Is the magnetic vecor field always solenoidal?

Question

In general magnetic field vector field B is solenoidal because $\nabla \cdot \mathbf{B} = \mathbf{0}$. What is the essential feature of a solenoidal field geometrically? Is there any condition when B ceases to be solenoidal?

Answer 1

No, $\mathbf{B}$ is never not purely solenoidal. That is, $\mathbf{B}$ is always solenoidal. The essential feature of a solenoidal field is that it can be written as the curl of another vector field, $$\mathbf{B} = \nabla \times \mathbf{A}.$$ Doing this guarantees that $\mathbf{B}$ satisfies the "no magnetic monopoles" equation from Maxwell's equation. This is all assuming, of course, that we never discover any magnetic monopoles.

In terms of field geometry, one consequence of this is that the magnetic field lines all have to form closed loops without beginnings or ends. Another consequence is that line integrals of the magnetic field are only path independent when changes to the path are confined to regions where there is no current and no changing electric fields. For example, if I want to calculate $\int \mathbf{B} \cdot \operatorname{d}\mathbf{\ell}$ in the neighborhood of a DC current carrying wire, then the answer will only change if we move the path across the wire in some way.

For more information, see the Helmholtz decomposition theorem.

Answer 2

By "solenoidal" you presumably mean "the magnetic field has everywhere zero divergence".

Let's draw an analogy between the field lines of $\mathbf{B}$ and the flow lines of a tub of water. At a point of nonzero divergence, we have "new" flow lines entering or exiting the tub. In other words, at points where water is flowing into the tub we get positive divergence, and at points where it is flowing out we get negative divergence.

For the electric field we have $\nabla \cdot \mathbf{E} = \rho$ where $\rho$ is the charge density. We can thus roughly think of positive charge as "sources" of electric field: $\mathbf{E}$ will point radially outwards from a positive point charge. Conversely, negative charges act like "sinks" and $\mathbf{E}$ will point radially inwards towards negative point charges. Of course, the assignment of which kind of charge is the sink and which the source is arbitrary.

To get analogous behaviour for the magnetic field, there would need to be some kind of "magnetic charge density" $\rho_B$. We would then have $\nabla \cdot \mathbf{B} = \rho_B$ for the magnetic Gauss law.

Current, of course, sources magnetic fields, but it acts as a dipole (to leading order), and dipoles involve closed field loops (and thus no divergence). To the best of our knowledge, such "magnetic monopoles" do not exist, at least not fundamentally. So yes, $\mathbf{B}$ will always be divergence free.