# Intuition differential ampere's law

Ampere's differential law states that -

$$\nabla \times {\bf B} = \frac{4 \pi \, {\bf J}}{c}$$

I know to derive amperes integral form from special relativity, and to use stokes theorem in order to get to this equation.

However I am confused about using and understanding this law in the "local" form.

I have read that curl is just microscopic rotation, like if we put a ball on a stick in water with curl it'll rotate on the stick (got this from http://mathinsight.org/curl_components).

How do we approach at looking on curl of B? Is the microscopic intuition working here?

If we take a certain point (x,y,z) in space and have zero J throught it and on the otherside (x,y,z+1) has current (J), according to this equation at (x,y,z) there will be no curl which is obviously not true.

• What makes you think it is obviously not true? If there is no current, then the curl of B is 0. Note that this does not necessarily mean that B is 0. Note also that this formula is only valid for electrostatics, thus only when the currents and fields are not changing in time. – Crimson Jun 22 '16 at 13:12
• @Crimson because we have current at point (x,y,z+1), so there must be magnetic field at point (x,y,z). However from the formula the J(x,y,z+1) = 0 = const* B(x,y,z+1). Maybe im getting something wrong. – The Capacitor Jun 22 '16 at 13:58

We might think of the integral form of this $$\oint\vec H\cdot d\vec l~=~\int\int\vec J\cdot d\vec s$$ We have two integrals, on a line integral and the other an area integral. With each unit of line $d\vec l$ this is projected onto the mangetic field $\vec H$. Each $\delta l$ length is associated with a pie shaped region with area $ds~=~\frac{1}{2}rdl$ $=~\frac{1}{2}r^2d\theta$. This just comes from the elementary equation for the area of a triangle.
Now consider $\Delta s^{-1}\oint\vec H\cdot d\vec s$ and we equate this to $$\sum_i\hat n_i\cdot \vec V~=~\Delta s^{-1}\oint\vec H\cdot d\vec s$$ or $$\sum_i\hat n_i\cdot \vec V\Delta s~=~\oint\vec H\cdot d\vec s$$ such that $\hat n_i$ is a normal vector in the direction of the vector $\vec V$ at each small pie slice unit of area. For circular motion it is known that the trangent velocity vector is accompanied by the angular velocity vector normal to the plane of rotation. The small unit vectors of area $\hat n_i\Delta s$ are perpendicular to tangent velocity vectors, and for the projection onto the magnetic field we requires $\vec V~=~\nabla\times\vec H$. Another way to think of this is that this is the mean value theorem of calculus that evaluates the boundary. In this way we then require $$\int\int\nabla\times\vec H\cdot d\vec s~=~\int\int\vec J\cdot d\vec s$$