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Induced EMF on a moving rod next to a line of current.

  1. Apr 23, 2012 #1
    1. The problem statement, all variables and given/known data

    I'm currently reviewing some of the old E&M material that I haven't seen in a while and have gotten stuck on this problem. I'm sure I'm just making a simple mistake but I can't seem to make any headway.

    Here's the problem: Figure 35-32 shows a copper rod moving with velocity v parallel to a long straight wire carrying a current i. Calculate the induced emf in the rod, assuming v = 5.0m/s, i = 100A, a = 1cm and b = 20cm. Answer: 3.0X10-4V


    3. The attempt at a solution

    The magnetic field of the wire a distance y away is [tex]B = \frac{\mu _{0}i}{2\pi y}[/tex]. Thus the total magnetic field across the length of the rod is [tex]B = \frac{\mu_{0}i}{2\pi}\int_{a}^{b}\frac{dy}{y} = \frac{\mu_{0}i}{2\pi}ln\frac{b}{a}[/tex]. Now the induced emf is [tex]\varepsilon = -\frac{d\Phi _{B}}{dt}=-B\frac{dA}{dt}=-B(b-a)\frac{dx}{dt}=-B(b-a)v[/tex]. Then the total equation is [tex]\varepsilon = \frac{-\mu_{0}iv}{2\pi}(b-a)ln\frac{b}{a}[/tex], the only problem being that this is apparently not the correct answer.

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  2. jcsd
  3. Apr 23, 2012 #2

    tiny-tim

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    hi grindfreak! :smile:
    i don't understand this at all :confused:

    yes, the area swept out in time t is A = (b-a)vt

    but the magnetic flux through that area is ∫ab B(y)vt dy,

    ie your previous integral times vt

    how did you get that (b-a) factor?
    whatever is "total magnetic field"?? :confused:

    if you mean the flux, then say so, and don't call it "B" !

    your strange nomenclature seems to have confused you
     
  4. Apr 23, 2012 #3
    Yes I think your first integral is probably the right way to go with it, the b-a factor is the length of the rod, so that (b-a)*x is the area swept out by the rod in time t. As for total magnetic field, I really should have said magnetic field on the rod by the line of current (sorry if this sounds a bit poorly worded, I've been working all day).
     
  5. Apr 23, 2012 #4

    tiny-tim

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    sorry, but you're not making much sense

    get some sleep :zzz:, and it'll be clearer in the morning :smile:
     
  6. Apr 23, 2012 #5
    I checked my equation again and you're correct, I don't know why I used BA for the magnetic flux.
     
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