WhatsApp Image 2025-03-31 at 17.55.44.jpeg

# Chapter 14-1 ## 14.1 The vector potential ### 14.1.1 Definition of Vector Potential $\qquad$Since $\nabla \cdot \mathbf{B}=0$ always, we can define a vector field $\mathbf{A}(\mathbf{r})$ such that $\qquad \mathbf{B}=\nabla \times \mathbf{A} \qquad (14.1)$ $\mathbf{A}$ is called the *vector p...

# Chapter 14-1 ## 14.1 The vector potential ### 14.1.1 Definition of Vector Potential $\qquad$Since $\nabla \cdot \mathbf{B}=0$ always, we can define a vector field $\mathbf{A}(\mathbf{r})$ such that $\qquad \mathbf{B}=\nabla \times \mathbf{A} \qquad (14.1)$ $\mathbf{A}$ is called the *vector potential*. $\qquad$Eq.(14.1) does not uniquely define $\mathbf{A}$. If we add to $\mathbf{A}$ the gradient of any scalar function $\lambda$, then $\mathbf{B}$ will be unchanged because the curl of a gradient vanishes. $\qquad \mathbf{A}^{\prime}=\mathbf{A}+\nabla \lambda \qquad (14.2)$ $\qquad \nabla \times \mathbf{A}^{\prime}=\nabla \times \mathbf{A}+\nabla \times \nabla \lambda=\nabla \times \mathbf{A}=\mathbf{B}$ $\qquad$Transformations of $\mathbf{A}$ of the form of Eq.(14.2) are called *gauge transformations*. We can exploit this *gauge invariance* to choose a convenient $\mathbf{A}$. $\qquad$From vector calculus we have $\qquad \nabla \times(\nabla \times \mathbf{A})=\nabla(\nabla \cdot \mathbf{A})-\nabla^{2} \mathbf{A} \qquad (14.3)$ $\qquad$Using $\mathbf{B}=\nabla \times \mathbf{A}$, we obtain $\qquad \nabla \times \mathbf{B}=\nabla \times(\nabla \times \mathbf{A})=\nabla(\nabla \cdot \mathbf{A})-\nabla^{2} \mathbf{A}$ $\qquad$Ampere's Law states $\qquad \nabla \times \mathbf{B}=\mu_{0} \mathbf{J}$ $\qquad \nabla(\nabla \cdot \mathbf{A})-\nabla^{2} \mathbf{A}=\mu_{0} \mathbf{J} \qquad (14.4)$ $\qquad$We can simplify Eq.(14.4) by choosing a gauge such that $\nabla \cdot \mathbf{A}=0$. This choice is called the *Coulomb gauge.* Then Eq.(14.4) becomes $\qquad \nabla^{2} \mathbf{A}=-\mu_{0} \mathbf{J} \qquad (14.5)$ $\qquad$Eq.(14.5) is three Poisson equations, one for each Cartesian component of $\mathbf{A}$. $\qquad \nabla^{2} A_{x}=-\mu_{0} J_{x}$ $\qquad \nabla^{2} A_{y}=-\mu_{0} J_{y}$ $\qquad \nabla^{2} A_{z}=-\mu_{0} J_{z}$ $\qquad$The solution to Poisson's equation $\nabla^{2} \phi=-\rho / \epsilon_{0}$ is $\phi(\mathbf{r})=\frac{1}{4 \pi \epsilon_{0}} \int \frac{\rho\left(\mathbf{r}^{\prime}\right)}{R} d^{3} r^{\prime}$ where $R=\left|\mathbf{r}-\mathbf{r}^{\prime}\right|$. Therefore, the solution to Eq.(14.5) is $\qquad \mathbf{A}(\mathbf{r})=\frac{\mu_{0}}{4 \pi} \int \frac{\mathbf{J}\left(\mathbf{r}^{\prime}\right)}{R} d^{3} r^{\prime} \qquad (14.6)$ $\qquad$where $\mathbf{r}^{\prime}$ is the point in the source and $\mathbf{r}$ is the field point. $\qquad$In magnetostatics, we have the continuity equation $\qquad \nabla \cdot \mathbf{J}=0$