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Magnetic Potential

  1. Feb 18, 2009 #1
    A long straight wire of radius R carries a unifrom current density [itex]\mathbf{J}[/itex] inside it.

    In the first part of the question I worked out that the magnetic field inside the wire was

    [itex]\mathbf{B}=\frac{\mu_0 I r}{2 \pi R^2} \mathbf{\hat{\phi}}[/itex]

    I am now asked to get the vector potential insde the wire and give the hint of taking the curl in cylindrical polars. So far I have:

    [itex]\mathbf{B}=\nabla \wedge \mathbf{A}[/itex]
    But [itex]\mathbf{A(r)}=\frac{\mu_0}{4 \pi} \int dV' \frac{\mathbf{J(r-r')}}{|\mathbf{r-r'}|}[/itex]
    so clearly it's parallel to [itex]\mathbf{J}[/itex] which is in the z direction so we conclude that
    [itex]A_r=A_{\phi}=0[/itex] and that [itex]A_z[/itex] is non zero.

    Taking the curl in cylindrical polars,

    [itex]\nabla \wedge \mathbf{A}=\left(\frac{1}{r} \frac{\partial{A_z}}{\partial{\phi}}-\frac{\partial{A_{\phi}}}{\partial{z}} \right) \mathbf{\hat{r}} + \left(\frac{\partial{A_r}}{\partial{z}}-\frac{\partial{A_z}}{\partial{r}} \right) \mathbf{\hat{\phi}} + \frac{1}{r} \left(\frac{\partial}{\partial{r}}(rA_{\phi})-\frac{\partial{A_r}}{\partial{\phi}} \right) \mathbf{\hat{z}}[/itex]
    which whittles down to:
    [itex]\nabla \wedge \mathbf{A}=\frac{1}{r} \frac{\partial{A_z}}{\partial{\phi}} \mathbf{\hat{r}} - \frac{\partial{A_z}}{\partial{r}} \mathbf{\hat{\phi}}[/itex]

    We have that this must be equal to [itex]\mathbf{B}[/itex] which is given above, so by comparing terms we get

    [itex]-\frac{\partial{A_z}}{\partial{r}}=\frac{\mu_0 I r}{2 \pi R^2} \Rightarrow -\int dA_z=\frac{\mu_0 I}{2 \pi R^2} \int r dr \Rightarrow A_z=-\frac{\mu_0 I}{4 \pi R^2}r^2 + const[/itex]
    and the otehr term gives:
    [itex]\frac{\partial{A_z}}{\partial{\phi}}=0 \Rightarrow A_z[/itex] is a constant, but in the line above, we found it to have radial dependence - something isn't quite right here, can anybody help me?
  2. jcsd
  3. Feb 19, 2009 #2


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    The 'const' in this equation doesn't have to be a constant in all variables, it just can't have any [itex]r[/itex] dependence (otherwise the partial derivative w.r.t. [itex]r[/itex] of that "constant" term would be non-zero) It can however depend on [itex]\phi[/itex] and [itex]z[/itex] (For example, [itex]\frac{\partial}{\partial r} 5z^2\cos\phi=0[/itex] )

    [tex]\implies A_z=-\frac{\mu_0 I}{4 \pi R^2}r^2+ g(\phi,z)[/tex]

    Where [itex]g[/itex] is some unknown function of [itex]\phi[/itex] and [itex]z[/itex]

    Again, your constant need not be a constant, it just can't depend on [itex]\phi[/itex]

    [tex]\implies A_z=h(r,z)[/tex]

    Put the two conditions together and you find that [tex]A_z=-\frac{\mu_0 I}{4 \pi R^2}r^2 +f(z)[/tex] satisfies both equations simultaneously for any function [itex]f(z)[/itex]....You are free to choose any [itex]f(z)[/itex] you like; as is typical since the vector potential is defined by a first order differential (the curl in this case) and so is only unique up to a 'constant' (in this case, your constant can depend on [itex]z[/itex] without affecting B)
  4. Feb 19, 2009 #3
    ok. thanks.

    next i'm asked to show explicitly that this satisfies Poisson's equation, i.e.

    [itex]\nabla^2 \mathbf{A}=-\mu_0 \mathbf{J}[/itex]

    and also that [itex]\nabla \cdot \mathbf{A}=0[/itex].

    For the first part I have:
    [itex]\nabla^2 \mathbf{A}=\nabla(\nabla \cdot \mathbf{A})-\nabla \wedge \nabla \wedge \mathbf{A}[/itex] (taking the Laplacian of a vector.

    then it makes more sense to show the divergence is 0 now to eliminate that term from the expansion of the laplacian.

    [itex]\nabla \cdot \mathbf{A}=\frac{\partial{A_z}}{\partial{z}}=\frac{\partial{f(z)}}{\partial{z}} \neq 0 \forall f(z) \neq 0[/itex]
    bit confused here???

    however assuming that works, we get [itex]\nabla^2 \mathbf{A}=-\nabla \wedge (\nabla \wedge \mathbf{A})=-\nabla \wedge \mathbf{B}[/itex]

    [itex]\nabla \wedge \mathbf{B}=-\frac{\partial{B_{\phi}}}{\partial{z}}\mathbf{\hat{r}} + \frac{1}{r} \frac{\partial}{\partial{r}} (r B_{\phi}) \mathbf{\hat{z}}[/itex] as [itex]B_r=B_z=0[/itex]

    now [itex]\frac{\partial{B_{\phi}}}{\partial{z}}=0[/itex] as [itex]\mathbf{B}=\frac{\mu_0 I r}{2 \pi R^2} \mathbf{\hat{\phi}}[/itex]

    so [itex]\nabla \wedge \mathbf{B}=\frac{1}{r} \frac{\partial}{\partial{r}}(\frac{\mu_0 I r^2}{2 \pi R^2}) \mathbf{\hat{z}}=\frac{\mu_0 I}{\pi R^2}\mathbf{\hat{z}}[/itex]

    [itex]\Rightarrow\nabla^2 \mathbf{A}=-\mu_0 \frac{I}{\pi R^2}\mathbf{\hat{z}}[/itex]

    now my trouble is explaining why [itex]\mathbf{J}=\frac{I}{\pi R} \mathbf{\hat{z}}[/itex]. Clearly the direction is fine and the untis seem ok as we have amperes/metre^2 - but there must be some definition i'm meant to use that i'm just missing in my notes???
    Last edited: Feb 19, 2009
  5. Feb 19, 2009 #4


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    Requiring [tex]\vec{\nabla}\cdot\vec{A}=0[/tex]

    corresponds to a specific gauge choice (it removes some of the freedom you normally have in chossing your 'constants'); so if you want to choose [tex]\vec{A}[/tex] in a way that makes this true, just choose an [itex]f(z)[/itex] with zero divergence (the simplest choice is just [itex]f(z)=0[/itex])

    Well, what is the total current passing through a cross-section of the wire if the volume current [tex]\vec{J}[/tex] is uniform and runs in the z-direction?....since the total current is required to be [itex]I[/itex] equate the two expressions and solve for [itex]J[/itex].
  6. Feb 20, 2009 #5
    [itex]I=\int_S \mathbf{J} \cdot \mathbf{dA} \Rightarrow I=|\mathbf{J}| \pi R^2 \Rightarrow |\mathbf{J}|=\frac{I}{\pi R^2}[/itex]

    cheers m8.
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