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}

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}

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}
\displaystyle  2\pi rH \displaystyle  = \displaystyle  \frac{r^{2}}{a^{2}}I
\displaystyle  \mathbf{H} \displaystyle  = \displaystyle  \frac{rI}{2\pi a^{2}}\hat{\boldsymbol{\phi}}

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}}

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}}

The magnetization is

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

so the free currents are

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

The total bound current is

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

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