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Charge moving from a wire

  1. Jun 18, 2013 #1
    1. The problem statement, all variables and given/known data
    A charge q+ 4.5 x 10^-9 is located a distance of 7.0 mm to the right of a wire and is moving directly away from the wire with a velocity of v =3.0 x 10^4 m/s, as shown in the drawing (drawing shows charge moving to the right and current moving upwards). The wire carries a current I= 2.2 A. What is the force (magnitude and direction) felt by the charge due to the wire?


    2. Relevant equations

    B=F/ (|q|vsinθ)

    3. The attempt at a solution

    According to the right hand rule, the Force would be pointing downward.
     
  2. jcsd
  3. Jun 19, 2013 #2

    rude man

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    That is correct.

    So how about using your 'relevant equation'?
     
  4. Jun 19, 2013 #3
    B=F/ (|q|vsinθ)

    B= F / (4.5x10^-9)*(3.0x10^4)*(sin 90)

    I am not sure how to find the magnetic field. I would use the current I = 2.2 A? And is sin90 correct?
     
  5. Jun 19, 2013 #4

    rude man

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    Sin(90) is correct.
    How about using Ampere's law?
     
  6. Jun 19, 2013 #5

    TSny

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    Are you saying that the force would be in the opposite direction of the current in the wire? I don't think that's correct.
     
  7. Jun 19, 2013 #6

    rude man

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    The charge is moving in the +x direction and the current is flowing in the +y direction, so at x > 0 the B field is in the -z direction:

    v x B =+i x (-k) = +j rats! TSny is right, it flows same direction as the current. OP take note ...
     
  8. Jun 20, 2013 #7
    So, If I use ampere's law:

    ΔB||*Δl=μ0I

    What do I use for Δl? 7.0mm?
     
  9. Jun 20, 2013 #8

    rude man

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    You should look up ampere's law. No, it's not 7mm.
     
  10. Jun 20, 2013 #9
    That's the equation I have in my textbook.
     
  11. Jun 20, 2013 #10

    rude man

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    I don't think so.
    Ampere's law integrates around a closed path, not along a radius.
     
  12. Jun 20, 2013 #11
    I don't have radius in the equation, I'm confused.
     
  13. Jun 20, 2013 #12

    rude man

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    Yes you do. 7mm is the radius of a circle surrounding the wire. You're supposed to integrate around the circle.

    BTW that's the problem with your other post so I will not answer that one again until you're clear on this point.
     
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