[NOTES/EM-07009]-Magnetic Field of a Current Loop at Large Distances

The vector potential \(\vec{A}\) for static magnetic field satisfies \begin{equation}\label{EQ20} \nabla^2 \vec{A} = -\mu_0 \vec{j}. \end{equation} \FigHere The figure shows a current loop \(\gamma\) with a line element \(\overrightarrow{dl}\) at position \(\vec{r}{'}\). The field point \(P\) has the position vector \(\vec{r}\). The magnetic vector potential of the current loop is given by \begin{equation} \vec{A} =\frac{\mu_0 I}{\pi}\oint \frac{\overrightarrow{dl}}{|\vec{r}-\vec{r}{'}|}. \end{equation} \MkGBx{How do you get the above equation?}{em-agbx-07003} We expand the denominator \(|\vec{r}-\vec{r}{'}|\equiv R\) in powers of \(\frac{r{'}}{r}\) and for large \(r\) we shall keep first two terms. \begin{eqnarray}\nonumber R &=& \{|\vec{r}-\vec{r}{'}|\}^{-1/2}\\\nonumber &=&\{r^2 -2\vec{r}\cdot\vec{r}{'} + r^{{'}\,2}\}^{-1/2}\\\nonumber &=&\frac{1}{r}\left\{ 1- \frac{2\vec{r}\cdot\vec{r}{'}}{r^2} + \frac{r^{\,{'}\, 2}}{r^2}\right\}^{-1/2}\\ &=&\frac{1}{r}\left( 1+ \frac{1}{2}\times \frac{2\vec{r}\cdot\vec{r}{'}}{r^2} + O\Big(\frac{{r}^{\,{'}\,2}}{r^2}\Big)\right). \end{eqnarray} Thus the vector potential for large \(r\) takes the form \begin{equation} \vec{A} = \frac{\mu_0 I }{4\pi} \oint_\gamma \left(\frac{\overrightarrow{dl}}{r} + \frac{1}{2}\times \overrightarrow{dl} \frac{2\vec{r}\cdot\vec{r}{'}}{r^3} \right) + \cdots \end{equation} In first term in the bracket \(\frac{1}{r}\) can be pulled out of the integration sign and \(\oint \overrightarrow{dl}\) vanishes for a closed loop. Thus we get \begin{equation} \vec{A}=\frac{\mu_0}{4\pi r^3}\oint I \frac{1}{2} (\vec{r}\cdot\vec{r}{'})\overrightarrow{dl}. \end{equation} The vector potential of the current loop for large distances can be cast in the form \begin{equation} \vec{A} =\frac{\mu_0}{4\pi}\frac{\overrightarrow{m}\times\vec{r}}{r^3} \end{equation} where we have used the identity \begin{equation}\label{EQ06} \oint \frac{1}{2} (\vec{r}\cdot\vec{r}{'})\overrightarrow{dl}= \frac{1}{2} \oint \vec{r}{'}\times \vec{dr}{'}. \qquad \mbox{\HighLight{Proof given at the end }} \end{equation} and defined the expression in \eqRef{EQ06} as the magnetic moment \(\overrightarrow{m}\) of the current loop. A proof of this last step is given at the end. The quantity \(\overrightarrow{m}\) \begin{eqnarray} \overrightarrow{m} &=& \dfrac{I}{2}\displaystyle \oint \vec{r}\,{'}\times \vec{dr}{'} . \end{eqnarray} is called the magnetic moment of the loop. The magnetic field due to magnetic dipole is found to be \begin{equation}\label{EQ08} \vec{B}= \frac{\mu_0}{4\pi}\frac{\big[ \, 3(\overrightarrow{m}\cdot\hat{r})\,\hat{r}\big] -\overrightarrow{m}\,\big]}{r^3}. \end{equation} Here \(\hat{r} = \vec{r}/r\) is unit vector along \(\vec{r}\). \MkGBx{Prove \eqRef{EQ08}}{em-agbx-07001}. For a planar loop the