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# 9.2: The Magnetic Vector Potential

[ "article:topic", "authorname:tatumj" ]

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.