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Magnetostatics - Magnetic field of a nonuniform current slab

  1. Nov 28, 2017 #1
    1. The problem statement, all variables and given/known data
    A thick slab in the region [itex] 0 \leq z \leq a [/itex], and infinite in [itex] xy [/itex] plane carries a current density [itex] \vec{J} = Jz\hat{x} [/itex]. Find the magnetic field as a function of [itex]z[/itex], both inside and outside the slab.

    2. Relevant equations
    Ampere's Law: [tex]\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{\text{enc}}=\mu_0\int_S\vec{J}\cdot d\vec{a}[/tex]
    Biot-Savart Law: [tex]\vec{B}(\vec{r}) = \frac{\mu_0}{4\pi}I\int_C d\vec{l}'\times\frac{\vec{r}-\vec{r}'}{\left|\vec{r}-\vec{r}'\right|^3}[/tex]

    3. The attempt at a solution
    My first approach was to start with the Ampere's Law. If the current density was uniform ([itex]\vec{J}=J\hat{x}[/itex]), by symmetry we would have [itex]B=0[/itex] at [itex]z=\frac{a}{2}[/itex] and find the magnetic field:
    [tex]Bl=\mu_0l(z-a/2)J \quad \Rightarrow \quad \vec{B_{\text{in}}} = -\mu_0J(z-a/2)\hat{y}[/tex]
    [itex]\dots[/itex]
    In this case though, with a non-uniform current density, I can't see any symmetry or a way to find a proper Amperian loop to use. Am I missing something here?

    Next I thought about using Biot-Savart Law,
    [tex]\vec{B}(\vec{r})\stackrel{?}{=}\frac{\mu_0}{4\pi}\int_V \vec{J}d\tau'\times\frac{\vec{r}-\vec{r}'}{\left|\vec{r}-\vec{r}'\right|^3}[/tex]
    which I think is a bit of a ...stretch, since [itex]V[/itex] is infinite.
     
  2. jcsd
  3. Nov 28, 2017 #2

    Charles Link

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    Suggestion is to use Ampere's law and find ## d \vec{B} ## for each wire of infinite length (as a function of z), that is located at height ## z=z' ##. (Each ## d \vec{B} ## will have x and y components, but no z component). ## \\ ##The current carried by each ## dz ## will be ## dI=Jz \, dz ##. Incidentally, the current is a surface current, so that it is a current density per unit length, and they would have been better to say that surface current per unit length ## \vec{K}=J \, z \, \hat{x} ## , where ## J ## is a constant with units of current/unit area.
     
  4. Dec 2, 2017 #3

    rude man

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    I suggest considering a thin sheet within the slab, at z=z. Make the x extent of the slab infinite but restrict the y direction to a length = 2L, making L arbitrarily large but finite.

    Now consider an amperian loop extending parallel to the sheet in the y direction & wrapping around x = +/- L above and below the sheet. Can you assume ∫B⋅dl = - 2L J dz along the top and 2L J dz along the bottom of the sheet, ignoring ∫B⋅dl along the z direction above to below the sheet at x= + and -L?

    Then sum all the thin sheets' B fields by integration.
     
  5. Dec 2, 2017 #4

    rude man

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    The current is traveling along the x direction so no x component of dB.
     
  6. Dec 2, 2017 #5

    Charles Link

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    I got the geometry momentarily confused, but otherwise my comments are applicable.
     
  7. Dec 3, 2017 #6

    rude man

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    I should have assumed a typo anyway.

    I agree that J should have been written J = Jz z i
    where Jz = constant = dJ/dz and J is the usual current density in amp/m2.

    Other than that I'm not sure what your "wire" is. What is its cross-section?

    Not trying to be difficult, just trying to help the OP.
     
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