# No magnetic fields?

1. Feb 24, 2008

### jackxx

$$\nabla \cdot B=0,$$
so $$\int\nabla \cdot B dv=0,$$
then $$\int B \cdot \widehat{n}da=0,$$
let $$B=\nabla \times A,$$
then $$\int\nabla \times A \cdot \widehat{n}da=0,$$
thus $$A=\nabla\varphi[/itex], thus [tex]B=\nabla \times A=\nabla \times\nabla\varphi=0$$

I know something is wrong with that but im not sure what it is any ideas?

Last edited by a moderator: Feb 25, 2008
2. Feb 24, 2008

The problem per se arises when you define A as a gradient of some scalar. This would be incompatible with Ampere's law mainly because the curl of a gradient is always zero. So in this case you are specifically setting B = 0 from the start essentially.

Last edited by a moderator: Feb 25, 2008
3. Feb 25, 2008

### Lojzek

A field is conservative if it's integral over any closed curve is zero. This condition is satisfied if the integral of field's rotor over area limited by the curve is zero. In your case you choose whole surface (of a volume) for the area: the surface is not an area limited by a curve, so you can't use the fact that integral of rotor over the surface implies that integral of field over a curve is zero. In fact the integral of rotor over a surface of a volume always give zero (for any vector field), because you can split the surface on two parts with a curve and the integrals of rotor over two parts of surface give +/- integral of field over curve, so they cancel each other out.
So the assumption that A=grad(fi) in not justified.

Last edited: Feb 25, 2008
4. Feb 26, 2008

### pam

You just confused an integral over a closed surface with the integral around a closed loop defining a conservative field.
We are lucky you are wrong. Otherwise all electric motors would instantly stop.