# 9.2: The Magnetic Vector Potential

- Page ID
- 5467

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.