, Now when I solve the equation I get K~sqrt(sigma), so indeed if sigma equals zero then also K=0 zero cause otherwise the particle will cross the board situated at x=a.

    Again thanks for your help with this question.
  7. Aug 7, 2010 #6
    Re: electromagnetism.

    Logically this is not the reason. We haven't had anything to verify our solution yet, since the only hint (the answer) is different from ours. I only pointed out that the statement dmin =... in the question is logically wrong based on another reasoning, not on our result K~sqrt(sigma).
  8. Aug 7, 2010 #7


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    Re: electromagnetism.

    But it doesn't make sense, even if there isn't an electric field, the particle would feel the force from the B-field, unless there is no also a magnetic field.
  9. Aug 7, 2010 #8
    Re: electromagnetism.

    Initially v=0, so if there is no electric field, even if there is B-field, the particle won't move at all.
  10. Aug 7, 2010 #9


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    Re: electromagnetism.

    I see your point, thanks for clearing this out.
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