Magnetic Flux Density and Vector Potential
Electrostatics had a scalar shortcut — the potential V. Magnetism cannot: with no monopoles, B has no scalar potential. Instead it earns a richer helper, the vector potential A, whose curl is the field and whose own equation echoes Poisson's.
- Magnetic flux and why \(\nabla\cdot\vec{B}=0\) means flux lines always close.
- Why \(\vec{B}\) admits no scalar potential in regions carrying current.
- The magnetic vector potential \(\vec{A}\) and the relation \(\vec{B}=\nabla\times\vec{A}\).
- The vector Poisson equation \(\nabla^2\vec{A}=-\mu_0\vec{J}\) and its solution.
- How to get magnetic flux straight from a line integral of \(\vec{A}\).
- The complete set of Maxwell's equations for static fields.
Magnetic Flux Revisited
Recall the magnetic flux through a surface from Chapter 11 — the surface integral of \(\vec{B}\), measured in webers:
Because there are no magnetic charges, every flux line that enters a closed surface must leave it: the net flux through any closed surface is zero, \(\oint_S\vec{B}\cdot d\vec{S}=0\), i.e. \(\nabla\cdot\vec{B}=0\). This single fact — that \(\vec{B}\) is solenoidal — is the seed from which the vector potential grows.
Why No Scalar Potential
In electrostatics we wrote \(\vec{E}=-\nabla V\) because \(\vec{E}\) was irrotational (\(\nabla\times\vec{E}=0\)) — and only a field with zero curl can be the gradient of a scalar. The magnetic field fails this test: Ampère's law says \(\nabla\times\vec{H}=\vec{J}\), which is non-zero wherever current flows. So in general \(\vec{B}\) cannot be written as \(-\nabla V_m\).
A scalar magnetic potential \(V_m\) can be defined, but only in current-free regions where \(\nabla\times\vec{H}=0\). There \(\vec{H}=-\nabla V_m\) holds, just as in electrostatics — useful for the air gaps of magnetic circuits, but not a general tool.
The Vector Potential A
Since \(\nabla\cdot\vec{B}=0\) and the divergence of any curl is zero, there must exist a vector field \(\vec{A}\) — the magnetic vector potential (Wb/m) — whose curl produces \(\vec{B}\):
Unlike the electric scalar potential, \(\vec{A}\) points in the same direction as the current that creates it. For a current element, \(\vec{A}\) takes a form strikingly parallel to the electrostatic potential of a charge element — the \(1/R\) kernel is identical, only the source changes from \(dQ\) to \(I\,d\vec{l}\):
Poisson's Equation for A
Substituting \(\vec{B}=\nabla\times\vec{A}\) into Ampère's law, and choosing the convenient Coulomb gauge \(\nabla\cdot\vec{A}=0\), the vector identity for the curl of a curl collapses the result into a clean Poisson equation — one for each component of \(\vec{A}\):
Every method from Chapter 9 carries straight over. The scalar potential's source is charge density \(\rho_v\); the vector potential's source is current density \(\vec{J}\). Solve the same equation, component by component, with the same techniques.
The freedom to add any gradient to \(\vec{A}\) without changing \(\vec{B}\) (since \(\nabla\times\nabla\chi=0\)) is called gauge freedom; the Coulomb gauge simply fixes that freedom to make the equation tidy.
Flux from A
The vector potential offers a slick route to magnetic flux. Applying Stokes' theorem (Chapter 3) to \(\vec{B}=\nabla\times\vec{A}\) turns the flux surface integral into a line integral of \(\vec{A}\) around the boundary:
Often the loop integral is far easier than the surface integral — especially when \(\vec{A}\) has a simple closed form. This trick is the everyday way to compute the flux linking a coil, and hence its inductance in Chapter 14.
Static Maxwell's Equations
With magnetostatics complete, all four of Maxwell's equations for static fields are now in hand. They split cleanly: two for the electric field, two for the magnetic, with divergence laws naming the sources and curl laws describing the circulation.
| Law | Point form | Integral form |
|---|---|---|
| Gauss (electric) | \(\nabla\cdot\vec{D}=\rho_v\) | \(\oint\vec{D}\cdot d\vec{S}=Q\) |
| Gauss (magnetic) | \(\nabla\cdot\vec{B}=0\) | \(\oint\vec{B}\cdot d\vec{S}=0\) |
| Faraday (static) | \(\nabla\times\vec{E}=0\) | \(\oint\vec{E}\cdot d\vec{l}=0\) |
| Ampère (static) | \(\nabla\times\vec{H}=\vec{J}\) | \(\oint\vec{H}\cdot d\vec{l}=I\) |
Worked Examples
Problem. Given \(\vec{A} = (x^2 y)\,\hat{a}_z\ \text{Wb/m}\), find \(\vec{B}\).
