Why can't we use Ampere's Law?

[Kyle Nemeth]

We are asked to find the magnetic field at point P, all of the quantities in the figure are known values and the current density is uniform. One way to solve this problem is by modeling the sheet as a collection of infinitely long wires, with each wire contributing an amount of magnetic field dB and then integrating to find the total magnetic field. Why is it that this approach must be used and not an approach involving Ampere's Law directly?

 

[Charles Link]

See https://www.physicsforums.com/threads/magnetic-field-of-an-infinite-sheet-of-current.475918/ It is apparent that Ampere's law works for this case.

 

[Kyle Nemeth]

I am familiar with this problem, the Amperian Loop is chosen as a rectangle within the slab with a height smaller than the thickness of it (to find B at a point within the slab) and can then be chosen as a rectangle with a height larger than the thickness of the slab (to find B at a point above the slab) and this is okay because the magnetic field is perpendicular to the length elements along the height, so the integral on the left side of Ampere's Law is 0. The plane slab actually extends infinitely in two dimensions rather than just one as in my problem, I apologize I should have specified that the sheet is "thin" so that it has no thickness.

 

[Charles Link]

My mistake: I see you have a finite width $w$ to the sheet. You don't have enough symmetry on the integral path for Ampere's law to supply the answer. The $B$ in the integrand is non-uniform.

 

[Kyle Nemeth]

Awesome, thank you, I've thought about this also, so if we choose a circle as our loop, or even a rectangle (so that the entire cross-section of the sheet is enclosed) it is true that both would be anti-symmetric with the geometry of the magnetic field formed by the sheet?

 

[Charles Link]

Awesome, thank you, I've thought about this also, so if we choose a circle as our loop, or even a rectangle (so that the entire cross-section of the sheet is enclosed) it is true that both would be anti-symmetric with the geometry of the magnetic field formed by the sheet?

Not anti-symmetric, but asymmetric. If it were simply anti-symmetric, then $\oint B \cdot dl=0$. When it is asymmetric, $B$ is non-uniform, and can't be removed from the integral in any fashion.

 

[Kyle Nemeth]

ASYMMETRIC, okay, well understood, thank you for your response.