Stokes Theorem with Ampere's Law

Question

Whilst researching Maxwell's Equations (here), I found (effectively) the following pieces of logic:

$$\int_S \left(\nabla \times \boldsymbol{H}\right) \cdot d\boldsymbol{S} = \oint \boldsymbol{H} \cdot d\boldsymbol{L}$$ $$\oint \boldsymbol{H} \cdot d\boldsymbol{L} = \int_S \boldsymbol{J}\cdot d\boldsymbol{S}=I_{enclosed}$$

Taking the surface to be that of a circle, the author says:

$$2\pi r H=I_{enclosed}$$

Which he rearranges to:

$$H=\frac{I_{enclosed}}{2\pi r}$$

However, at this point the author notes that the $H$ field is in fact a vector field, and as such, he makes the addendum that:

The direction of the H-field is everywhere tangential to the imaginary loops.

And he then says that this is defined by the right hand rule. So me, that would imply something like the following formula:

$$2\pi \boldsymbol{H} \times \boldsymbol{r} = \boldsymbol{I}$$

However, I can think of no way to prove this formula, and the proof of the author's statement about the direction is simply glossed over. I have spent some time researching vector calculus identities, but have yet to find anything which might help explain what looks to me like a small logical jump.

Thank you for your time, and my apologies if this is in some way a stupid question!

Please see if this could help you a little before even considering your "formula". https://en.wikipedia.org/wiki/Axis%E2%80%93angle_representation — Global, Jul 15 '18 at 19:19
I just read the page, however I am not entirely certain how it helps with this problem... Are you questioning the choice of $2\pi$, somehow? That page seems to me to be on a method of describing rotation - I'm not certain how that bears on my question... — DoublyNegative, Jul 15 '18 at 19:36
That's a fair point - but $J$ is a vector, so a vector with the units of $A$ would presumably still make logical sense? — DoublyNegative, Jul 16 '18 at 19:10
Related : (1) Magnetic field due to a single moving charge.(2) A generalization of the Biot-Savart law. — Frobenius, May 29 '20 at 12:39

score 1 · Answer 1 · answered Jul 16 '18 at 19:04

There is a convention linked to the right hand set of xyx-axes which is used when evaluating integrals of the type which occur in one of the equations that you have quoted.

$$\oint_C \boldsymbol{H} \cdot d\boldsymbol{L} = \int_S \boldsymbol{J}\cdot d\boldsymbol{S}$$

The diagram shows a current density $\boldsymbol J$ passing through an area $dS$ whose normal is $\boldsymbol{n}$ so $\boldsymbol{J}\cdot \boldsymbol{n} \,dS= J\,dS$ and when this is integrated over the whole area $S$ this gives the enclosed current.

And he then says that this is defined by the right hand [grip] rule.

The sign of the line integral will change depending on the direction of the path (anticlockwise or clockwise).
The convention is that in this case the anticlockwise direction (looking from the top) is chosen using the right hand grip rule the reference vector direction being the direction of the normal to the surface $\boldsymbol n$.

So it is $\boldsymbol B \cdot d\boldsymbol L = B\, dL$ which is integrated taking the anticlockwise path $(+B\,2 \pi r)$.

Sorry, but what I don't understand is how we end up with a vector field, bearing in mind that $BdL$ is not a vector field. Perhaps I wasn't overly clear in my question? Or is the logic simply that, given an infinitesimal step along $L$, the $B$ field too increases infinitesimally, and it's that gradient I am thinking of when I think direction? — DoublyNegative, Jul 16 '18 at 19:17

Stokes Theorem with Ampere's Law

1 Answers1