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Homework Help: Trouble understanding this hw problem

  1. Jul 19, 2005 #1


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    An electron is moving in a plane that also contains a long straight current-carrying wire. The electron is heading at a 45° angle toward the wire, with a speed of 3.8 106 m/s, when it is 43 cm away. The electron reaches only as close as 1 cm, before being repelled away, always moving in the same plane. What is the current in the wire?

    I could picture it well but i have no idea how to solve this problem. All i knwo is that its velocity should be constant because its in the plane perpendicular to the magnetic field and the force i think.. Can anyone tell me how to do it or give me any hints that would help me think better.
    Your help would be greatly appreciated!
  2. jcsd
  3. Jul 19, 2005 #2
    First, have you drawn a picture yet? Second, be careful in your choice of terms. Is it really the velocity that is constant? Review the force law for moving charges and think about what is required for uniform circular motion. I think these hints will change your thinking.
  4. Jul 19, 2005 #3


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    I can't see any easy way to do this. I'm not sure this will work out, but the only thing that comes to mind so far is to use the constant speed condition to find the force as a function of the distance from the wire in terms of the unknown current, and the known relationship between field strength and distance for a long straight wire. At any distance from the wire the electron's path will have a radius of curvature determined by that force and the speed (centripetal force). Perhaps the relationship between radius of curvature and coordinate derivatives combined with the constant velocity will lead you to the answer


    The parametric form of the radius (eq 4) seems to hold some promise. It looks to me like constant speed should lead to a constant numerator. Can you relate the acceleration components to the velocity components? If this leads you to an expression for the acceleration at any point (such as the point of closest approach) you should be able to then find the current.
    Last edited: Jul 19, 2005
  5. Jul 20, 2005 #4
    This was a tricksy problem.

    If we assume the current is lying along the y axis and headed
    in the y direction the B field from the wire is
    [tex] \vec{B} = \frac{-\mu_0I}{2 \pi x} \hat{z}[/tex]
    We can take the coordinates so the particle has an initial position (x_0, 0).

    The force on the particle is given by
    [tex] \vec{F} = q \dot{\vec{x}} \times \vec{B} = m \ddot{\vec{x}} [/tex]
    working that out gives two coupled ODE
    [tex] \ddot{x} = -\frac{c}{x} \dot{y} \; , \; \ddot{y} = \frac{c}{x} \dot{x} [/tex]
    [tex] c = \frac{q \mu_0 I}{2 \pi m} [/tex]

    Since the force is always perpendicular to the motion
    the energy is conserved. (Mathematically you get
    this by dividing the first by the second, cross multiplying
    denominators and recognizing the derivative.)
    [tex] \dot{x}^2 + \dot{y}^2 = const = v^2 [/tex]
    where v is the initial velocity.

    We can solve this for y-dot, and plug into the x-equation.
    [tex] \ddot{x} = \pm \frac{c}{x}\sqrt{v^2 - \dot{x}^2} [/tex]

    now we use the standard trick:
    [tex] \ddot{x} = \frac{d}{dt} \dot{x} =\frac{d \dot{x}}{dx} \frac{dx}{dt} = \dot{x} \frac{d \dot{x}}{dx} [/tex]

    This leaves us a seperable equation. We can integrate that
    using the initial conditions. To make things easier let's
    decompose the velocity into x and y components so
    [tex] \vec{v} = (-v \cos(\theta), v \sin(\theta)) [/tex]
    The unstandard use of minus signs allows theta = 45 deg
    for the original problem. (theta measures from the -x axis
    o the y axis.)

    then the integral gives us
    [tex] \pm (\sqrt{v^2 - \dot{x}^2} - v\sin(\theta_0)) = c \ln{\frac{x}{x_0}} [/tex]

    This should hold for all time t, specifically, it should hold at the
    closest point. but at the closest point x-dot is 0. and x is x_c.
    (that is, the particle is moving parallel to the wire).

    [tex] \pm v\left(1-\sin \theta_0 \right) = c \ln{\frac{x_c}{x_0}} [/tex]

    We can solve this for I.

    [tex] I = \pm \frac{2 \pi m v(1- \sin( \theta_0))}{q \mu_0 \ln{ \frac{x_c}{x_0}} } [/tex]

    We take the positive (by choice of the coordinate system I is in
    the +y direction I>0). That means we had to take the negative root
    clear back in the beginning.

    Now we're done.

    for the original problem putting in the mass/charge of an
    electron and all the pertinent info gives
    I = 8.43e-6 amps
    using that I and all the problem info, I had maple do a DEplot
    of the solution of the ODE system.
    The attachment shows the graph from t=0.18 to 0.20 seconds.

    Attached Files:

  6. Jul 20, 2005 #5
    oops. i didn't mean to do the homework problem
    for the op. (i got carried away.)
  7. Jul 22, 2005 #6


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    Well done. Did you explore a second solution? For the same initial velocity, if the current is in the other direction the electron's path will curve toward the wire. It will eventually get back to the problem you solved, but with a smaller "initial" distance from the wire when at 45 degrees again. A larger current should still result in the 1cm closest approach.

    I'm curious about the full trajectory of your solution. Initially I thought the path might be parabolic, but the possibility of a second solution makes me think otherwise. With a force that is always perpendicular to the velocity, diminshing as the distance from the wire increases, I don't see any possibility of multiple approaches to the wire, but I suspect that the y component of the velocity can retrograde one time for an approach, something along the lines of this diagram. Does something like this fit the solution?

    Attached Files:

    Last edited: Jul 22, 2005
  8. Jul 22, 2005 #7
    yeah, i think the solutions are more loopy than parabolic,
    which makes sense. it's easier just to play around with
    the DE system using a nice numerical solve/plot package.

    changing conditions from the original problem to
    vx_0 = -3*cos(Pi/4)
    vy_0 = 3*cos(Pi/4)
    x_0 = 0.5
    y_0 = 0

    I = 1e-4 gives the first graph (t=0..3)
    [inverting the equation for I = f(x_c) i plotted x_c given
    the I, it's the red vertical line.]

    and I = -1e-4 gives the second graph (t=0..1)
    [this is weird, the two solutions for x_c are plotted,
    but the negative solution gives nothing, and the positive
    solution gives the farthest point. i'm not really sure
    what's happened here. ]

    notice the loops are tighter for a current down the wire
    and the direction of motion changes.

    if we go with a positive current (which is the only
    one i trust.) and increase it. the solutions become
    more and more circle like. this makes sense because
    the field is better and better approximated by a const
    linear field.

    the third drawing is for i = 1amp. t = 0..0.001 (or something,
    i forgot to write the time down before i exported it.)

    Attached Files:

  9. Jul 23, 2005 #8


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    I don't have any tools handy to do numerical solutions. Apparently I was wrong about the electron not making multiple passes near the wire. It makes sense now that I think about it again. All that is required is that the radius of curvature be proportional to the distance from the wire. Your diagrams appear to have a smaller radius at the maximum distance than at some closer point, but I'm quite sure that is just a scaling issue. Any chance you could draw those so that the axes are to the same scale to better represent the shape of the trajectories?
    Last edited: Jul 23, 2005
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