Understanding Griffiths' Example 5.8: Ienc = KL?

[joeyjj3]

I'm having a hard time understanding example 5.8 in Introduction to Electrodynamics by Griffiths. Why, exactly, is Ienc = KL? It makes sense intuitively, but I don't see how to get this result explicitly -- shouldn't the line integral be along the y-axis, and since $$\bar{K}$$=K$$\hat{x}$$, shouldn't the dot product be equal to zero? If it's not along the y-axis, how is it that the integral can be from L to 0? If anybody could explicitly do the dot product and integral that finds the enclosed current in this example (5.8 Griffiths), that would be really helpful.

Thanks for any info.

 

[gabbagabbahey]

You don't take the dot product between $\vec{K}$ and $\vec{dl}$. You take the dot product of the magnetic field with $\vec{dl}$.

The magnetic field $\vec{B}$ points in the y-direction and is uniform (due to symmetry), so the integral of $\vec{B} \cdot \vec{dl}$ over the path of the Amperian loop simply gives you 2Bl (If you go counterclockwise around the loop; and -2Bl if you go clockwise)

The current enclosed by the amperian loop is not $$\oint\vec{K} \cdot \vec{dl}$$!

It is instead given by the flux of current through the surface bounded by the Amperian loop. This surface has a normal in the x-direction (if you go counterclockwise around the loop when determining the integral of B dot dl) , so the flux of current is

$$\int_{\mathcal{S}} \vec{K} \cdot \vec{da}=\int_{\mathcal{S}} (\delta(z)K)dydz=\int_0^l Kdy=Kl$$

Where the dirac delta is used since the surface current is zero when $z\neq 0$.

Follow?

 

[joeyjj3]

Yes, that makes perfect sense. Thanks!