# Is the curl of a B field at a point on a wire infinity?

According to this text, curl is defined as a limit of the surface area approaching zero. If the point where the curl is evaluated at is near the wire, B is evaluated at a decreasing r value, and divided by a decreasing s value, it seems that the curl will approach infinity. Is this a correct approach in evaluating the curl of a B field?

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• Next time, it's better to ask this question at physics.stackexchange.com. – 比尔盖子 Jun 12 at 21:03
• This is a question I have related to an engineering electromagnetics textbook definition. – vishayp Jun 12 at 21:08
• Electromagnetic questions are fine here as long as you're not talking about anything on the quantum level like tunneling or individual electron diffusion. To better understand what you're asking, are you asking about if the contour path $C$ of the magnetic field $B$ is an infinitely endless orbit around a current at a single point of the wire? – KingDuken Jun 12 at 21:22
• Before you evaluate the limit, what does integrating the magnetic field around the contour give you? It's not infinity. – laptop2d Jun 12 at 21:36
• Consider looking here for a qualitative point worth considering, as well. – jonk Jun 12 at 23:05

From the mathematics alone, yes, this does go to infinity.

But notice that this assumes an infinitesimally thin wire with infinite current density; while the approximation of a thin wire is often useful, it breaks down, as you noticed, when you start dealing with magnetism in anything more than a cursory manner.

If the wire has finite size, the integration contours that are located inside the wire also have less current enclosed, so the singularity disappears.

The area $$\\Delta s\$$ in the left half of equation 3.103 is bounded by the contour $$\C\$$ over which the integral is taken. So as $$\\Delta s\$$ goes to zero, so does the integral.

Remember that $$\\nabla \times B\$$ is equal to the current (and displacement current) density. You can't get infinite curl without infinite current density -- and infinite current density is kinda hard to achieve in the real world.

According to first Maxwell's equation in derivative form: The curl, or rotational, of the magnetic field H is:

$$\nabla \times \mathbf{H}=\mathbf{J} + \frac{\partial \mathbf{D}}{\partial t}$$

The relationship with the magnetic flux density B is:

$$\mathbf{H}= \frac{\mathbf{B}}{\mu}$$

So,

$$\nabla \times\frac{\mathbf{B}}{\mu}=\mathbf{J} + \frac{\partial \mathbf{D}}{\partial t}$$

or

$$\nabla \times \mathbf{B} = \mu \mathbf{J} + \mu \frac{\partial \mathbf{D}}{\partial t}$$

Also (involving the electric field E):

$$\nabla \times \mathbf{B} = \mu \mathbf{J} + \mu \varepsilon \frac{\partial \mathbf{E}}{\partial t}$$

In time invariant case:

$$\nabla \times \mathbf{B} = \mu \mathbf{J}$$

In this case, the curl of flux density (B) in any point is proportional to the current density (J) at that point. In a general, it's not infinity.

EDIT

Regarding the convergence of the limit involving the division, there is a similar reasoning: Considering the expression of the definition of velocity vector in a point, $$\ \mathbf{v}(t) = \lim_{\Delta t \to 0} \frac{\Delta \mathbf{x}}{\Delta t} \$$, one could argue that since the time interval tends to zero, then the magnitude of the velocity would tend to infinity. The limit is because we are interested in the value of a measure at a well-defined point.