9.2: The Magnetic Vector Potential

Although we cannot express the magnetic field as the gradient of a scalar potential function, we shall define a vector quantity $\textbf{A}$ whose curl is equal to the magnetic field:

$$\textbf{B} = \textbf{curl A} = \nabla \times \textbf{A}.\label{9.2.1}$$

Just as $\textbf{E} = −\nabla V$ does not define $V$ uniquely (because we can add an arbitrary constant to it), so, similarly, Equation $\ref{9.2.1}$ does not define $\textbf{A}$ uniquely. For, if $ψ$ is some scalar quantity, we can always add $\nabla ψ$ to $\textbf{A}$ without affecting $\textbf{B}$, because $\nabla \times \nabla ψ = \textbf{curl grad }ψ = 0$.

The vector $\textbf{A}$ is called the magnetic vector potential. Its dimensions are $\text{MLT}^{−1}\text{Q}^{−1}$. Its SI units can be expressed as $\text{T m, or Wb m}^{−1}\text{ or N A}^{−1}$.

It might be briefly noted here that some authors define the magnetic vector potential from $\textbf{H = curl A}$, though it is standard SI practice to define it from $\textbf{B = curl A}$. Systems of units and definitions other than SI will be dealt with in Chapter 16.

Now in electrostatics, we have $\textbf{E}=\frac{1}{4\pi \epsilon}\frac{q}{r^2}\hat{\textbf{r}}$ for the electric field near a point charge, and, with $\textbf{E} = −\textbf{grad} V$, we obtain for the potential $V=\frac{q}{4\pi\epsilon r}$. In electromagnetism we have $\textbf{dB}=\frac{\mu I}{4\pi r^2}\hat{\textbf{r}}\times \textbf{ds}$ for the contribution to the magnetic field near a circuit element $\textbf{ds}$. Given that $\textbf{B} = \textbf{curl A}$, can we obtain an expression for the magnetic vector potential from the current element? The answer is yes, if we recognize that $\hat{\textbf{r}}/r^2$ can be written $-\nabla (1/r)$. (If this isn't obvious, go to the expression for $\nabla ψ$ in spherical coordinates, and put $ψ = 1/r$.) The Biot-Savart law becomes

$$\textbf{dB}=-\frac{\mu I}{4\pi}\nabla (1/r)\times \textbf{ds}=\frac{\mu I}{4\pi}\textbf{ds}\times \nabla (1/r).\label{9.2.3}$$

Since $\textbf{ds}$ is independent of $r$, the nabla can be moved to the left of the cross product to give

$$\textbf{dB}=\nabla \times \frac{\mu I}{4\pi r}\textbf{ds}.\label{9.2.4}$$

The expression $\frac{\mu I}{4\pi r}\textbf{ds}$, then, is the contribution $\textbf{dA}$ to the magnetic vector potential from the circuit element $\textbf{ds}$. Of course an isolated circuit element cannot exist by itself, so, for the magnetic vector potential from a complete circuit, the line integral of this must be calculated around the circuit.