# 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?

• 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. – Voltage Spike 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 basic definition of curl is given in the textbook. It uses the cylindrical coordinate system (in general orthogonal curvilinear system) for the representation. Now, first of all let me introduce you to the basic definition of the derivative: $$$$ For a differentiable function $$\f(x)\$$ its derivative $$\f'(x)\$$ is defined as $$$$f'(x)=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$$$$ This is the same method used in the curl. Basically, for a cylindrical coordinate system, the the curl definition can be unfolded as (taking the plane (surface) of constant radius (it yields the same result in other two planes too)): $$$$(\nabla\times\textbf{A})_r=\lim_{\Delta s\rightarrow 0}\frac{\oint_C \textbf{A} \cdot \textbf{dl}}{\Delta s}\\=\lim_{\Delta\phi\rightarrow 0\\\Delta z\rightarrow 0}\frac{[A_\phi]_{(r,z)}rd\phi+[A_z]_{(r,\phi+d\phi)}dz-[A_\phi ]_{(r,z+dz)}rd\phi-[A_z]_{(r,\phi)}dz}{rd\phi dz}\textbf{a}_r\\=\Bigg(\frac{1}{r}\lim_{\Delta\phi\rightarrow 0}\frac{[A_z]_{(r,\phi+d\phi)}-[A_z]_{(r,\phi)}}{d\phi}\times rd\phi+\lim_{\Delta z\rightarrow 0}\frac{[A_\phi]_{(r,z)}-[A_\phi]_{(r,z+dz)}}{dz}\times dz\Bigg)\textbf{a}_r\\=\Bigg(\frac{1}{r}\frac{\partial A_z}{\partial\phi }- \frac{\partial A_\phi}{\partial z}\Bigg)\textbf{a}_r$$$$ This means that the curl is the rate of change of the altitude component $$\A_z\$$ with respect to the differential arc length (arc length is defined as product of radius and the angle in radians) minus the change of the angular component with respect to to the altitude. For better visualization look at the two pictures below:

Image credit: Google Similarly, $$$$(\nabla\times \textbf{A})_\phi= \lim_{\Delta r\rightarrow 0\\\Delta z\rightarrow 0}\frac{[A_z]_{(r+dr,\phi)}dz+[A_r]_{(z,\phi)}dr-[A_z]_{(r,\phi)}dz-[A_r]_{(z+dz,\phi)}dz}{drdz}\textbf{a}_\phi\\=\Bigg(\lim_{\Delta r\rightarrow 0} \frac{[A_z]_{(r+dr,\phi)}-[A_z]_{(r,\phi)}}{dr}+\lim_{\Delta z\rightarrow 0}\frac{[A_r]_{(z,\phi)}-[A_r]_{(z+dz,\phi)}}{dz}\Bigg)\textbf{a}_\phi\\=\Bigg(\frac{\partial A_z}{\partial r}-\frac{\partial A_r}{\partial z}\Bigg)\textbf{a}_\phi$$$$ Similarly, $$$$(\nabla \times \textbf{A})_z=\lim_{\Delta r\rightarrow 0\\\Delta z\rightarrow 0}\frac{[A_r]_{(\phi,z)}dr+[A_\phi]{(r+dr,z)}-[A_r]_{(\phi+d\phi,z)}dr-[A_\phi]_{(r,z)}}{rd\phi dz}\textbf{a}_z\\=\Bigg(\frac{\partial{A_\phi}}{\partial r}-\frac{1}{r}\frac{\partial A_r}{\partial \phi}\Bigg)\textbf{a}_z$$$$ So, the complete curl equation will be: $$$$\nabla\times\textbf{A}=(\nabla\times\textbf{A})_r -(\nabla\times\textbf{A})_\phi + (\nabla\times\textbf{A})_z$$$$ Note that $$\\textbf{a}_\phi=\textbf{a}_z\times\textbf{a}_r\$$. But I have taken the reverse way while taking the curl. That is why the minus sign is implemented.

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 posible infinite current density (finite current over a filamentary current - or, in other words, an impossible zero diameter wire) causing the B blowing, consider the discussion on Why can't magnetic flux be no longer defined for non-zero diameter wire?