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Homework Help: Griffiths 6.5

  1. Apr 29, 2008 #1
    [SOLVED] griffiths 6.5

    1. The problem statement, all variables and given/known data
    A uniform current density [itex]\mathbf{J} = J_0 \hat{\mathbf{z}}[/itex] fills a slab straddling the yz plane x=-a to x=+a. Find the magnetic field inside the slab.
    This is part of Griffiths Problem 6.5.

    2. Relevant equations

    [tex]\nabla \times \mathbf{B} = \mu_0 \mathbf{J}[/tex]
    [tex]\int \mathbf{B} \cdot d\mathbf{l} = \mu_0 I_{enc}[/tex]

    3. The attempt at a solution
    I first need to find the direction of B. What can I use to do that? The right-hand corkscrew rule does not seem to help here!
     
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  3. Apr 29, 2008 #2

    Ben Niehoff

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    Sure it does. Why do you think it doesn't?
     
  4. Apr 29, 2008 #3
    The right-hand corkscrew rule gives you the B-field around a wire. How do you use it to get the B-field of a slab? Why do you think it does help?
     
  5. Apr 29, 2008 #4

    Hootenanny

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    I would approach this question by first noting that,

    [tex]\underline{J}=\sigma\underline{E}[/tex]

    And then using the differential form of Faraday's law,

    [tex]\text{curl}\left(\underline{E}\right) = -\frac{\partial\underline{B}}{\partial t}[/tex]
     
  6. Apr 29, 2008 #5
    Wait, but that is not always true. What makes you think Ohm's law applies here?
     
  7. Apr 29, 2008 #6

    Hootenanny

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    Sorry, I was looking at the wrong question.

    The direction of the magnetic field can definitely be found using the RH 'corkscrew' rule, as you would for a wire. Take your right-hand and place your thumb in the direction of the current density, your curling fingers indicate the direction of the magnetic field.
     
  8. Apr 29, 2008 #7
    OK. But in the case of the wire the B-field is circumferential. Why is it not circumferential here? Obviously there will be some cancellation here...but how do we know that after all the cancellation is done the B-field will only be in the y-direction?
     
  9. Apr 29, 2008 #8
    Never mind. In Griffith's Example 5.8, he proved that the magnetic field of an infinite uniform surface current [itex]K\hat{\mathbf{x}}[/itex] flowing over the xy plane produces a magnetic field that is only in the y-direction. This is just a stack of those! We can replace x by z here due to symmetry I think.
     
    Last edited: Apr 29, 2008
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