[NOTES/EM-07003]-Biot Savart Law from Maxwell's equations

Starting form Maxwell's equations for magnetostatics, vector potential is introduced and the Biot Savart Law is derived

1. Biot Savart Law from Maxwell's equations

For magnetostatics we have the Maxwell's equation $$ \vec{\nabla}\cdot\vec{B}=0~,~~{\nabla}\times\vec{B}=\mu_0\vec{J}\,. $$ The equation ${\nabla}\cdot\vec{B}=0$ implies that there exists a vector field $\vec{A}(x)$ such that $$ \bar{B}=\nabla\times\bar{A} $$ Given $\vec{B},\vec{A}$ is not unique. This can be seen as follows consider $\vec{A}$ and $\vec{A}^\prime=\vec{A}+{\nabla}\Gamma$ where $\Lambda(x)$ is an arbitrary function of $\vec{x}$. Then \begin{align*} \nabla\times\bar{A}^\prime = & \nabla\times A+\nabla\times (\nabla\Gamma)\\ = & \nabla\times\bar{A} \end{align*} Thus given magnetic field $\vec{B}$, there are solutions $\vec{A}^\prime$ related by ${\nabla}\Lambda$. To make $\vec{A}$ unique we have to supply a condition on $\vec{A}$, called \textit{gauge condition}. In magnetostatics it is convenient to choose $$ {\nabla}\cdot\bar{A} =0 $$ as it simplifies equation for $\vec{A}$. The equation $$ \nabla\times\bar{B}=\mu_0\bar{J} $$ leads to $$ \nabla\times(\nabla\times\vec{A})=\bar{\mu}_0\vec{J} \Rightarrow \nabla(\nabla\cdot\bar{A})=\nabla^2A=\mu_0\vec{J}~~\text{or}~~\nabla^2\vec{A} = -\mu_0\vec{J} $$ \begin{align*} \nabla^2A_1 = & -\mu_0J_1\\ \nabla^2A_2 = & -\mu_0J_2\\ \nabla^2A_3 = & -\mu_0J_3 \end{align*} Using Green function for Poisson equation the solution for the vector potential can be written down as \begin{align*} A_1 = &\frac{\mu_0}{4\pi} \int \frac{J_1(\bar{x}^\prime)}{|\vec{x}-\vec{x}^\prime|} ~d^3x^\prime\\ A_2 = &\frac{\mu_0}{4\pi} \int \frac{J_2(\bar{x}^\prime)}{|\vec{x}-\vec{x}^\prime|} ~d^3x^\prime\\ A_3 = &\frac{\mu_0}{4\pi} \int \frac{J_3(\bar{x}^\prime)}{|\vec{x}-\vec{x}^\prime|} ~d^3x^\prime\\ \vec{A} = & \frac{\mu_0}{4\pi} \int \frac{\vec{J}(\bar{x}^\prime)} {|\vec{x}-\vec{x}^\prime|}~d^3x^\prime \end{align*} Therefore the magentic field is given by \begin{align*} \vec{B}= & {\nabla}\times\vec{A} = \frac{\mu_0}{4\pi}~\nabla \times \int~\frac{J(\vec{x}^\prime)}{|\vec{x}-\vec{x}^\prime|}~d^3x^\prime. \end{align*} The first component of the magnetic field \begin{equation} \vec{B}=\nabla\times\vec{A}. \end{equation} is given by \begin{align*} B_1=&\frac{\partial}{\partial x_2}~A_3 - \frac{\partial}{\partial x_3}~A_2\\ =&\frac{\mu_0}{4\pi}\int\left[J_3(\vec{x}^\prime)\frac{\partial}{\partial x_2}~\frac{1}{|\vec{x}-\vec{x}^\prime|} - J_2~\frac{\partial}{\partial x_3}~\frac{1}{|\vec{x}-\vec{x}^\prime|}\right]d^3x\\ \frac{\partial}{\partial x_2}\,\frac{1}{|\bar{x}-\bar{x}^\prime|}=&\frac{(x_2-x_2^\prime)} {\sqrt{(x-x_1^\prime)^2+(x_2-x_2^\prime)^2 + (x_3-x_3^\prime)^2}},\\ =&\frac{(x_2-x_2^\prime)}{((x-x^\prime_1)^2+(x_2-x_2^\prime)^2+ (x_3-x_3^\prime)^2)}\\ =&\frac{x_2-x_2^\prime}{|\vec{x}-\vec{x}^\prime|^3}. \end{align*} Using the notation \(|\vec{R}=(\vec{x}-\vec{x}^\prime)|\), we get \begin{align*} \frac{\partial}{\partial x_2}\frac{1}{R} = & \frac{R_2}{R^3}, \end{align*} and therefore \begin{align*} \vec{B} = & \frac{\mu_0}{4\pi}\int \frac{\vec{J}\times\vec{R}}{R^3}~ d^3x^\prime\\ \vec{B} = & \frac{\mu_0}{4\pi} \int \frac{I\overrightarrow{dl}\times \vec{R}}{R^3},\qquad\qquad \mbox{\HighLight[MyGray]{For a thin wire~$\vec{J}dV=I\overrightarrow{dl}$}}. \end{align*} due to a small line element \begin{align*} d\vec{B} = &\frac{\mu_0}{4\pi} ~\frac{I\overrightarrow{dl}\times\vec{R}}{R^3}, \qquad \text{where } \vec{R} = \vec{x}-\vec{x}^\prime. \end{align*} This is the Biot-Savart Law for magnetic field of a current line element.