Magnetic field strength infinitely long wire

[wellmax]

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

Homework Statement

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).

Homework Equations

Local Maxwell equation for H-field

Curl(H) = J + δD/δt

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

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)]u_z (U are unit step functions)

Now for the Maxwell equation

Curl(H) = [1/r δ(r H_θ)/δr]u_z

Working this whole equation out by integrating over r gives me

H = H_θ*u_θ = [-rI/(π b²)U(r-b)]u_θ

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

 

[rude man]

What exactly are you looking for?

 

[wellmax]

What exactly are you looking for?

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

 

[rude man]

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

The H field where?

 

[wellmax]

The H field where?

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.

 

[rude man]

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.

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?

 

[DrZoidberg]

To get the average H field along a closed line you integrate the curl over the entire area inside that line and divide 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.

 

[rude man]

π

To get the average H field along a closed line you integrate the curl over the entire area inside that line and divide 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.

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 H_θ which I offhand don't know how to solve, even if I knew the boundary condition. Of course, I do know H_θ 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!

 

[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)u_θ

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

 

[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 + zu_z
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.

please help

 

[rude man]

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 + zu_z
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.

please help

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?

 

[wellmax]

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?

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

H_θ1 = 1/2 r₁J -> H₁ = 1/2 J [cross] r₁
H_θ1 = 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))(-r_yu_x + r_xu_y)

 

[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.)

 

[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.

 

[rude man]

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.

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.

 

[wellmax]

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.

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.

 

[rude man]

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.

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.

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.

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.

 

[wellmax]

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.

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 H_θ component of H
should I split this problem up into r<a a<=r<=b r>b??

 

[rude man]

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

I don't think it makes any difference. Once you have figured out how to do the hole field the rest is pretty much the same thing.

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 H_θ component of H
should I split this problem up into r<a a<=r<=b r>b??

Whew, one at a time ...

In these last two problems, is there still a hole off-center somewhere? Because if the configurations are perfectly symmetrical w/r/t theta then just use Ampere directly.

 

[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

 

[rude man]

Yes I see I can just use Ampères law again

Yes, the last 2 problems are a piece of cake compared to the one with the hole.

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

We may be on the same page on this. I suggest you proceed with your solution & show me what you came up with.

Here's doing it my way:
frame #1 (unprimed): origin at wire center, hole center on x axis, wire direction z axis, cylindrical coord:
H1A = 1/2 J x r = Jr/2 θ , 0 < r <= b
H1B = Jb²/2r θ, r > b
θ is the theta unit vector

frame #2 (primed): origin at hole center (x = +r0 in #1 frame), otherwise same as frame #1:
H2A = -Jr'/2 θ' , 0 < r' <= h
H2B = -Jh²/2r' θ' , r' > h.

Hole field H = H1A + H2A.

As I said, the remaining problem is to translate H2A into the #1 frame so it can be added to H1A. I would take your instructor's hint and use cartesian coordinates to make the transition. I haven't done the summation myself but I suspect it's the best way.

One thing you should discover is that the H field inside the hole is constant from its left to right edge, going along the x-axis from x = r0 - h to r0 + h. Using the θ vector notation it's easy to show both the magnitude and direction of H along this diameter.

 

[wellmax]

Yes, the last 2 problems are a piece of cake compared to the one with the hole.We may be on the same page on this. I suggest you proceed with your solution & show me what you came up with.

Here's doing it my way:
frame #1 (unprimed): origin at wire center, hole center on x axis, wire direction z axis, cylindrical coord:
H1A = 1/2 J x r = Jr/2 θ , 0 < r <= b
H1B = Jb²/2r θ, r > b
θ is the theta unit vector

frame #2 (primed): origin at hole center (x = +r0 in #1 frame), otherwise same as frame #1:
H2A = -Jr'/2 θ' , 0 < r' <= h
H2B = -Jh²/2r' θ' , r' > h.

Hole field H = H1A + H2A.

As I said, the remaining problem is to translate H2A into the #1 frame so it can be added to H1A. I would take your instructor's hint and use cartesian coordinates to make the transition. I haven't done the summation myself but I suspect it's the best way.

One thing you should discover is that the H field inside the hole is constant from its left to right edge, going along the x-axis from x = r0 - h to r0 + h. Using the θ vector notation it's easy to show both the magnitude and direction of H along this diameter.

Alright, I just considered r and r' to be vectors pointing from the two origins to a certain point in the H-field (inside the hole). So r points from the middle of the wire to a certain point inside the hole and r' points from the center of the hole to the same certain point inside the hole.
Now r₀ is the vector from the center of the wire to the center of the hole so from this it follows that r₀ = r - r'.
And because the current density is uniform the H-field will be like this

H = H + H' = 1/2 J x r₀

I don't think I can give a better explanation of my solution than this, the thing I'm not so sure about is if I may consider r and r' to be Cartesian vectors?

 

[rude man]

I have good news! I just computed the net H field inside the hole also and guess what, I got
H = (J/2)*r0 j where j is the unit vector in the y diection.

Not only is the H field the same along the diameter I mentioned - it's constant everywhere within the hole, both in magnitude and direction! And that is exactly what you got also, since the vector r0 = r0 i in the first frame (origin at wire center). So 1/2 J x r0 = J*r0/2 j !

Certainly, r and r0 are cartesian vectors when expressed as r = x i + y j and the way I did it, r0 = r0 i.

I owe you one for pointing out H = 1/2 J x r since I never saw Ampere's law written that way. Athough, it's limited to uniform J within the circle described by r. It could be esily extended to cases where the current density is a functon of r but not θ: H(r) = (1/r)∫₀^r J(x)x dx which reduces to J(r)r/2 if J is uniform.

 

[wellmax]

I have good news! I just computed the net H field inside the hole also and guess what, I got
H = (J/2)*r0 j where j is the unit vector in the y diection.

Not only is the H field the same along the diameter I mentioned - it's constant everywhere within the hole, both in magnitude and direction! And that is exactly what you got also, since the vector r0 = r0 i in the first frame (origin at wire center). So 1/2 J x r0 = J*r0/2 j !

Certainly, r and r0 are cartesian vectors when expressed as r = x i + y j and the way I did it, r0 = r0 i.

I owe you one for pointing out H = 1/2 J x r since I never saw Ampere's law written that way. Athough, it's limited to uniform J within the circle described by r. It could be esily extended to cases where the current density is a functon of r but not θ: H(r) = (1/r)∫₀^r J(x)x dx which reduces to J(r)r/2 if J is uniform.

Thank you very much, this cleared up a lot of my misconceptions