Magnetic field of an infinite cyliner of fixed magnetization

[gabbagabbahey]

Okay, so using r''=\sqrt{r^2+r'^2-2rr'cos(\phi)+(z-z')^2} you'll still want to integrate over \phi first. divide the integrals up to treat the x and y components separately. The integrand of the y-component terms are even function and are integrated over a full period, so they will be zero. The x-component terms will have a sin(phi) in the numerator, which is the derivative of cos(phi) so use a substitution like u=cos(phi).

 

[insynC]

Don't apologise, it is only with some pushes in the right direction I've been able to make any progress whatsoever.

Integration approach:

I used the cosine rule, letting the point A be (r, phi, z) and the general point of integration (say some volume element) be (r', phi', z'), then I defined phi' from the line of the center of the circle to A and got:

|r''|^2 = (z-z')^2 + r^2 + |r'|^2 -2rr'cos(phi')

This arrangement gives \hat{}phi = sin(phi)\hat{}y - cos(phi)\hat{}x

This seems reasonable, but it then needs to be used in the following integral:

Integral(-k\hat{}phi/(r'')dV') + Integral(kR\hat{}phi/(r'')dS)

Both of these integrals involve a z integration from - to + infinity, which doesn't work out nicely. But the alternative is to do the phi integration first.

On the top line I broke up the sin(phi) y, and the cos(phi) x so there are two separate integrals. The sin one of these I can do, but the cos one becomes (simplifying non phi based terms to constants a & b):

Integral from 0 to 2pi (cos(phi)/SQRT(a-2bcos(phi)) dphi)

This I cannot calculate and I am not sure how to proceed.

Ampere's Law approach:

How did you go from:

Curl(B) = mu Curl(H) + mu Curl(M)

to B = mu M ?

All I can show is that Curl(H) = 0, so Curl(B) = mu Curl(M) = -k mu \hat{}phi

From this I struggle to determine what B is in the z/r plane.

 

[granpa]

curl(H)=free current?
curl(M)=bound current?

 

[gabbagabbahey]

The integral with the cosine in the numerator will be zero (see my previous post)

Use the integral version of Ampere's Law:

\oint \vec{H} \cdot d\vec{l}=I_{free_{enclosed}}

Since there is no free current anywhere in the problem, the integral will be zero for any Amperian loop. The only way that can happen is if \vec{H}=0.

So using, \vec{H}=\frac{1}{\mu_0}\vec{B}-\vec{M}=0, you get my above answer.

 

[insynC]

gabbagabbahey - I haven't done a math's course for a while, I'm pretty sure your right that cos(phi)/SQRT(a-bcos(phi)) is even in phi, but how do I show this?

Also is the reason = 0 when integrated from (0,2pi) because due to spherical symmetry you could also integrate from (-pi, pi)?

 

[gabbagabbahey]

Yes, you could integrate from -pi to pi instead.

 

[insynC]

But how do you know the function is even?

 

[gabbagabbahey]

cos(-x)=cos(x)

 

[gabbagabbahey]

I think that to deal with the z integration you will have to integrate from -L to L and take the limit of \vec{A} as L approaches infinity.

 

[insynC]

Ah... woops.

So then the integral cos(phi)/SQRT(a-bcos(phi)) goes to 0.

I just did the other integral - namely sin(phi)/SQRT(a-bcos(phi)), and I also found that when integrated from (0,2pi) it also goes to 0.

Therefore both terms go to 0, and so A = 0 => B = Curl(A) = 0.

But from Ampere's law didn't we just find out that B wasn't 0?

I'm probably just getting confused now, been working on this problem for about a week.

 

[granpa]

I think you guys are going to kick yourselves. it appears to me that there's a trivial solution. just think of magnetization as an infinitesimal loop of current. the cylinder then becomes a bundle of infinitesimal solenoids. the field in each one proportional to the magnetization at that point.

field=r

apparently the field inside a solenoid depends only on the current not the radius of the solenoid.

(but magnetization=current*area of loop)? I must have done something wrong again.

no wait. magnetization is per unit volume so area of loop=1. I did it right. whew.

 

[insynC]

Granpa - Even if that is so, the question still wants you to find it using the bound volume currents and bound surface currents, so still need to use that expression for A.

Also second part of the question wants you to find it using Ampere's law.

But at least you've shown that B isn't 0! This leaves me worrying about what's gone wrong with my integrals though.

 

[insynC]

Also Granpa, does that mean you would get the solution B = kr mu in the z direction?

 

[granpa]

I'm pretty sure its proportional to r and in the z direction. the other symbols I don't know enough about.

but they are constants, right?

 

[insynC]

Thanks for all the help, I've certainly made progress on this question. I think I'll set out what's been achieved and what is still not clear.

What we know for sure:

granpa's argument clearly shows that B is proportional to r\hat{}z for r < R and B=0 for r > R.

As gabbagabbahey pointed out, using Ampere's law leads to the result that:

\vec{}B = k \mu_o \hat{}z

So it seems almost certain that this is the answer for B for r < R.

