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Question regarding current & magnetic field/force

  1. Oct 28, 2015 #1
    1. The problem statement, all variables and given/known data
    An electron moves in a straight line at a speed of [itex]6.0 \times 10^7[/itex]m/s. Calculate the magnitude and direction of the magnetic field at a position 0.005m behind the electron and 0.015m below its line of motion.

    2. Relevant equations
    [itex]F=qv \times B[/itex]

    3. The attempt at a solution
    image.jpg

    I'm not really even sure if I drew the scenario correctly.. I figured that since there is a flow/movement of an electron, that it will current some sort of current in that direction (well technically, in the opposite direction to that of the electron's velocity vector). So then I modeled the field as if there is a current in a wire. The way I answered the questions (which need checking...) was by using the diagram that I drew..

    a). I said that [itex]F=qvB sin(\theta)[/itex] but I'm not sure what the angle is.. If the electron produces a magnetic field, won't it extend almost forever to the left? So that would mean that the angle is 180 between v and B?

    b). [itex]B=\frac{\mu_0 I}{2 \pi r}[/itex]

    [itex]=\frac{(4\pi \times 10^{-7})I}{2\pi (0.015)} =1.33 \times 10^{-5}I[/itex]



    It's all probably wrong.. If someone could give a detailed explanation as to what's going on and what how I may proceed, that would be greatly appreciated. Thanks!
     
  2. jcsd
  3. Oct 28, 2015 #2

    gneill

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    Staff: Mentor

    In this instance you're not looking for the force on a moving charge in a magnetic field, nor the field that produces a given force on a moving charge. There are no forces involved.

    The "other" handy law dealing with magnetic fields is the Biot-Savart law which tells you the contribution of a small current element (moving charges!) to the magnetic field at some location away from the current element.

    With a little "physicist's calculus" manipulation of differential elements you should be able to convince yourself that the ##\vec{I}~dL## in that law can be converted to ##dq~\vec{v}##.
     
  4. Oct 28, 2015 #3
    Okay, that makes sense.

    So how does that relationship allow me to obtain the magnetic fields?
     
  5. Oct 28, 2015 #4

    gneill

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    Staff: Mentor

    Because you have a charge moving with some velocity and the Biot-Savart rule will then tell you the magnetic field it produces at a given location... which is what the question is asking for.
     
  6. Oct 28, 2015 #5
    Okay, so if I understand what you're saying, then that means that

    [tex]B=\frac{\mu_0}{4\pi}\int \frac{\vec{v}|r| \sin(\theta)}{|r|^3}dq= (\frac{\mu_0}{4\pi})\frac{v \sin(\theta)}{|r|^2}q[/tex]
     
  7. Oct 28, 2015 #6

    gneill

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    Staff: Mentor

    I think you want the magnitude of the velocity vector in your first integration, but the result looks good for giving you the magnitude of the field.

    You could go with the vector version and get a vector result. The "dB" in the Biot-Savart law is a vector element, and integrating over the dq leaves everything else as a constant if you're dealing with a point charge. That would leave a ##\vec{v} \times \vec{r}## cross product in the result:

    $$\vec{B} = \frac{\mu_o}{4 \pi} q \frac{\vec{v} \times \vec{r}}{|r|^3}$$
     
  8. Oct 28, 2015 #7
    Okay! And what are the limits for the integral? Are they in terms of r or in terms of q since I integrated dq? Or do I not need limits?
     
  9. Oct 28, 2015 #8

    gneill

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    Staff: Mentor

    No limits required. It's an indefinite integral in this case. On the left hand side dB becomes B (vectors) and on the right the dq becomes q. Everything else is a constant.
     
  10. Oct 29, 2015 #9

    rude man

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    Homework Helper
    Gold Member

    There really is no integral, the Biot-Savat law is used in its differential form:
    dB = (μ0/4π) i dl x r / |r|2
    with i dl replaced by q v. This is valid since v dt = dl so i dl becomes (dq/dt)(v dt) = v dq = qv. The electron is here approximated as a differential amount of charge.
    I think this is essentially what gneill was saying but I thought I'd paraphrase it anyway.
     
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