# Magnetic field at points on the circuit

I know magnetic field lines due to a circuit always form closed loops. Therefore $$\nabla \cdot \vec{B}=0$$ everywhere (even at points on the circuit). However due to singularity, magnetic fields are not defined at points on the circuit. Then how does it make sense to say $$-$$ divergence of "magnetic field at points on the circuit"?

• What do you mean by "However due to singularity"? Why do you think magnetic fields are not defined somewhere? Perhaps an example would clarify the question. Nov 29 '18 at 10:58
• Magnetic field reduces as inverse square of the distance $\dfrac{1}{r^2}$. While computing the field at a point on the circuit due to a current element at that point, $r=0$. Therefore field at that point becomes infinity. This is what I mean by "magnetic fields are not defined at points on the circuit".
– Joe
Nov 29 '18 at 11:14

The magnetic field strength outside a (long) wire falls off as $$r^{-1}$$. The gist of your question seem to be what happens as $$r \rightarrow 0$$?

The answer is that the $$r^{-1}$$ dependence is only true outside the wire. Inside the wire you would need to use Ampere's law with a finite current density to work out what current was encircled by a chosen loop. e.g. For a uniform current density the magnetic field scales as $$r$$ inside the wire and $$B \rightarrow 0$$ as $$r \rightarrow 0$$.

EDIT: You are asking about mathematical abstractions (1-dimensional currents) rather than physical situations; this is how to proceed.

Ampere's law (in magnetostatics) says $$\oint \vec{B}\cdot d\vec{l} = \int \vec{J}\cdot d\vec{A}$$ If we consider an infinitely long wire defined by the z-axis, then taking a circular loop around the z-axis, the enclosed current is always the same.The B-field is therefore $$\propto r^{-1}$$ and would become infinite when $$r=0$$.

However, if we say instead that we have a uniform current density $$\vec{J}$$ that occupies a cylinder of radius $$a$$, then this treatment only applied for $$r>a$$.

If we allow $$r then Ampere's law gives $$2\pi r B = \pi r^2 J$$ For any finite current density, then as $$r \rightarrow 0$$ then the right hand side goes to zero faster than the left hand side and $$B \rightarrow 0$$.

If instead you allow the current density to be infinite, so that a 1-d wire can carry a current, then do not be surprised that you get an infinite B-field! (You also need an infinite E-field because $$J = \sigma \vec{E}$$.)

• What if my wire is a line ($1$ dimensional) wire in space passing through $(1,4,7)$ and we wish to find the magnetic field at $(1,4,7)$.
– Joe
Nov 29 '18 at 13:12
• The question would be ill-posed, as one-dimensional wires don't exist in reality. It's an abstraction we use because it makes things simpler, but it relies on approximations that break down when we get too close to the line. Nov 29 '18 at 13:32
• @Joe If your wire is "1 dimensional" then it cannot conduct a current. Ampere's law is $\oint \vec{B}\cdot d\vec{l} = \int \vec{J} \cdot d\vec{A}$, where $\vec{J}$ is the current per unit area. A 1-dimensional wire has no area to integrate over. Nov 29 '18 at 13:57
• @lr1985: I agree that one-dimensional wires don't exist in reality. But if we are only concerned with mathematics, we can consider a one-dimensional wire. Since magnetic field always forms closed loops, we can exploit Gauss divergence theorem and show that $\nabla \cdot \vec{B}=0$ everywhere (even at points on the circuit). But is there a way to find $\vec{B}$ at points on the mathematically constructed one dimensional wire?
– Joe
Nov 29 '18 at 13:58
• Is it valid to apply the limit method to Biot-Savart law in order to find $\vec{B}$ at points on the mathematically constructed one dimensional wire.
– Joe
Nov 29 '18 at 14:30