Magnetic field strength infinitely long wire

1. May 5, 2013

wellmax

I'm new to posting on this forum so excuse me if I'm using the wrong symbols
underlined := vector/vectorfield
δ := derivative
uvar := unit vector

1. The problem statement, all variables and given/known data
Consider a finitely conducting wire of infinity length with radius b. The wire is centered about the z-axis. Current is flowing through the wire in opposite z-direction and may be characterized by stationary uniform current density with magnitude |J|=I/(π b2).

2. Relevant equations
Local Maxwell equation for H-field

Curl(H) = J + δD/δt

δD/δt = 0 because D does not change over time

3. The attempt at a solution
I described J using cylindrical coordinates (r, θ and z)

J = -I/(π b2)[U(r+b)-U(r-b)]uz (U are unit step functions)

Now for the Maxwell equation

Curl(H) = [1/r δ(r Hθ)/δr]uz

Working this whole equation out by integrating over r gives me

H = Hθ*uθ = [-rI/(π b2)U(r-b)]uθ

but this cannot be right because the H-field should decrease in strength by 1/r :(
though the dimensions of the H-field are right

Also need to give the H-field in Cartesian coordinates and have no idea how to do that

Last edited: May 5, 2013
2. May 5, 2013

rude man

What exactly are you looking for?

3. May 5, 2013

wellmax

Thank you for the quick reply.
I am looking for a derivation of the H-field when the current density is given.

4. May 5, 2013

rude man

The H field where?

5. May 5, 2013

wellmax

For every point with distance r to the wire.
I'm just trying to analytically derive an equation for H with a given J, using the differential form of the Maxwell-Ampère equation.

6. May 5, 2013

rude man

H inside the wire, or outside, or both?
For r > b, current = 0.

Regardless, why don't you just use Ampere's law? Are you told to solve the Maxwell relation?

7. May 5, 2013

DrZoidberg

To get the average H field along a closed line you integrate the curl over the entire area inside that line and devide the result by the line's length. But outside the wire the current density and therefore the curl is zero. So if you draw a circle around a wire the integral of the curl in the entire circle is equal to the integral of the curl inside the wire. In other words it's equal to the current.

8. May 5, 2013

rude man

π
Easy way to find H in this case is Ampere's law: ∫ H*ds = I integrated over a closed path, the area of which is cut by I. You have perfect symmetry with an infinitely long wire and the current density is uniform within the wire, so we get 2πrH = I. This is valid inside and outside the wire. Of course, inside the wire I < I0 where I0 is the total wire current.

OK, but you want to solve curl H = J instead. I have looked at this and I wind up with a nonlinear 1st-order ODE in Hθ which I offhand don't know how to solve, even if I knew the boundary condition. Of course, I do know Hθ from Ampere's law so I could cheat. But I assume you don't want to.

Which is why God gave us Ampere's law!

9. May 5, 2013

wellmax

Yes, I have used the integral form of the Ampère law but for less obvious problems than this one the integral will become very hard to solve (I think) so I want to know what I'm doing wrong when using the differential form.
They should yield the same outcome but for me they don't.

I found using the integral form

H = -I/(2 π r)uθ

to be the H-field outside the wire but when I use the differential form I get the equation in my original post.

And what would be the difference between the H-field inside the wire and outside the wire? The H-field can't be zero outside the wire right?

Also along the z-axis the H-field equals zero.

EDIT: Thanks rude man :)
When you say it like that I see what I did wrong with the differential form (I didn't get the ODE at first, dumb mistake).
Thanks for helping everybody :D

10. May 6, 2013

wellmax

I have one (or maybe two) more question on the same subject and setting so I just post them in this thread.
The next exercise goes like this
"
Subsequently, let us consider a new situation in which there is a cylindrical hole of radius h centred about the line
r0 + zuz
parallel to the z-axis. Here, r0
is a position vector in the xy-plane with |r|0 + h < b, such that the hole
does not extend to the boundary of the ﬁnitely conducting cylinder (in which the current density distribution is still considered to be uniform). This situation is depicted in the ﬁgure in the middle.
Determine the magnetic ﬁeld H(r, t) inside the hole. Hint: employ the superposition principle, in cartesian coordinates.
"

I find it hard to visualize how the superposition will be used in this problem.
I already tried to use Ampère's law again and just subtract the current that is not flowing through the hole anymore from the total current and work out the integral but it leads me nowhere.
Also subtracting the H-field caused by the current that used to be flowing through the hole from the previously found H-field doesn't work.

