[NOTES/EM-07010]-Magnetic Field of a Current Distribution at a Large Distances

For a volume distribution of current an expression for magnetization density, or the magnetic moment per unit volume is obtained. An expression for the magnetic field at large distances, in terms of magnetization density, is derived. 

1. Vector potential

The equation \(\nabla\cdot\vec{B} =0\) leads to introduction of vector potential \(\vec{A}\) How?{em-agbx} such that \begin{equation} \vec{B} =\nabla \times \vec{A}. \end{equation} The gauge freedom in choosing the vector potential allows us to impose a gauge condition on the vector potential, all physical quantities being independent of the choice of gauge condition. Selecting the gauge condition as \(\nabla\cdot\vec{A}=0\), the Maxwell's equation \begin{equation}\label{EQ01} \nabla \times \vec{B} =\mu_0\vec{j}, \end{equation} gives the Poisson equation for the components of the vector potential. \begin{equation}\label{EQ02} \nabla^2 \vec{A}= -\mu_0\vec{j}. \end{equation} This equation has solution \begin{equation}\label{EQ03} \vec{A}(\vec{r}\,) =\frac{\mu_0}{4\pi} \iiint_V \frac{\vec{j}(\vec{r}{'})}{|\vec{r}- \vec{r}{'}|} \, d^3r{'} \end{equation}

