Magnetic Vector Potential

The magnetic vector potential, denoted by A, is a vector field, just like the magnetic field B.  It is related to B by a simple equation:

\[ \mathbf{B} = \nabla \times \mathbf{A} \]

Therefore, B can be found by taking the curl of A. 

In some situations, A  is easier to calculate than B. This is especially true when dealing with complex current distributions. Also, A plays a key role in certain areas of physics, like electromagnetic waves and quantum mechanics.

Unit of A

Since the curl operation does not change the basic unit of a quantity apart from involving spatial derivatives, A must have a unit that gives the unit of B when taking the curl. B has the tesla (T) unit in the SI system. It turns out that the unit of A is:

A = T·m = Wb/m

Scalar Potential Analogy

The magnetic vector potential is analogous to the electric field E. E can be written as the negative gradient of a scalar potential ɸ:

\[ \mathbf{E} = −\nabla \phi \]

The scalar potential helps us understand electric fields, just as the vector potential helps us understand magnetic fields. However, unlike ϕ, A is a vector quantity. It means that A has direction and magnitude at every point in space, which is denoted by three components (A_x, A_y, A_z) in a three-dimensional space.

Similar to how the electric potential simplifies electrostatic computations, the magnetic vector potential facilitates solving magnetostatic problems.

Biot-Savart law for steady currents is applied to calculate the magnetic vector potential.

The general expression for A is:

\[ \mathbf{A}(\mathbf{r}) = \frac{\mu_0}{4\pi} \int \frac{\mathbf{J}(\mathbf{r’})}{|\mathbf{r} – \mathbf{r’}|} \, d^3r’ \]

Where:

– A(𝐫): Magnetic vector potential at the field point r

– μ₀: permeability of free space (= 4π x 10^-7)

– J(𝐫′): Current density at the source point r′, which gives the amount of current flowing per unit area at that point

– |𝐫 – 𝐫’|: Distance between the source point (where the current is) and the field point (where A is calculated)

– d³r′: Small volume element at the source

This equation tells us that A at a certain point in space depends on the current. The contribution from each current piece gets weaker the farther that point is from the wire, due to the 1/|r – r’| term.

Current-Carrying Wire

Consider a long, straight wire carrying a steady current I along the z-axis. Since the current is uniform and extends infinitely, we expect that the magnetic vector potential A will have a circular symmetry around the wire.

We start with the general expression for A in magnetostatics:

\[ \mathbf{A}(\mathbf{r}) = \frac{\mu_0}{4\pi} \int \frac{\mathbf{J}(\mathbf{r’})}{|\mathbf{r} – \mathbf{r’}|} \, d^3r’ \]

Since the current is concentrated along a thin wire, the current density J can be expressed using a line current representation. Instead of a volume integral, we use:

\[ \mathbf{A}(\mathbf{r}) = \frac{\mu_0}{4\pi} \int \frac{I \, d\mathbf{l}’}{|\mathbf{r} – \mathbf{r’}|} \]

Where dl′ is an infinitesimal element of the current-carrying wire.

Since the wire is along the z-axis, we use cylindrical coordinates (r,ϕ,z), where:

– The current I flows in the z-direction.
– The position vector r is the field point where we want to compute A.
– The source point along the wire is r′ = (0, 0, z′).
For a field point at (r, ϕ, z), the distance between the source and field point is:

\[ |\mathbf{r} – \mathbf{r’}| = \sqrt{r^2 + (z – z’)^2} \]

Since the current flows only in the z-direction, we express dl′ as:

\[ d\mathbf{l’} = \hat{\mathbf{z}} \, dz’ \]

Substituting into the equation for A, we get:

\[ \mathbf{A} = \frac{\mu_0 I}{4\pi} \int_{-\infty}^{\infty} \frac{\hat{\mathbf{z}} \, dz’}{\sqrt{r^2 + (z – z’)^2}} \]

This integral is a standard result in electromagnetism, and it evaluates to:

\[ \mathbf{A} = \frac{\mu_0 I}{2\pi} \ln \left( \frac{1}{r} \right) \hat{\mathbf{z}} \]

Since only differences in A are physically meaningful, we can redefine the constant and write:

\[ \mathbf{A} = \frac{\mu_0 I}{2\pi} \ln(r) \hat{\mathbf{z}} \]

This shows that A is directed along the z-axis, meaning the magnetic field B must form circular loops around the wire.

Now, we calculate B using:

\[ \mathbf{B} = \nabla \times \mathbf{A} \]

Since A has only a z-component, taking the curl in cylindrical coordinates gives:

\[ \mathbf{B} = \frac{1}{r} \frac{dA_z}{dr} \hat{\boldsymbol{\phi}} \]

Substituting \( A_z = \frac{\mu_0 I}{2\pi} \ln(r) \), we differentiate:

\[ \frac{dA_z}{dr} = \frac{\mu_0 I}{2\pi r} \]

Thus,

\[ \mathbf{B} = \frac{\mu_0 I}{2\pi r} \hat{\boldsymbol{\phi}} \]

This result matches Ampère’s Law, confirming that the magnetic field circles around the wire with magnitude:

\[ B = \frac{\mu_0 I}{2\pi r} \]