Sign Conventions for Paths and Surfaces for Electromagnetic Calculations

[Jessehk]

Hello all.

I'm trying to figure out how to determine the correct sign of paths and surfaces defined for calculating quantities in electromagnetic problems.

For example, say there's a wire in the shape of a rectangular prism along the z-axis with some current density, $$\vec{J}$$.

Then the current through the wire is $$I = \int_S { \vec{J} \cdot \textrm{d} S }$$. Now, $$\textrm{d} S = \textrm{d} x \textrm{d} y \hat{z}$$ but how can we know the sign of the normal to the cross-section? Is it in the -z direction (ie, $$-\hat{z}$$) or the +z direction?

Similarly, the potential difference in an electrostatic field is $V_{12} = - \int_2^1 \vec{E} \cdot \textrm{d} \gamma$ but how do we define the sign of $\textrm{d} \gamma$? For example, if both points are in the same axis, is the direction of the path from (1) to (2) or from (2) to (1)?

Is there a general rule? I apologize if this question is not clear, and thanks in advance. :)

 

[vanhees71]

There are only sign problems if you consider the integral form of Maxwell's Equations. Take, e.g., Faraday's Law, which reads in differential form

$$\vec{\nabla} \times \vec{E}=-\partial_t \vec{B}.$$

If you integrate this equation over a fixed (i.e., time independent) surface $$F$$ with boundary $$\partial F$$. You can choose the orientation of the surface as you like or as is most convenient for you, but then you like to apply Stokes integral theorem, and then you have to orient the boundary in the same way as the surface. By convention, this is defined according to the right-hand rule: Point the thumb of your right hand in direction of the chosen surface normal (defining the orientation of the surface). Then the orientation of the boundary path in Stokes theorem is given by the direction of the fingers. With this convention, the integral form (for time-independent surfaces and boundaries) reads

$$\int_{\partial F} \mathrm{d} \vec{x} \cdot \vec{E}=-\frac{\mathrm{d}}{\mathrm{d} t} \int_F \mathrm{d}^2 \vec{F} \cdot \vec{B}$$.

BTW: For moving surfaces we recently had a big discussion in this forum.