2. Vector potential at large distances
For points far away from volume \(V\) where the current is present, i.e. for \(r> r{'}\) we use the expansion \begin{eqnarray} \frac{1}{|\vec{r}-\vec{r}{'}|} &=& \frac{1}{\sqrt{r^2-2 \vec{r}\cdot\vec{r}{'} + r^{{'}\,2}}}\\\nonumber &=& \frac{1}{r} \Big( 1-\frac{2\vec{r}\cdot\vec{r}{'}}{r^2} + \frac{r^{{'}\,2}}{r^2}\Big)^{-1/2}\qquad \mbox{\HighLight{What is the formula form $(1+x)^\alpha$?}}\\ &=& \frac{1}{r}\Big( 1+\frac{\vec{r}\cdot\vec{r}{'}}{r^2} +\cdots \Big) \end{eqnarray} Substituting this in \eqRef{EQ03} gives \begin{equation}\label{EQ08} \vec{A} = \frac{\mu_0}{4\pi r}\iiint_V \vec{j}(r{'})d^3r{'} + \frac{\mu_0}{4\pi r^3}\iiint_V \vec{j}(r{'})(\vec{r}\cdot\vec{r}{'})d^3r{'} +\cdots \end{equation} with similar expressions for \(A_y\) and \(A_z\). The expressions for the vector potential components will be rewritten in a convenient form by making use of equation of continuity. \begin{equation} \pp[\rho]{t} + \nabla\vec{j} = 0. \end{equation} For static case, the current and charge distributions do not change with time. Hence, we get \begin{equation}\label{EQ10} \nabla\cdot\vec{j} = 0. \end{equation} For two arbitrary scalar functions \(f,g\) we have \begin{eqnarray} \iiint_V (\nabla \vec{j}) f\, g\, d^3r =0 \\ \iint_S (\hat{n}\cdot\vec{j}) (f\,g)\, dS - \iiint_V \vec{j} \cdot \nabla(f\, g)\, d^3r =0 \end{eqnarray} where \(S\) is the surface enclosing the volume \(V\). The current is assumed to vanish outside and on the surface \(S\). Thus for arbitrary functions \(f,g\) we get \begin{eqnarray}\label{EQ13} \iiint_V f\, (\vec{j} \cdot \nabla g) + g\, (\vec{j} \cdot \nabla f) d^3r =0. \end{eqnarray} We will first use the identity \eqRef{EQ13} with \(f=1, g=\vec{r}\). This gives \begin{equation} \iiint _V \vec{j}(\vec{r}{'})\, d^3 r{'} =0. \end{equation} Next, in \eqRef{EQ13} we take \(f=\vec{a}\cdot\vec{r}\) and \(g=\vec{b}\cdot \vec{r}\) to get \begin{equation}\label{EQ15} \iiint_V \big\{(\vec{a}\cdot\vec{r})(\vec{j}\cdot\vec{b}) + (\vec{b}\cdot\vec{r})(\vec{j}\cdot\vec{a})\, \big\}d^3r =0. \end{equation} Therefore, we can write \begin{eqnarray}\nonumber \iiint_V (\vec{a}\cdot\vec{j})(\vec{b}\cdot\vec{r})\, d^3r &=& \iiint_V \big(\vec{a}\cdot\vec{j})(\vec{b}\cdot\vec{r}) - \frac{1}{2} \left(\iiint_V \big\{(\vec{a}\cdot\vec{j})(\vec{b}\cdot\vec{r}) + (\vec{a}\cdot\vec{r})(\vec{b}\cdot\vec{j})\big\} d^3r\right)\\ &=& \frac{1}{2} \iiint_V \big\{(\vec{a}\cdot\vec{j})(\vec{b}\cdot\vec{r}) - (\vec{a}\cdot\vec{r})(\vec{b}\cdot\vec{j})\big\} d^3r \\ &=& \frac{1}{2} \iiint_V \vec{b}\cdot\big((\vec{a}\cdot\vec{j})\vec{r}- \vec{j}(\vec{a}\cdot\vec{r} ) \big) \, d^3r \\ &=&\frac{1}{2} \iiint_V \vec{a}\cdot\big( \vec{b}\times(\vec{r}\times\vec{j})\, d^3r.\label{EQ18} \end{eqnarray} where the second term in the first step vanishes on account of \eqRef{EQ15}. In \eqRef{EQ18}, change the integration variable to \(\vec{r}{'}\) \begin{eqnarray} \iiint_V \big\{\vec{a}\cdot\vec{j}(\vec{r}{'})\big\}(\vec{b}\cdot\vec{r}{'})\, d^3r{'} &=& \frac{1}{2} \iiint_V \vec{a}\cdot\big\{ \vec{b}\times(\vec{r}{'}\times\vec{j}(\vec{r}{'})\big\}\, d^3r{'}. \end{eqnarray} Next we choose \(\vec{b} =\vec{r}\) to get \begin{eqnarray} \iiint_V \big\{\vec{a}\cdot\vec{j}(\vec{r}{'})\big\}(\vec{r}\cdot\vec{r}{'})\, d^3r{'} &=& \frac{1}{2} \iiint_V \vec{a}\cdot\big\{ \vec{r}\times(\vec{r}{'}\times\vec{j}(\vec{r}{'})\big\}\, d^3r{'}. \end{eqnarray} Since vector \(\vec{a}\) is arbitrary, we get \begin{eqnarray}\label{EQ21} \iiint_V \vec{j}(\vec{r}{'})(\vec{r}\cdot\vec{r}{'})\, d^3r{'} &=& \frac{1}{2} \iiint_V \big\{ \vec{r}\times(\vec{r}{'}\times\vec{j}(\vec{r}{'})\big\}\, d^3r{'}. \end{eqnarray} Using \eqRef{EQ10} and \eqRef{EQ21} in \EqRef{EQ08} gives \begin{equation} \boxed{\vec{A}(\vec{r}) = \frac{\mu_0}{4\pi r^3}\frac{1}{2} \iiint_V \vec{r}\times\big[\vec{r}{'} \times \vec{j}(\vec{r}{'})\big]\,d^3 r{'}.} \end{equation} The expression \begin{equation} \vec{M}(\vec{r}\,) = \frac{1}{2}\vec{r} \times \vec{j}(\vec{r}\,) \end{equation} is called magnetization density, or magnetic moment per unit volume. Its integral over the entire current distribution gives the magnetic moment \begin{equation} \vec{m} = \frac{1}{2}\iiint_V \vec{r}{'} \times \vec{j}(\vec{r}{'})\, d^3r{'} \end{equation} of the current distribution. Written in terms of the magnetic moment, the vector potential takes the form \begin{equation} \boxed{\vec{A}(\vec{r}\,) = \frac{\mu_0}{4\pi r^3}\big(\vec{r}\times\vec{m}\big)}. \end{equation}