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Electromagnetism help: Find magnetic flux through a loop

  1. Nov 7, 2015 #1
    1. The problem statement, all variables and given/known data
    Very large conductor with DC current is in vacuum. Find magnetic flux through a loop.
    Given parameters: [itex]I,a,\alpha[/itex]

    2. Relevant equations
    [itex]\Phi=\int_S B\mathrm dS[/itex] - basic equation for magnetic flux
    [itex]B=\frac{\mu_0I}{2\pi x}[/itex] - electromagnetic induction created by very long
    linear conductor at distance [itex]x[/itex]

    3. The attempt at a solution
    After assuming the orientation of the loop, we can find direction of electromagnetic induction
    by the right-hand rule. The flux through the loop is equal to the flux through some arbitrary shaped
    surface which lies on the loop (see attachment). I don't know how to derive equation for [itex]dS[/itex].

    Elementary flux through the infinitely small surface:
    [itex]d\Phi=BdS\cos(B,n)=BdS[/itex]

    In my books solution it says that [itex]d\Phi=\frac{\mu_0I}{2\pi x}\cdot 2z\cos\theta dx[/itex], where
    [itex]z=a\sin\theta, x=a(1-\cos\theta)[/itex]

    Flux through the loop is
    [tex]\frac{\mu_0Ia}{\pi}\int_\alpha^{\pi} {(1+\cos\theta)}\mathrm d\theta=\frac{\mu_0Ia}{\pi}(\pi-\alpha-\sin\alpha)[/tex]

    Could someone explain how to derive equation for [itex]dS[/itex] and how to set the limits of integration?
     

    Attached Files:

  2. jcsd
  3. Nov 7, 2015 #2

    Simon Bridge

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    Use your coordinate system to determine dS ... limits of integration are related.
    ie. If dS lies in the x-y plane of a rectangular coordinate system you use dS = dx.dy
     
  4. Nov 8, 2015 #3

    rude man

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    Since the center of your circle on the x axis is not given, the problem can't be solved.
     
  5. Nov 8, 2015 #4
    The radius of imaginary circle is [itex]a[/itex].
     
  6. Nov 8, 2015 #5

    rude man

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    I know. That doesnt define the center of the "imaginary circle", just its radius.
     
  7. Nov 9, 2015 #6

    Simon Bridge

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