Solution. Take the curl:
Problem. Check whether \(\vec{A} = y\,\hat{a}_x - x\,\hat{a}_y\) satisfies the Coulomb gauge.
Solution. Compute \(\nabla\cdot\vec{A}\):
Problem. If \(\vec{A} = \tfrac{1}{2}B_0(-y\,\hat{a}_x + x\,\hat{a}_y)\), find the flux through a circle of radius \(a\) in the \(xy\)-plane.
Solution. On the circle \(\vec{A}\cdot d\vec{l} = \tfrac{1}{2}B_0 a\,(a\,d\phi)\):
This equals \(B_0\) times the area — confirming \(\vec{A}\) describes a uniform field \(B_0\,\hat{a}_z\).
Problem. In a current-free region \(\vec{H}=-\nabla V_m\). If \(V_m = 5x\ \text{A}\), find \(\vec{H}\).
Solution. Take the negative gradient, valid since \(\vec{J}=0\):
Problem. An infinite wire carries \(I\). Find the flux through a rectangle of height \(L\) spanning \(\rho=a\) to \(\rho=b\).
Solution. Integrate \(B=\mu_0 I/2\pi\rho\) across the strip:
Problem. A vector potential satisfies \(\nabla^2\vec{A} = -\mu_0\vec{J}\). If \(\vec{A}=A_0(1-\rho^2/a^2)\,\hat{a}_z\), find \(\vec{J}\) (use the cylindrical Laplacian).
Solution. Apply \(\nabla^2 A_z = \tfrac{1}{\rho}\tfrac{d}{d\rho}(\rho\,dA_z/d\rho)\):
Chapter Summary
\(\Psi_m=\int\vec{B}\cdot d\vec{S}\) (Wb); \(\nabla\cdot\vec{B}=0\) makes flux lines close.
\(\nabla\times\vec{H}=\vec{J}\neq0\), so \(\vec{B}\neq-\nabla V_m\) except in current-free regions.
\(\vec{B}=\nabla\times\vec{A}\); \(\vec{A}\) is parallel to its source current.
\(\nabla^2\vec{A}=-\mu_0\vec{J}\); the magnetic mirror of \(\nabla^2 V=-\rho_v/\varepsilon\).
\(\Psi_m=\oint_L\vec{A}\cdot d\vec{l}\) — often easier than the surface integral.
Four equations; \(\vec{E}\) and \(\vec{B}\) independent until time enters in Part 4.
Problems
For each item, decide whether you need \(\vec{B}=\nabla\times\vec{A}\), the Poisson equation, or the flux line-integral, then proceed. Difficulty rises down the list.
- Given \(\vec{A} = 3yz\,\hat{a}_x\ \text{Wb/m}\), find \(\vec{B}\).
- Check whether \(\vec{A} = x\,\hat{a}_x + y\,\hat{a}_y\) satisfies the Coulomb gauge.
- For \(\vec{A} = \tfrac12 B_0(-y\,\hat{a}_x + x\,\hat{a}_y)\), confirm \(\vec{B}=B_0\,\hat{a}_z\).
- Find the flux of a uniform field \(B_0\,\hat{a}_z\) through a square of side \(L\) using \(\oint\vec{A}\cdot d\vec{l}\).
- An infinite wire carries \(20\ \text{A}\). Find the flux through a \(1\ \text{m}\times(2\to6\ \text{cm})\) rectangle beside it.
- In a current-free region \(V_m = 10xy\). Find \(\vec{H}\).
- Given \(\vec{A}=A_0\sin(kx)\,\hat{a}_z\), find \(\vec{B}\) and then \(\vec{J}\) from the Poisson equation.
- Show that adding \(\nabla\chi\) to \(\vec{A}\) leaves \(\vec{B}\) unchanged (gauge invariance).
- Derive \(\nabla^2\vec{A}=-\mu_0\vec{J}\) from Ampère's law using the curl-of-curl identity and the Coulomb gauge.
- Compute the flux linking a single circular loop of radius \(a\) placed coaxially in a solenoid field \(B_0\).
- Explain why a scalar magnetic potential becomes multivalued if its path encircles a current.
- Summarise, in your own words, why electrostatics gets a scalar potential but magnetostatics needs a vector one — citing the two null identities.