Last edited: May 6, 2013
11. May 6, 2013

rude man

You can't use Ampere's law in this case directly because there is no symmetry to give you a constant H field around any closed contour of your choice.

Big hint: consider analyzing by replacing the hole with a current density equal in magnitude to the given one but in the opposite z direction, and leaving the original distribution as is, then use superposition as suggested by analyzing the two distributions separately & then adding the result. Does that give you the hotly-desired symmetry in each case so you can use Ampere?

12. May 6, 2013

wellmax

Alright I think I got it but not sure if it's correct.

I replaced the magnitude of the initial current density by

|J| = I/(πh2-πb2)

and I got two H-fields

Hθ1 = 1/2 r1J -> H1 = 1/2 J [cross] r1
Hθ1 = 1/2 r2J -> H2 = 1/2 J [cross] r2

where r1 and r2 are the distances to the centers of the two surfaces.
Now adding these two H-fields gives me

H = 1/(2π(b2 - h2)) I [cross] r0

with r0 the position-vector of the center of the hole

What I'm not sure about is if this is in Cartesian coordinates or not?
In Cartesian coordinates it will look like this

H = I/(2π(b2 - h2))(-ryux + rxuy)

Last edited: May 6, 2013
13. May 6, 2013

rude man

Does not make sense to me.

Your current density J does not change for either computation (H1 and H2), except in the hole the DIRECTION is reversed.

Step 1: pretend the hole isn't there and come up with H1 using Ampere's law for both the inside & outside of the wire.
Step 2: now pretend the wire current density is zero but the hole current density is the negative of the current density of step 1. Repeat step 1 to get a second field H2. Again, H2 is valid for both inside & outside the wire.

At this point you should be able to appreciate the fact that the SUM of steps 1 and 2 give you the actual situation: a wire carrying current density J with a hole of radius h running parallel to its axis but off-center by a distance r0.

Now just add H1 and H2. Your big job here is to come up with an expression for the sum of H1 and H2. I'm not sure I would use cartesian coordinates but I haven't done it. Thing is, H1 will be in terms of polar coordinates with the origin at the wire axis whereas H2 will be defined by polar coordinates with origin at the hole's center. Your job is to convert the summation referenced to just one coordinate system, be it cartesian or polar.

(Note: I say 'polar' instead of 'cylindrica' because the H field of course will not have a z component.)

14. May 7, 2013

wellmax

I think that is exactly what I did. I just subtracted the area of the hole from the area of the wire and did the entire equation using this current density.
The thing I was not sure about was the cross-product I used to turn the scalar into a vector.

15. May 7, 2013

rude man

I don't think this is what I had in mind.

There is no need to take any cross-products. You simply apply Ampere's law twice, once for the entire cylinder current as though the hole wasn't there, and once for the hole current as though the cylinder wasn't there. Then you add the two H fields.

16. May 7, 2013

wellmax

Ok but how do I know the direction of the total H-field? I can't use cylindrical coordinates because I would have two bases.
So I take the cross-product of the current density and the position-vector of the point I want to know the H-field of (with respect to the center of the hole or wire) and then I add the two resulting H-fields.

17. May 7, 2013

rude man

I said your big job will be to translate the second (the hole current) H field into the coordinate system of the first H field so they can then be added together.

This is a problem in analytic geometry, purely.

In the region outside the wire that would compute to zero since J = 0 there, yet the H field there is not zero! You are required to find H everywhere, inside the wire, outside the wire, and inside the hole.

And anyway I don't see how you're evading the problem of joining H fields with different coordinate systems this way.

18. May 7, 2013

wellmax

Yes but I only had to find the H-field inside the hole so I thought this was sufficient?

Now I have the same question but with two coaxial cylindrical shells with equal current and thus different current density. I thought I would be able to solve this one after solving the one with the hole in it but no :<

"
Finally, let us consider the case of two thin ﬁnitely conducting concentric cylindrical shells. In the inner shell, occupying the region a-d/2<=r<=a+d/2, the uniform current density points in the positive z-direction and has
magnitude |J| =I/(2πad) throughout. In the outer shell, occupying the region b-d/2<=r<=b+d/2, the uniform
current density points in the opposite z-direction and has magnitude |J|=I/(2πbd) throughout. This situation is
depicted in the ﬁgure on the right.
Determine H' for 0 < r < Infinity.
"
I tried the same method as in the previous problem but it doesn't give me an Hθ component of H
should I split this problem up into r<a a<=r<=b r>b??

19. May 7, 2013

rude man

20. May 7, 2013

wellmax

Yes I see I can just use Ampères law again

Now I still don't get what I'm doing wrong with the problem about the hole in the wire