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Griffiths Example 5.8

  1. Dec 4, 2008 #1
    I'm having a hard time understanding example 5.8 in Introduction to Electrodynamics by Griffiths. Why, exactly, is Ienc = KL? It makes sense intuitively, but I don't see how to get this result explicitly -- shouldn't the line integral be along the y-axis, and since [tex]\bar{K}[/tex]=K[tex]\hat{x}[/tex], shouldn't the dot product be equal to zero? If it's not along the y-axis, how is it that the integral can be from L to 0?

    If anybody could explicitly do the dot product and integral that finds the enclosed current in this example (5.8 Griffiths), that would be really helpful.

    Thanks for any info.
  2. jcsd
  3. Dec 4, 2008 #2


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    You don't take the dot product between [itex]\vec{K}[/itex] and [itex]\vec{dl}[/itex]. You take the dot product of the magnetic field with [itex]\vec{dl}[/itex].

    The magnetic field [itex]\vec{B}[/itex] points in the y-direction and is uniform (due to symmetry), so the integral of [itex]\vec{B} \cdot \vec{dl}[/itex] over the path of the Amperian loop simply gives you 2Bl (If you go counterclockwise around the loop; and -2Bl if you go clockwise)

    The current enclosed by the amperian loop is not [tex]\oint\vec{K} \cdot \vec{dl}[/tex]!

    It is instead given by the flux of current through the surface bounded by the Amperian loop. This surface has a normal in the x-direction (if you go counterclockwise around the loop when determining the integral of B dot dl) , so the flux of current is

    [tex]\int_{\mathcal{S}} \vec{K} \cdot \vec{da}=\int_{\mathcal{S}} (\delta(z)K)dydz=\int_0^l Kdy=Kl[/tex]

    Where the dirac delta is used since the surface current is zero when [itex]z\neq 0[/itex].

    Last edited: Dec 4, 2008
  4. Dec 6, 2008 #3
    Yes, that makes perfect sense. Thanks!
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