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Fun times with wires

  1. Sep 29, 2011 #1
    1. The problem statement, all variables and given/known data
    The metal rod in the figure moves with speed v parallel to a straight wire that is carrying current I. The distance between the two wires is d, as shown in the figure. 34.CP81.jpg

    If the end of the rod closest to the wire is chosen to be at zero potential, what is the potential at the end of the rod furthest from the wire?
    Give your answer in terms of I, v, d, l, and constants u0 and pi.

    2. Relevant equations

    N/A

    3. The attempt at a solution

    I don't even know where to start with this question.
     
  2. jcsd
  3. Sep 29, 2011 #2
    consider some charge on the metal rod, now metals have electrons as conduction charges, but since conventional direction of the current is that of positive charges, consider positive charge, q. the direction of [itex]\vec{v}\times \vec{B}[/itex] at any point on the rod is into the paper. now according to lorentz force law,

    [tex]\vec{F}=q(\vec{v}\times \vec{B})[/tex]

    while talking about the potential difference we talk about unit charges, so let q = 1 C.
    now

    [tex]V_B-V_A=-\int_A^B \vec{E}\bullet\vec{dl}[/tex]

    since q is 1C, the above equation can be written as

    [tex]V_B-V_A=-\int_A^B \vec{F}\bullet\vec{dl}[/tex]

    can you work this now , going from nearest point to farthest
     
  4. Oct 1, 2011 #3
    the answer has to be in terms of ΔV=
    i'm stuck on the exact same question.
     
  5. Oct 1, 2011 #4

    rude man

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    I would go it another way:

    If an element of wire dL moves in a B field orthogonal to dL and with velocity v orthogonal to both B and dL, what is the induced emf?
     
  6. Oct 1, 2011 #5
    I don't know if this is going against the rules of the physics forum, but can some one please just post the answer, i've been doing the same problem over and over and getting no where and now it is due tomorrow. So if you know how to work it out, please can you just tell me.

    Thanks
     
  7. Oct 1, 2011 #6

    SammyS

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    I hope not, because that would be against the rules of these forums.

    From what I've seen this is only your second post in this thread. In the first, you only gave the problem and showed no work.

    Now after two there have been two posts with suggestions, you still haven't shown any attempt to work on the problem.

    Show us what you have tried & where you are stuck.
     
  8. Oct 2, 2011 #7
    alrighty,

    probably the due date of your HW is over, but with the hints you should solve this
    problem. Now with the direction of v and B, where do you think is the direction of the force
    F on unit charge q ?

    you have to do some efforts to answer the hints given here.........
     
  9. Oct 2, 2011 #8
    Hi!

    VB−VA=−∫F⃗∙dl⃗

    For a wire, B= μ0I/2(pi)r

    VB-VA=−∫F⃗∙dl⃗ = -∫μ0I/2(pi)rv.dl, integrating from d to (l+d)

    This gives me an answer: μ0I/2(pi)v log((l+d)/d)

    Am I heading in the right direction?
     
  10. Oct 2, 2011 #9
    starshine, yes , i think negative sign is missing..... V(d)=0 as given in the problem
    so V(l+d)-V(d) = -(what you got) ...... -------> delta V =-(what you got)
     
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