expression \begin{equation}\label{EQ11} \frac{1}{2} \oint \vec{r}{'}\times \vec{dr}{'} \end{equation} equals the vector area \( \vec{A}\) of the loop, Thus the magnetic moment takes a particularly simple form \begin{eqnarray} \overrightarrow{m}&=& I \vec{A} \end{eqnarray} for a planar loop. Here \(\vec{A}\) is the vector area of the loop. Show that the area of a planar loop is given by \(|\vec{A}|\) where \(\vec{A}\) is given by \Eq.(10}. % For a planar current loop we will have \(\vec{j} d^3r \to I % \overrightarrow{dl}\). % \begin{equation} % \vec{m} = \frac{1}{2}\oint \vec{r}{'} \times \big(I \overrightarrow{dl}\big) % = I \vec{A} % \end{equation} % where\(\vec{A}\) is the vector area of the loop having magnitude equal to the % area of a loop and the direction is given by the right hand rule. % For a loop in the plane of paper the direction of area is out of the paper % ,\(\odot\), if the current flows anticlockwise. It is into the plane of % paper, \(\otimes\), if the current is clockwise. \subsubsection{Magnetic moment and angular momentum} For point charges it is easy to see that the magnetic moment is related to the angular momentum. Note that \begin{equation} \vec{j}(\vec{r})=q_k \vec{v}_k\delta(\vec{r}-\vec{r}_k). \end{equation} Therefore the magnetic moment is given by \begin{equation} \vec{m} = \sum_k \frac{q_k}{2M_k}\big\{ \vec{r}_k\times \vec{p}_k\big\} = \sum_k \frac{q_k}{2M_k}\vec{\ell}_k. \end{equation} Here the following notation has been used for \(k^\text{th}\) charged particle.\\ \(q_k=\)charge; \(M_k=\)mass; \(\vec{v}_k=\)velocity; \(\vec{p}_k =\) momentum; \(\vec{\ell}_k\) angular momentum. \subsubsection*{Proof of \eqRef{EQ06}:} Let \(\vec{e}\) be numerical vector. since we have \begin{equation} d [(\vec{e}\cdot \vec{r}{'})\vec{r}{'}] = [(\vec{e}\cdot d\vec{r}{'})\vec{r}{'} + (\vec{e}\cdot \vec{r}{'})d\vec{r}{'}] \end{equation} and \[\oint d[ (\vec{e}\cdot \vec{r}{'})\vec{r}{'}]=0,\] it follows that \begin{equation} \oint [ (\vec{e}\cdot d\vec{r}{'})\vec{r}{'} + (\vec{e}\cdot \vec{r}{'})d\vec{r}{'}] =0. \end{equation} Thus \begin{equation} \oint (\vec{e}\cdot d\vec{r}{'})\vec{r}{'} =- \oint (\vec{e}\cdot \vec{r}{'})d\vec{r}{'}. \end{equation} Therefore we get \begin{eqnarray} \oint (\vec{e}\cdot d\vec{r}{'})\vec{r}{'} &=& \frac{1}{2}\left(\oint (\vec{e}\cdot d\vec{r}{'})\vec{r}{'} - \oint (\vec{e}\cdot \vec{r}{'})d\vec{r}{'}.\right)\\ &=&\frac{1}{2} \oint \vec{e}\times( \vec{r}{'}\times d\vec{r}{'})\\ &=&\frac{1}{2} \vec{e}\times\oint ( \vec{r}{'}\times d\vec{r}{'}) \end{eqnarray} For a line integral \(\overrightarrow{dl}=d\vec{r}{'}\), replacing \(\vec{e}\) with \(\vec{r}\) we get the desired result \eqRef{EQ06}.

References

• {Jack} Sec 5.6 Magnetic Field of a Localized Current Distribution\\ J. D. Jackson, {|it Classical Electrodynamics}, John Wiley and Sons, New York (1998) ×