Magnetic field within a current-carrying wire

References: Griffiths, David J. (2007) Introduction to Electrodynamics, 3rd Edition; Prentice Hall – Problem 6.17.

Another example of using the auxiliary field to find the magnetic field. This time, we have a long wire of radius {a} carrying a uniform current {I}. The wire is made of a linear metal so that the magnetization is directly proportional to {\mathbf{H}}.

\displaystyle  \mathbf{M}=\chi_{m}\mathbf{H} \ \ \ \ \ (1)

where {\chi_{m}} is the magnetic susceptibility. Given this proportionality, the field is also directly proportional to {\mathbf{H}}:

\displaystyle  \mathbf{B}=\mu_{0}\left(\mathbf{H}+\mathbf{M}\right)=\mu_{0}\left(1+\chi_{m}\right)\mathbf{H}\equiv\mu\mathbf{H} \ \ \ \ \ (2)

where {\mu} is the magnetic permeability.

From the symmetry of the setup, {\mathbf{H}} is circumferential, so we can take a circular integration path to get

\displaystyle   \oint\mathbf{H}\cdot d\boldsymbol{\ell} \displaystyle  = \displaystyle  I_{f}\ \ \ \ \ (3)
\displaystyle  2\pi rH \displaystyle  = \displaystyle  \frac{r^{2}}{a^{2}}I\ \ \ \ \ (4)
\displaystyle  \mathbf{H} \displaystyle  = \displaystyle  \frac{rI}{2\pi a^{2}}\hat{\boldsymbol{\phi}} \ \ \ \ \ (5)

From this we get the field for {r<a}

\displaystyle  \mathbf{B}=\frac{\mu_{0}\left(1+\chi_{m}\right)rI}{2\pi a^{2}}\hat{\boldsymbol{\phi}} \ \ \ \ \ (6)

Outside the wire, {\chi_{m}=0} and the enclosed free current is just {I}, so for {r>a}

\displaystyle  \mathbf{B}=\frac{\mu_{0}I}{2\pi a}\hat{\boldsymbol{\phi}} \ \ \ \ \ (7)

The magnetization is

\displaystyle  \mathbf{M}=\frac{\chi_{m}rI}{2\pi a^{2}}\hat{\boldsymbol{\phi}} \ \ \ \ \ (8)

so the bound currents are

\displaystyle   \mathbf{J}_{b} \displaystyle  = \displaystyle  \nabla\times\mathbf{M}\ \ \ \ \ (9)
\displaystyle  \displaystyle  = \displaystyle  \frac{\chi_{m}I}{\pi a^{2}}\hat{\mathbf{z}}\ \ \ \ \ (10)
\displaystyle  \mathbf{K}_{b} \displaystyle  = \displaystyle  \mathbf{M}\times\hat{\mathbf{n}}\ \ \ \ \ (11)
\displaystyle  \displaystyle  = \displaystyle  -\frac{\chi_{m}I}{2\pi a}\hat{\mathbf{z}} \ \ \ \ \ (12)

The total bound current is

\displaystyle  \mathbf{I}_{b}=\pi a^{2}\mathbf{J}_{b}+2\pi a\mathbf{K}_{b}=0 \ \ \ \ \ (13)

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Comments

  • Anonymous  On Thursday, 8 May 2014 at 08:25

    i think you meant to say bound current where you said “so the free currents are”

    • growescience  On Friday, 9 May 2014 at 13:50

      Quite right; fixed now.

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