Biot-savart law question for an infinite sheet

[Artori]

Homework Statement

A filamentary conductor carrying current I in the az direction extends along
the entire negative z axis. At z=0 it connects to a copper sheet that fills the
x>0,y>0 quadrant of the xy plane. (a) Set up the Biot-Savart law and
find H everywhere on the z axis; (b) repeat part (a), but with the copper sheet
occupying the entire x y plane (Hint: express aφ in terms of ax and ay and
angle φ in the integral).

Homework Equations

dH=(Idl X ar)/(4pi*r^2)

The Attempt at a Solution

Since it wants the H on the z axis I am going to ignore the filament since it is on the z axis.

The current from the filament "I" goes into the sheet so the surface current "K" will be I/(pi/2) since φ ranges from 0 to pi/2 being on the positive xy axis.

H=∫∫(I/(pi/2)ap X (zaz-pap)/(z^2+p^2)^(1/2))/(4pi*p^2)pdpdφ

H=I/(2pi^2)∫∫z/(p(z^2+p^2)^(1/2))aφ with φ=0 to pi/2 and p=0 to infinity for the bounds of the integrands

I did the integral with wolfram alpha and it obviously isn't right. The answer is supposed to be H=I/((2pi^2)z) (ax-ay) A/m for part a and 0 for part b, any help would be greatly appreciated.

 

[anathema]

Your approach seems correct, but there are a few mistakes in your integration. First, the integral should be in terms of z instead of p, as z is the variable along the z-axis. Also, the limits of integration for z should be from -∞ to ∞, as the sheet extends along the entire negative z-axis. Finally, the surface current should be K=I/(pi/4), as the sheet occupies the entire x>0, y>0 quadrant.

So, the correct integral should be:

H=∫∫(I/(pi/4)ap X (zaz-pap)/(z^2+p^2)^(1/2))/(4pi*p^2)dzdφ

H=I/(2pi^2)∫∫z/(p(z^2+p^2)^(1/2))aφ with φ=0 to pi/2 and p=0 to infinity for the bounds of the integrands

With these corrections, you should get the correct answer of H=I/((2pi^2)z) (ax-ay) A/m for part a. For part b, the sheet now occupies the entire xy plane, so the surface current should be K=I/(pi), and the integral would be:

H=∫∫(I/(pi)ap X (zaz-pap)/(z^2+p^2)^(1/2))/(4pi*p^2)dzdφ

H=I/(2pi^2)∫∫z/(p(z^2+p^2)^(1/2))aφ with φ=0 to pi/2 and p=0 to infinity for the bounds of the integrands

This integral would give you H=0, as the contributions from the sheet cancel out in all directions.