[NOTES/EM-07008]-Magnetic Vector Potential

The vector potential is introduced using the Maxwell's equation \(\nabla \times \vec{B}=0\) and the equation \( \nabla \times \vec{B} = \mu_0 \vec{j}\) is derived. The expression for the magnetic field is obtained as volume integral, the Biot Savart law, is derived. The expressions for the magnetic field for the surface current and the line current are given.

The equation \begin{equation} \nabla\cdot\vec{B}=0 \end{equation} holds in all situations. A theorem in vector calculus implies that there exists a vector field \(\vec{A}\), call the vector potential, such that \begin{equation} \vec{B}=\nabla\times\vec{A}. \end{equation} Given magnetic field, one cannot determine the potential uniquely. This is because \(\vec{A}\) and \(\vec{A}{'}\) defined by \begin{equation} \vec{A} \rightarrow\vec{A}{'} = \vec{A} + \nabla \Lambda, \end{equation} give rise to the same magnetic field. Here \(\Lambda\) is an arbitrary function of \(vec{r}\): \begin{equation} \nabla \times \vec{A}{'} = \nabla \times \vec{A} + \nabla \times( \nabla \Lambda ) = \nabla \times \vec{A}. \end{equation} Thus we have \begin{equation} \nabla \times \vec{A} =\vec{B} \text{ also } \nabla \times \vec{A}{'} =\vec{B}. \end{equation} Two different potentials \(\vec{A},\vec{A}{'}\) correspond to the same magnetic field. To make \(\vec{A}\) unique we may impose a condition on \(\vec{A}\), for example, \(\nabla\cdot\vec{A}=0.\) The a magnetic field will have a unique vector potential.\\ The transformation \begin{equation} \vec{A} \rightarrow\vec{A}{'} = \vec{A} + \nabla \Lambda, \end{equation} is called a gauge transformation. The change in vector potential under a gauge transformation does not the magnetic field and does not affect the force on charges and currents. Thus a gauge transformation does not have any physical observable effect. Thus observable quantities do not change under gauge transformation of vector potential. We say that they are gauge invariant.

Equation for vector potential

Substituting \(\vec{B}=\nabla \times \vec{A}\) in Maxwell's equation \begin{equation} \nabla \times \vec{B} = \mu_0 \vec{j}, \end{equation} gives \begin{eqnarray} \nabla\times(\nabla\times\vec{A})=\mu_0\vec{j}\\ \nabla(\nabla\cdot\vec{A})- \nabla^2\vec{A}= \mu_0\vec{j}. \end{eqnarray} We shall assume, the Coulomb gauge, \(\nabla\cdot\vec{A}=0\). \MkGBx{What is a gauge?}{em-agbx-07004}\\ Thus we get \begin{equation} \nabla^2\vec{A}=-\mu_0\vec{j}. \end{equation} These are three equations, one for each component of \(\vec{A}\): \begin{equation} \nabla^2A_x=-\mu_0 j_x, \quad \nabla^2A_y=-\mu_0 j_y, \quad \nabla^2A_z=-\mu_0 j_z. \end{equation} Thus the equation for each component of the vector potential is similar to the equation \begin{equation} \nabla^2 \phi= -\frac{\rho}{\epsilon_0} \end{equation} which has solution \begin{equation}\label{EQ13} \phi(\vec{r}) =\frac{1}{4\pi\epsilon_0}\iiint \frac{\rho(\vec{r}{'})}{|\vec{r}-\vec{r}{'}|}, d^3r{'}. \end{equation} 
Question 1 : Do you recognise  EQ13 ? Have you encountered similar equation in electromagnetic theory? If YES, what was its solution? Can you relate that solution to the one given below in EQ14 ? Connect with electrostatics!  Therefore we can write the solution for the vector also in the form \begin{equation} \vec{A} =\frac{\mu_0}{4\pi} \iiint \frac{\vec{j}(\vec{r}{'})}{|\vec{r}-\vec{r}{'}|}, d^3r{'}. \label{EQ14} \end{equation} From the above equation we can arrive at the Biot Savart law. In fact computation of \(\vec{B}\) as \(\nabla \times\vec{A}\) leads to \begin{equation}\label{EQ15} \vec{B}(\vec{r})=\frac{\mu_0}{4\pi}\iiint_V\frac{\vec{j}(\vec{r}{'})\times \vec{R}}{R^3}\,d^3r{'}, \quad \text{where}\quad \vec{R}=\vec{r}-\vec{r}{'}. \end{equation} The expression for the magnetic field, obtained above, of course satisfies \({\nabla}\cdot\vec{B}=0\). The \eqRef{EQ15}, can be written down for line and and surface currents. \begin{equation}\label{EQ16} \vec{A}(\vec{r}) = \frac{\mu_0}{4\pi}\int_\gamma\frac{I\overrightarrow{dl}}{R}, \qquad \vec{A}(\vec{r}) = \frac{\mu_0}{4\pi}\iint_S\frac{\vec{K}dS}{R} \end{equation} where \(I\) is the current in circuit \(\gamma\) and \(\vec{K}\) is the surface current density and \(\vec{R}=\vec{r}-\vec{r}{'}\). The line and surface integrals are over the wire and the area of surface carrying the current respectively. The integrations are over the variable \(\vec{r}{'}\) which denotes the position of line and surface elements \(\overrightarrow{dl}\), \(dS\)in respective integrals. The corresponding expressions for the magnetic field are \begin{equation}\label{EQ17} \vec{B}(\vec{r}) = \frac{\mu_0 I}{4\pi}\int_\gamma\frac{\overrightarrow{dl}{'}\times \vec{R}}{R^3}, \qquad \qquad \vec{B}(\vec{r}) = \frac{\mu_0}{4\pi}\iint_S\frac{\vec{K}(\vec{r}{'})\times \vec{R}}{R^3}dS{'} \end{equation} 

References

• Sec 14-2 The vector potential of known currents
• Sec 14-6 The vector potential of a circuit
• See also, Sec 14-3 A straight wire R. P. Feynman, Robert B. Leighton and Mathew Sands Lectures on Physics, vol-II, B.I. Publications (1964)