# Relationship between line current & surface current density

1. Jul 24, 2017

### QuantumDuality

1. The problem statement, all variables and given/known data
In a configuration having axial symmetry about the z axis, a line current I
flows in the −z direction along the z axis. This current is returned at the
radii a and b, where there are uniform surface current densities Kza and
Kzb , respectively. The current density is zero in the regions 0 < r < b, b <
r < a and a < r.

(a) Given that Kza = 2Kzb , show that Kza= I/π(2a + b).
(b) Show that H is:
-I/2πr for 0<r<b
-Ia/πr(2a+b) for b < r < a

2. Relevant equations

I = lim|J|→∞ A→0S J ⋅da (1)
2 π r K⋅ in≈ ∫S J⋅da (2)
cH⋅ds = ∫S J ⋅da (3)
Kza = 2Kzb

3. The attempt at a solution

To get Kza, I used ∫S J ⋅da on the region a < r.
Because there is not current density on this region, the only contribution to the integral comes form the surface current densities, using (2):

S J ⋅da = 2π( b Kzb + aKza )

Substituting the value of Kza:

S J ⋅da = 2πKzb( b + 2 a)

As the problem is talking about Line current and surface current densities, I am assuming is a tick wire, but that is length is much bigger than its thickness. Therefore, (I think) I can equate (1) and (2)

2πKzb( b + 2 a) = I

Kzb = I/2π(2a + b)
Kza = I/π(2a + b)

For answer (b), On the region r < b, the only contribution to the magnetic field comes from -I, therefore:
cH⋅ds = - I
H = -I/2πr i for r < b

For the region b < r < a, the contribution to the integral should come from the surface current density for r = b and I, therefore:

H = (bKzb/r -I/2πr) iφ for b < r < a
H = (bI/2π(a+b)r -I/2πr) iφ for b < r < a

I don't understand why for b < r < a Kza should be taken into account, because r doesn't necessarily have to be near a

2. Jul 26, 2017

### rude man

Having a hard time seeing the setup.