# Energy In B field of infinite line current

1. Jul 20, 2011

### Onufer

My professor had an example in class the other day where we had an infinite line of current uniformly distributed over a circular wire with radius R. He wanted to look at the energy of the B field produced by this current. He did the normal thing and integrated 1/2u*B^2 over all space. It obviously diverges since it is an infinite wire. But the interesting thing is that it also diverges if you look the energy per unit length. (it ends up being some constants *(ln infinite - lnR) for the energy of the field outside the wire)

My intuition (which is often not right) felt that it shouldn't be infinite per unit length. I looked at how Griffiths derived the B^2 energy equation and it looks like it uses the assumption that the surface integral of AXB goes to zero as the radius of the surface goes to infinite...

I have two big questions:

1) which formula for energy of a mag field is more fundamentally right? (less assumptions about the nature of the problem)

(my proff said the 1/2u*B^2 is most fundamental but it seems like all the proofs i can find do integration by parts and assume the surface integral goes to zero as space goes to infinite to get that equation)

2) When you solve for A you get an integration constant in both parts. (inside and outside the wire) How do you decide what the integration 2nd constant should be?

Obviously A needs to be continuous a have a continuous derivative (since there are no surface currents) over all space. The 2 functions I got for A were an S^2 function and an ln(s) function The functions as calculated I had 2 constants and one boundary condition A (must be continuous) so there is another constant that we cant calculate. In electrostatics we often say V=0 at the surface for a problem like this but since A does not have much of a physical meaning. This A constant DOES matter though because of the equation W=.5*(intigral of A*J over all space) (Griffiths 7.31 pg 318) . J is our current density so it is constant and 0 outside of R. But if A has an arbitrary integration constant this integral is completely dependent on that constant which makes no sense since there would be different amounts of energy depending on an arbitrary integration constant.

Also I may as well show my equations for A and B:

u=mu not
I= total current
s is the radius for cylindrical coordinates

Inside:
A =-u*I*s^2/(4*pi*R^2)+C1 in the Z direction (assuming current is flowing up)
B = u*s*I/(2*pi*R^2) in the phi direction

Outside:
A = -u*I*ln(s)/(2*pi)+C2 in the z direction
B = u*I/(2*pi*s) in the phi direction

I hope that was reasonably clear.....
Thanks in advance for the help!

2. Jul 20, 2011

### Onufer

Btw all Griffiths equations mentioned and the derivation of B^2/(2u) were on page 318

3. Jul 20, 2011

### nnnm4

The formula that keeps the surface integral in correct in the general case for any magnetic field of any form. The formula that neglects it is only true if the field goes to zero at the boundaries (infinity). Which one is more fundamental? I don't know really what that would entail. I will say though that the latter formula is much nicer and is *true in any physically realizable problem in electrodynamics*.

4. Jul 21, 2011

### Onufer

Cool. That's what i thought too but wasn't sure. Applying that volume and surface integral equation i get:

W = 1/(2u) *(integral B^2 over inside volume + integral B^2 over outside volume - surface integral of AxB)

I'm gonna divide everything by z since i was interested in the energy per unit length.

the first volume integral ends up being (after dividing by z)

u^2*I^2/(8*pi)
Its kinda cool that it has no dependence on the radius of the wire...

For volume outside B^2 integral i will integrate from R the outside radius to some arbitrary s value which will be denoted by s'
This gives

u^2*I^2/(2*pi)(ln(s') - ln(R))

The last surface area integral I will do at the radius s' the enclosing the previous volume integral. It is important to note the top and bottom of this cylindrical area contributes zero since BxA points radially while da points up thus there dot product is zero. For all other spots the BxA lines up perfectly with da allowing to simply multiply their magnitudes while observing the sign. If A is negative (which it will be if we go out far enough) AxB points outward so the integral will be positive but there is a negative sign in front of it...

-u^2*I^2*ln(s)/(2*pi)+C2*I/2

notice that the first part here EXACTLY cancels out the first part of the outside volume integral. (the part that was diverging when we tried to integrate B^2 over all space). The bounds of the integral are irrelevant as long as we are outside the wire.

Summing these integrals gets

W/z (energy per unit length)= u*I^2/(16*pi) - u*I^2*ln(R)/(4*pi) + C2*I/2

Here is why I am interested in the integration constant of A... The energy seems oddly to depend on it. I am hoping there is a justifiable boundry condition on C2 that it would be a function of R and do some canceling with the middle term to get a reasonable answer (the problem goes crazy if you have a small radius)

Is there a boundary condition I am not thinking or have I taken the math too far out of the realm of physics... I have learned a lot from playing around with this problem but i would really like it to work out to something meaningful....

If anyone has any thoughts I would really appreciate it.

5. Jul 21, 2011

### Dickfore

If you are worried about these kinds of infinities, please calculate the total energy of the electrostatic field of a point charge.

6. Jul 21, 2011

### Onufer

I would say its zero because it takes zero energy to move a point charge in from infinite since its field does not effect itself and we generally consider the energy of a charge distribution to be the energy required to assemble the system. probably wrong though.

I'm not uncomfortable with infinities, its just that for a similar problem in electrostatics, with an infinite line of uniformly bound charge, i don't think the energy of that system goes to infinite per unit length. Thus why would this problem?

In the electrostatic problem we would generally make the assumption that V = 0 a the surface allowing us to define the integral. I was just wondering if there is a similar condition we can use for the magnetic vector potential.