2
$\begingroup$

Let $V\subset\mathbb{R}^3$ be an infinitely high solid cylinder of radius $R$, with its axis coinciding with the $z$ axis, entirely enclosed by the cylinder's lateral surface. Then, for any constant $\mu_0\in\mathbb{R}$ and any point of coordinates $\boldsymbol{r}\notin \bar{V}$ external to the cylinder, $$\frac{\mu_0}{4\pi}\int_V\frac{I\mathbf{k}\times(\boldsymbol{r}-\boldsymbol{x})}{\pi R^2\|\boldsymbol{r}-\boldsymbol{x}\|^3}d^3x=\frac{\mu_0 I \mathbf{k}\times\boldsymbol{r}}{2\pi\|\mathbf{k}\times\boldsymbol{r}\|^2}$$where $\mathbf{k}$ is the unit vector defyning the direction of the $z$ axis, which is the same as$$\frac{1}{2\pi}\int_V\frac{\mathbf{k}\times(\boldsymbol{r}-\boldsymbol{x})}{ R^2\|\boldsymbol{r}-\boldsymbol{x}\|^3}d^3x=\frac{\mathbf{k}\times\boldsymbol{r}}{\|\mathbf{k}\times\boldsymbol{r}\|^2}\quad (1)$$The result is the formula for the magnetic field generated by a steady current parallel to $\mathbf{k}$, having intensity $I$ and flowing through the cylinder.

My textbook, Gettys's Physics, derives the formula by using Ampère's law in the form $\oint_{\partial^+ \Sigma}\left(\frac{\mu_0}{4\pi}\int_{\mathbb{R}^3}\frac{\boldsymbol{J}(\boldsymbol{x})\times(\boldsymbol{r}-\boldsymbol{x})}{\|\boldsymbol{r}-\boldsymbol{x}\|^3}d\mu_{\boldsymbol{x}}\right)\cdot d\boldsymbol{r}=\mu_0\int_\Sigma \boldsymbol{J}\cdot\boldsymbol{N}_e \,d\sigma$, but I would like to understand a way to calculate the integral without using that law, since I have never seen a proof of it for a discontinuous density of current such as $\boldsymbol{J}=\frac{I\chi_V }{\pi R^2}\mathbf{k}$.

How can we proof the identity $(1)$? I have also tried with cylindrical coordinates $(x,y,z)=(r\cos\theta,r\sin\theta,z)$ with $r\in [0,R]$, $\theta\in[0,2\pi)$, $z\in(-\infty,+\infty)$, but I have got serious problem to manipulate the integral I get. I $\infty$-ly thank any answerer...

$\endgroup$
  • 1
    $\begingroup$ These integrals usually become extremely difficult to do using everyday methods (such as substitution). Using physical intuition and spherical/cylindrical harmonics is the only viable way most of the time. $\endgroup$ – grdgfgr May 23 '16 at 19:59
  • 1
    $\begingroup$ @grdgfgr: In this case, physical intuition provides the order of integration: We should only expect a nice result for the infinite cylinder, so the $z$ integration comes first; then since we know from Ampère's law that each cylindrical shell separately yields the same result (proportional only to the density of current in it, which is proportional to $r$), the integration over $\theta$ must be possible independent of $r$ and should thus be next, and then only the trivial integration over the cylindrical shells along $r$ remains. $\endgroup$ – joriki May 23 '16 at 20:42
2
$\begingroup$

Fix a point outside the cylinder, without loss of generality at $\mathbf r=(a,0,0)$. By symmetry, the magnetic field at $\mathbf r$ has only a $y$ component, and since $k$ only has a $z$ component, the only relevant component of $\mathbf r-\mathbf x$ is the $x$ component. The $y$ component of the left-hand side is

