I understand well enough how to calculate the radial and tangential components in spherical coordinates at a point due to a magnetic dipole field using the magnetic potential gradient ($\overrightarrow{B}=\nabla\times \overrightarrow{W}$). Is it possible to calculate radial ($B_r=\frac{2\mu_om\cos\theta}{4\pi r^3}$) and tangential ($B_\theta=\frac{\mu_om\sin\theta}{4\pi r^3}$) components in spherical coordinates without using the magnetic potential? I.e. I would like to know if it is possible to demonstrate these components in spherical coordinates starting out from the magnetic field due to a monopole ($B=\frac{\mu_oQ}{4\pi r^2}$).

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    $\begingroup$ Yes. Place two monopoles with magnetic charge $Q$ on the $z$ axis at distance $a/2$ above and below the origin. Compute their combined field in Cartesian coordinates. Take the limit as the separation $a$ goes to zero and $Q$ goes to infinity but their product stays a constant, $m=Qa$, the magnetic dipole moment. (In the process, only terms that are first order in $a$ will survive.) Transform the resulting field back into spherical coordinates. $\endgroup$ – G. Smith May 20 '19 at 17:12
  • $\begingroup$ @G.Smith That should be an answer, not a comment. $\endgroup$ – rob May 20 '19 at 17:43

$\def\m{\mu} \def\p{\pi} \def\th{\theta} \def\vm{\mathbf{m}} \def\vr{\mathbf{r}} \def\vs{\mathbf{s}} \def\VB{\mathbf{B}} \def\ur{\mathbf{\hat{r}}} \def\uz{\mathbf{\hat{z}}} \def\uth{\boldsymbol{\hat{\th}}} \def\rd{\mathrm{d}} \def\o{\cdot}$If we assume $$\VB = \frac{\m_0 g}{4\p r^3}\vr$$ for a monopole then the dipole field will be given by $$\VB_\rd = \frac{\m_0 g}{4\p r_+^3}\vr_+ - \frac{\m_0 g}{4\p r_-^3}\vr_-,$$ where $\vr_\pm = \vr\mp\vs/2$, $\vr$ points from the center of the dipole to the point of interest, $\vs$ points from the negative to the positive charge, and where $s/r\ll 1$. It is a straightforward exercise to expand in small $s$ with the result $$\VB_\rd = \frac{\m_0}{4\p r^3}(3(\vm\cdot\ur)\ur-\vm),$$ where $\vm = g\vs$ is the dipole moment. Letting $\vm = m\uz$ and noting that \begin{align*} (3(\vm\cdot\ur)\ur-\vm)\cdot\ur &= 2m\cos\th\\ (3(\vm\cdot\ur)\ur-\vm)\cdot\uth &= m\sin\th \end{align*} we find $$\VB_\rd = \frac{\m_0 m\cos\th}{2\p r^3}\ur + \frac{\m_0 m\sin\th}{4\p r^3}\uth$$ as claimed.


Some important steps for the expansion referred to above: \begin{align*} \frac{1}{r_+^3} &= (\vr_+\o\vr_+)^{-3/2} \\ &= ((\vr-\vs/2)\o(\vr-\vs/2))^{-3/2} \\ &= (r^2-\vr\o\vs+s^4/4)^{-3/2} \\ &= \frac{1}{r^3}(1-\ur\o\vs/r+s^4/(4r^2))^{-3/2} \\ &= \frac{1}{r^3}(1-(\ur\o\vs/r-s^4/(4r^2))^{-3/2} \\ &\approx \frac{1}{r^3}(1+3\ur\o\vs/(2r)). \end{align*} Likewise $$\frac{1}{r_-^3} \approx \frac{1}{r^3}(1-3\ur\o\vs/(2r)).$$ Here we use the generalized binomial theorem, $(1+x)^s \approx 1+s x$ (also known as the Taylor series of $(1+x)^s$ about $x=0$). Use these approximations to derive the final result.

  • $\begingroup$ $\def\vr{\mathbf{r}} \def\vs{\mathbf{s}} \def\ur{\mathbf{\hat{r}}}$@Rodrigues: I am glad to help. Note that $\ur/r^2 = \vr/r^3$. We define $\vs$ to point from the negative to the positive charge. The vector $\vr$ points from the center of the dipole to the point of interest. The vector pointing from the positive charge to the point of interest is given by $\vr_+ = \vr-\vs/2$ since $\vs/2+\vr_+=\vr$. Try sketching the situation if you are not convinced. Let me know if this doesn't clear things up. $\endgroup$ – user26872 May 29 '19 at 1:32
  • $\begingroup$ can you please expand the "straightforward" part for a non physicist? I'm struggling with that $\endgroup$ – Rodrigues May 29 '19 at 1:33
  • $\begingroup$ @Rodrigues: Do you know Taylor series? $\endgroup$ – user26872 May 29 '19 at 1:36
  • $\begingroup$ I'll learn if it is the only way. Cant it be solved with integrals? $\endgroup$ – Rodrigues May 29 '19 at 1:45
  • $\begingroup$ @Rodrigues: At some point a series expansion will need to be made. I have added something above. $\endgroup$ – user26872 May 29 '19 at 1:58

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