Magnetic Vector Potential

We saw in Equation (2.16) that

$$\frac{{\bf r}- {\bf r}' } {\vert{\bf r} - {\bf r}'\vert^3} = -\nabla\!\left(\frac{1}{\vert{\bf r} - {\bf r'}\vert}\right).$$ (2.249)

This equation can be combined with the generalized Biot-Savart law, (2.248), to give

$${\bf B}({\bf r}) = \frac{\mu_0}{4\pi}\int\nabla\!\left(\frac{1}{\vert{\bf r} - {\bf r'}\vert}\right)\times {\bf j}({\bf r}')\,dV'.$$ (2.250)

It follows that

$${\bf B} = \nabla\times {\bf A},$$ (2.251)

where

$${\bf A}({\bf r})= \frac{\mu_0}{4\pi}\int\frac{{\bf j}({\bf r}')}{\vert{\bf r}-{\bf r}'\vert}\,dV'.$$ (2.252)

(See Section A.24.) Here, the vector field ${\bf A}({\bf r})$ is known as the magnetic vector potential.

It is possible to prove that the magnetic vector potential defined in the previous equation is a divergence-free field. Note that

$$\frac{\partial}{\partial x}\! \left(\frac{1}{\vert{\bf r} - {\bf ... ...c{\partial}{\partial x'} \!\left(\frac{1}{\vert{\bf r} - {\bf r}'\vert}\right),$$ (2.253)

which implies that

$$\nabla\left( \frac{1}{\vert{\bf r} - {\bf r}'\vert}\right) = - \nabla' \left( \frac{1}{\vert{\bf r} - {\bf r}'\vert}\right),$$ (2.254)

where $\nabla'$ is the operator $(\partial/\partial x', \partial/\partial y', \partial/\partial z')$. (See Section A.19.) Taking the divergence of Equation (2.252), and making use of the previous relation, we obtain

$$\nabla\cdot{\bf A} = \frac{\mu_0}{4\pi} \int {\bf j}({\bf r'})\cd... ...\bf r'})\cdot \nabla'\left( \frac{1}{\vert{\bf r} - {\bf r}'\vert}\right)\,dV'.$$ (2.255)

Now,

$$\int_{-\infty}^{\infty} g \,\frac{\partial f}{\partial x}\,dx = \... ...infty}^{\infty} - \int_{-\infty}^{\infty} f\,\frac{\partial g}{\partial x}\,dx.$$ (2.256)

However, if $g\,f\rightarrow 0$ as $x\rightarrow\pm\infty$ then we can neglect the first term on the right-hand side of the previous equation, and write

$$\int_{-\infty}^{\infty} g\, \frac{\partial f}{\partial x}\,dx = - \int_{-\infty}^{\infty} f\, \frac{\partial g}{\partial x}\,dx.$$ (2.257)

A simple generalization of this result yields

$$\int {\bf g}\cdot\nabla f\,dV = - \int f\,\nabla\cdot {\bf g}\,dV,$$ (2.258)

provided that $g_x\,f \rightarrow 0$ as $\vert{\bf r}\vert \rightarrow\infty$, etc cetera. Thus, Equation (2.255) yields

$$\nabla\cdot{\bf A} = \frac{\mu_0}{4\pi}\int \frac{\nabla' \cdot {\bf j}({\bf r}') } {\vert{\bf r} - {\bf r}'\vert} \,dV',$$ (2.259)

provided that $\vert{\bf j}({\bf r})\vert$ is bounded as $\vert{\bf r}\vert \rightarrow\infty$. Now, the flux of electric charge out of a surface $S$, enclosing a volume $V$, is

$$\oint_S {\bf j}\cdot d{\bf S} = \int_V \nabla\cdot{\bf j}\,dV,$$ (2.260)

where use has been made of the divergence theorem. (See Section A.20.) However, for a steady current distribution, this flux must be zero, otherwise positive or negative electric charge would build up inside $V$. Moreover, the flux must be zero for all possible volumes, $V$, which implies that

$$\nabla\cdot{\bf j} = 0$$ (2.261)

for a steady current distribution. Hence, we deduce from Equation (2.259) that

$$\nabla\cdot{\bf A} = 0$$ (2.262)

for a steady current distribution.