\begin{align} \frac1{2\pi R^2}\int_V\frac{a-x}{\|\mathbf r-\mathbf x\|^3}\mathrm d^3x &= \frac1{2\pi R^2}\iiint_V\frac{a-r\cos\theta}{\left((a-r\cos\theta)^2+(r\sin\theta)^2+z^2\right)^{3/2}}\,\mathrm dz\,\mathrm d\theta\,r\mathrm dr \\ &= \frac1{\pi R^2}\int_0^R\int_0^{2\pi}\frac{a-r\cos\theta}{(a-r\cos\theta)^2+(r\sin\theta)^2}\,\mathrm d\theta\,r\mathrm dr \\ &= \frac1{\pi R^2}\int_0^R\int_0^{2\pi}\frac{a-r\cos\theta}{a^2-2ar\cos\theta+r^2}\,\mathrm d\theta\,r\mathrm dr \\ &= \frac a{\pi R^2}\int_0^{R/a}\int_0^{2\pi}\frac{1-\rho\cos\theta}{1-2\rho\cos\theta+\rho^2}\,\mathrm d\theta\,\rho\mathrm d\rho \\ &= \Re\frac a{\pi R^2}\int_0^{R/a}\int_0^{2\pi}\frac{1-\rho\mathrm e^{\mathrm i\theta}}{\left(1-\rho\mathrm e^{\mathrm i\theta}\right)\left(1-\rho\mathrm e^{-\mathrm i\theta}\right)}\,\mathrm d\theta\,\rho\mathrm d\rho \\ &= \Re\frac a{\pi R^2}\int_0^{R/a}\int_0^{2\pi}\frac1{1-\rho\mathrm e^{-\mathrm i\theta}}\,\mathrm d\theta\,\rho\mathrm d\rho \\ &= \Re\frac a{\pi R^2}\int_0^{R/a}\oint\frac1{1-\rho/z}\,\frac{\mathrm dz}{\mathrm iz}\,\rho\mathrm d\rho \\ &= \Re\frac a{\pi R^2}\int_0^{R/a}\frac1{\mathrm i}\oint\frac1{z-\rho}\,\mathrm dz\,\rho\mathrm d\rho \\ &= \Re\frac a{\pi R^2}\int_0^{R/a}2\pi\rho\,\mathrm d\rho \\ &= \frac1a\;, \end{align}

in agreement with the right-hand side. Note that the contour integral is proportional to the residue at the pole because $r\lt a$. A cylindrical shell of current outside the test point does not contribute to the magnetic field; in this case, $r\gt a$, the contour integral vanishes as it doesn't contain the pole.

My $\infty$ pleasure.

$\endgroup$
  • $\begingroup$ Thank you so much! A thing isn't clear to me: since my knowledge of complex analysis is almost null, I'm not sure what property is used to write the 5th row... $\endgroup$ – Self-teaching worker May 25 '16 at 13:13
  • 1
    $\begingroup$ @Self-teachingworker: This uses Euler's formula $\mathrm e^{\mathrm i\theta}=\cos\theta+\mathrm i\sin\theta$. The $\Re$ at the beginning takes the real part of the expression. The denominator is just rewritten; if you multiply it out and use $\mathrm e^{\mathrm i\theta}\mathrm e^{-\mathrm i\theta}=1$ you get the denominator from the line before. I added an imaginary term $-\rho\mathrm i\sin\theta$ to the numerator and wrote the $\Re$ to get rid of it again. Alternatively, the integral over the added term is zero by symmetry. $\endgroup$ – joriki May 25 '16 at 13:57
  • 1
    $\begingroup$ ...and then you used the residue theorem. Thank you very much! $\endgroup$ – Self-teaching worker May 25 '16 at 20:55
  • $\begingroup$ I've reflected more deeply on this integral: since the integral of the absolute values of all the components of $\frac{\mathbf{k}\times(\boldsymbol{r}-\boldsymbol{x})}{\| \boldsymbol{r}-\boldsymbol{x}\|^3}$ converge since they are $\le\frac{1}{\|\boldsymbol{r}-\boldsymbol{x}\|^2}$, whose integral converges (as shown here) provided that $\boldsymbol{x}\notin V$, [...] $\endgroup$ – Self-teaching worker Jun 28 '16 at 16:50
  • $\begingroup$ [...] the Lebesgue integral $\int_V\frac{\mathbf{k}\times(\boldsymbol{r}-\boldsymbol{x})}{\|\boldsymbol{r}- \boldsymbol{x}\|^3}d\mu_{\boldsymbol{x}}$ exists finite in the case that $\boldsymbol{x}\notin V$. Moreover if $\forall r\quad a>r$, the integral you wrote is always non-negative, therefore even if $\boldsymbol{x}$ is on the external surface of $V$, the $y$ component $\int_V\frac{a-x}{\|\mathbf r-\mathbf x\|^3}\mathrm d^3x$ converges as a limit as $R\to a$. [...] $\endgroup$ – Self-teaching worker Jun 28 '16 at 16:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.