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 r}\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)

7 thoughts on “Magnetic field within a current-carrying wire

  1. Haris

    Could you please check your curl of the volume current. I believe it should have a minus sign.
    Also what happened to the little r when you take the curl?

    Reply
    1. gwrowe Post author

      If you mean equation 10 (it’s really helpful if you quote equation numbers!), which is the curl of the magnetization, not the volume current, it’s correct. My {r} variable is what Griffiths calls {s} and if you look up the formula for the curl in cylindrical coordinates (inside the front cover of Griffiths, for example), the curl reduces to (since the only non-zero component of {\mathbf{M}} is {M_{\phi}}):

      \displaystyle  \nabla\times\mathbf{M}=\frac{\chi_{m}I}{2\pi a^{2}}\left[\frac{1}{r}\frac{\partial}{\partial r}\left(r^{2}\right)\right]\hat{\mathbf{z}}=\frac{\chi_{m}I}{\pi a^{2}}\hat{\mathbf{z}}

      Reply

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