Also in terms of part (a) of the question, the following seems pretty certain:

r'' = SQRT( (z-z')^2 + r^2 + (r')^2 -2rr'cos(\phi) )

and that for the x-axis defined the line |OA| and phi anticlockwise:

\hat{}\phi = cos(\phi) \hat{}x + sin(\phi) \hat{}y

Also from basic cylindrical coordinate ideas:

dV' = r'dr'd\phi'dz' and dS' = R d\phi' dz'

Finally the terminals of integration to contain the entire sphere can be:

r: (0,R)
\phi: (-\pi, \pi)
z: (-\infty, \infty)

I'm pretty sure we know all these things for certain.

What is not so clear:

\vec{}A = \mu/(4\pi) { \intdV' (-k\hat{}\phi/(r'') ) + \intdS (kR \hat{}\phi/(r'') ) }

Evaluating this integral, despite knowing everything above still is the problem. As stated in my previous posts I keep getting zero for everything, which can't be correct.

Furthermore at this stage nothing about r has been defined. There must be some technicality that means that for r > R it is all 0, but for r < R it isn't. NB: in r'' in the surface integral, r'=R.

So showing that this integral gives an A such that curl(A) equals the result we've already found for B is what I'm struggling with.

 

[gabbagabbahey]

I think the problem is that 1/r'' has a singularity at the origin, so that the volume integral cannot be explicitly evaluated as-is.

 

[granpa]

" There must be some technicality that means that for r > R it is all 0, but for r < R it isn't"

if you are asking why the field outside a solenoid is zero, I've spent the last 24 hours trying to figure that out.

look at a loop of current. the magnetic field outside the loop depends not just on the current on the near side but also on the current on the far side. these two currents produce opposite fields that partially cancel but since they are different distances and directions away they never completely cancel.

in a solenoid things are weird. the external field due solely to the near current would be inversely proportional to the distance not from the wire on the near side but from the center of the solenoid (same as an infinite hollow line of charge). symmetry forces it to do this. evidently the same holds true for the field due to the current on the far side so they completely cancel.

thats all I can figure out.

I hinted about this below.

 

[gabbagabbahey]

I found this problem in Griffith's Introduction to Electrodynamics, and the solution for part (a) that the Author gives, doesn't bother evaluating the integrals, but instead uses Ampere's law directly for B. I think that is all that is required for this question. Remember,

\oint \vec{B} \cdot d\vec{l}=\mu_0 I_{enc}

In this case, the current enclosed inside the cylinder is just \int \vec{J}_b \cdot d\vec{a} where d\vec{a} is the vector area element of the surface bounded by your Amperian loop and zero outside the cylinder.

 

[gabbagabbahey]

Okay, so I got it backwards on which terms integrate to zero. The sin(phi) terms have odd integrands and so they integrate to an even periodic function which is evaluated over a full period and hence zero.

The cos(phi terms do not integrate to zero, and are actually VERY difficult to integrate. You end up with a hyper-geometric function after doing the phi integral. I think if you then carried out the z and r integrations you would end up with a non-zero vector potential. However, you would have to be somewhat of a masochist to do this IMO since using Ampere's law directly on B is much easier.

 

[insynC]

Thanks for your help, I think that's right.

The lecturer who set the question has a reputation of asking questions without trying them himself. I wouldn't be surprised if he never checked if the integrals were actually manageable.

As far as I'm concerned that answers the question, so thanks again to all who helped.

 

[gabbagabbahey]

Just to be clear, I think you should calculate the field using Ampere's law in terms of B and J_b for part (a). And then calculate it again using Ampere's law in terms of H for part (b).

I think the point of the question is to illustrate that using Ampere's Law in terms of H is often quicker than using it in terms of B; since for the former you don't need to know the bound currents, while for the latter you do.

 

[insynC]

Yeah that makes sense. Unfortunately, our lecturer laboured the expression for A that I was using, so I'm not sure what exactly was wanted.

But I'll do what you said, thanks.

 

[Cyb]

I have been struggling with this problem. I can get the equivalent expressions for most everything, what I can't understand is the Ampere's loop of the B field using the current density J_b. The differential area vector would always be in the z direction if we wanted B to be constant over the line, but J_b points in the phi direction. Thus J_b o da' will be zero, and then B=0 even when r<R. Can someone please help me??

 

[gabbagabbahey]

I have been struggling with this problem. I can get the equivalent expressions for most everything, what I can't understand is the Ampere's loop of the B field using the current density J_b. The differential area vector would always be in the z direction if we wanted B to be constant over the line, but J_b points in the phi direction. Thus J_b o da' will be zero, and then B=0 even when r<R. Can someone please help me??

Hi cyb, welcome to PF!

What are you using as your Amperian loop and why?

 

[Cyb]

I think I figured it out. I was using circles, with centers at the axis of symmetry for the cylinder. On thinking about it more I came to a rectangle, whose long sides are parallel to the cylinder's axis, the top was outside of the cylinder and the bottom inside. Thus, B=0 outside, and since B has to be in the direction of the axis the only non-zero part of the loop is the length inside. Then using my current density and surface current for the intersected area I came up with the correct answer. Does that seem right?

 

[gabbagabbahey]

I think I figured it out. I was using circles, with centers at the axis of symmetry for the cylinder. On thinking about it more I came to a rectangle, whose long sides are parallel to the cylinder's axis, the top was outside of the cylinder and the bottom inside. Thus, B=0 outside, and since B has to be in the direction of the axis the only non-zero part of the loop is the length inside. Then using my current density and surface current for the intersected area I came up with the correct answer. Does that seem right?

Sounds good!