# Ampere's Law and two conducting parallel plates

• latentcorpse
In summary, the magnetic field inside and outside of the plates is due to the current flowing through the plates. Theargument that the field is zero outside of the plates does not hold up because the amperian loops should cross the plate and have 2 sides parallel to the plate.f

#### latentcorpse

hey sorry for the prolific posting but exam's are coming up and I'm just working through past papers.

two conducting parallel plates have uniform uniform current densities flowing through them parallel to their surface. Use Ampere's Law to find the magnetic field inside and outside the plates.

im quite confused as to where to take my loops here in order to find the field at all the various places I am reuqired to. any suggestions?

also why does the following argument not work here:
integrate round an amperian loop that doesn't intersect either plate and get
B(r)-B(r+dr)=0 as no enclosed current
gives B(r)=B(r+dr) implies B(r)=0 but B goes to 0 at infinity so B is zero outised the plates?

What direction do you think the field is in? Use symmetry.

surely it depends on the direction the two currents are flowing in - if they're opposite then they will cancel. I am not sure abou where to put the magnetic field lines when its a plane and not like a wire or something.

also why does my arhument about the field being 0 outside break down here - isn't that legitimiate for showing the field is 0 outside of a solenoid?

Use superposition. Find the field due to one plate, then the other (which will be either symmetric or anti-symmetric) and then add them up.

ok but where do i place my amperian loops

and why is the field not 0 outside the plates according to me argument at the end of post 1?

The ampere loops should be rectangles that cross the plate, and have 2 sides parallel to the plate.

I don't think I get what you mean by your previous argument.

ok so if i integrat round said loop for the top plate i get

int(B.dr) = mu_0 int(K.dA)
int(B.dr)= BL -0.L where L is the length of the loop

so B =mu_0/L int(K.dA)
hows that?

surely my previous argument proves there is no field outside a solenoid? why can't i use it similarly here?

This isn't a solenoid, it's a plate...

Uh. Int(K.dl) not dA since K is a surface current, you integrate over a line.

Draw the thing and see that in the loop, B moves to the right above the plate, and to the left below the plate. Thus B is parallel to dl through the top and bottom lines and perpendicular to dl in the left and right lines.

What do you get for that?

BL-B(-L)=mu_0 KL implies B=mu_0 K/2 from each plate presumably

two questions:
(i) are the field lines from these plates similar to those for a current loop magnetic dipole? i.e. they run parallel to the surface and then curve back round onto the other end of the surface?
(ii)we aren't told the directions of the two currents so how do we know whether they superimpose or cancel each other out?