Why isn't path integral of H-field 0?

[yosimba2000]

So for an infinite plane of current, current traveling in the X direction, the magnetic field everywhere above the plane is going clockwise, and the m. field below the plane is going counterclockwise.

So the path integral is Integral of H dot dl = Current Enclosed

Why, in this video, does the path integral NOT equal 0?



Isnt the top path going from Y = -L to L, and the bottom path going from Y = -L to L?

So you get ∫H ⋅ dl = ∫H_{ydirection⋅} ⋅ dy_ydirection + ∫H_{negativeydirection⋅} ⋅ dy_{negativeydirection}

Then you get Hy, y from -L to L, plus Hy, y from L to -L?

So you get H(L - -L) + H(-L - L), and you get 2HL-2HL = 0?

 

[NFuller]

So for an infinite plane of current, current traveling in the X direction, the magnetic field everywhere above the plane is going clockwise, and the m. field below the plane is going counterclockwise.

No, the magnetic field is parallel to the sheet of current both above and below the plane.

Why, in this video, does the path integral NOT equal 0?

Because the loop encloses a non-zero current.

Isnt the top path going from Y = -L to L, and the bottom path going from Y = -L to L?

Ampere's laws states that
$$\oint\mathbf{H}\cdot d\mathbf{l}=I_{enclosed}$$
Using $K$ to represent the surface current density, this becomes
$$H_{x}L+(-H_{x})(-L)=2H_{x}L=KL$$
So the magnitude of the field is
$$H_{x}=\frac{K}{2}$$

 

[yosimba2000]

I think I see it. I was trying to add the magnitudes without account for the directions.