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

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

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

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

• 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.

• Ollie  On Wednesday, 11 February 2015 at 20:31

I believe the B field outside the wire (r>a) is actually (Muo*I)/(2pi*r)…

• growescience  On Thursday, 12 February 2015 at 09:07

Fixed now. Thanks.