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I am reading through Introduction to Electrodynamics by David J. Griffiths and came across the following problem:

A steady current $I$ flows down a long cylindrical wire of radius $a$. Find the magnetic field, both inside and outside the wire if the current is distributed in such a way that $J$ is proportional to $s$, the distance from the axis.

If $J$ is proportional to the distance from the axis $r$, then we have:

$$\vec{J}(\vec{r})=kr\,\boldsymbol{\hat{z}}$$

We also have:

$$\iint_{\Sigma} \vec{J}(\vec{r})\cdot\:\mathrm{d}\vec{A}=I$$

We therefore have:

$$\int_{0}^{2\pi}\int_{0}^{a}kr^{2}\:\mathrm{d}r\:\mathrm{d}\theta=\frac{2\pi k a^{3}}{3}=I $$

And therefore we have:

$$\vec{J}(\vec{r})=\frac{3Ir}{2\pi a^{3}}\,\boldsymbol{\hat{z}}$$

Using Ampére's law we have:

$$2\pi r B = \mu_{0}I \implies \vec{B}(\vec{r})=\frac{\mu_{0} I}{2\pi r}\,\boldsymbol{\hat{\theta}}$$

If we take a Ampérian loop inside the cylinder, we have:

\begin{align} 2\pi r B &= \mu_{0} \iint_{\Sigma}\vec{J}(\vec{r})\cdot\:\mathrm{d}\vec{A} \\ &= \mu_{0}\int_{0}^{2\pi}\int_{0}^{r}\frac{3 I r'^{2}}{2\pi a^{3}}\:\mathrm{d}r'\:\mathrm{d}\theta \\ &= \frac{\mu_{0} I r^{3}}{a^{3}} \end{align}

And therefore we have:

$$\vec{B}(\vec{r})=\begin{cases}\frac{\mu_{0} I r^{2}}{2\pi a^{3}} \,\boldsymbol{\hat{\theta}} & r < a \\ \frac{\mu_{0} I}{2\pi r}\,\boldsymbol{\hat{\theta}} & r \geq a \end{cases}$$

Is this the correct magnitude and direction of the magnetic field?

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closed as off-topic by Rob Jeffries, Danu, ACuriousMind, Qmechanic Sep 9 '15 at 14:34

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Those answers are correct. A first check is to see if the units match. You can always check direction by the right hand rule. The part for outside the wire is the same as if the current were uniform, because the enclosed current is all that matters when you have enough symmetry for Ampére's Law. For the part inside the wire, check to see if the function makes sense: for a uniformly distributed current, the magnetic field grows linearly with the distance from the axis, so it makes sense that for this current it would grow like the square of the distance from the axis.

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  • $\begingroup$ Final check - continuity of the solution at the boundary $r=a$. Which complements the sanity check you did. $\endgroup$ – Floris Oct 7 '14 at 4:10

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