Magnetic field of an infinite cyliner of fixed magnetization

[insynC]

I'm taking a second year undergraduate course in electromagnetism and have a question that has arisen in the course of my studies.

I'm normally reluctant to seek help on such matters, but this question has caused me a lot of grief.

The question is as follows:

An infinitely long cylinder, radius R, has a fixed magnetization, parallel to the cylinder axis: $$\vec{M}$$ = kr $$\hat{z}$$ , where k is a constant, and r is the distance from the axis of the cylinder (the z-axis).

(a) Determine the bound volume currents and bound surface currents, and hence determine the magnetic field inside and outside the cylinder.

(b) Use Ampere's Law (for H and then determine B) to check the result of part (a).

Attempted solution:

In approaching (a) I reckon the right way is to use the following expression for the magnetic potential, which is derived from the most general expression:

A = (\mu)/(4\pi)*(integral(dV curl(M)/R) + integral(dV M\timesdS/R))

This is because the volume current is in the first integral and the surface current in the second.

I can determine R using a diagram of the cylinder, and by using cylindrical coordinate expressions I can convert everything into seemingly only an integral problem. Nonetheless my integrals keep diverging because of the integral over all z in both terms.

A possible way around this I thought was to integrate A from -z to z, and set the limit as z goes to infinity and then find B=curl(A), hoping the z's would disappear. No such luck and I'm rather stumped as to how to figure this out.

Ran into same integration problems in (b).

I'm sure the problem lies in a misunderstanding of the setup of the problem - I've found setting up the integrals in this electromag course to be very challenging.

Thanks to anyone who can help.

 

[granpa]

if M isn't the current and isn't the field then what is it? what does z hat stand for?

 

[gabbagabbahey]

@ granpa, $\vec{M}$ is the magnetization and $\hat{z}$ is the unit vector pointing in the z-direction.

@insync, Start with your expressions for the volume current and surface current, what do you get? What expression are you using for R?

 

[granpa]

Magnetic moment can be visualized as a bar magnet which has magnetic poles of equal magnitude but opposite polarity. Each pole is the source of magnetic force which weakens with distance. Since magnetic poles always come in pairs, their forces partially cancel each other because while one pole pulls, the other repels. This cancellation is greatest when the poles are close to each other i.e. when the bar magnet is short. The magnetic force produced by a bar magnet, at a given point in space, therefore depends on two factors: on both the strength p of its poles, and on the distance d separating them. The force is proportional to the product \vec{\mu}=\mathbf{p}\mathbf{d}, where \vec{\mu} describes the "magnetic moment" or "dipole moment" of the magnet along a distance R and its direction as the angle between R and the axis of the bar magnet. These equations are completely analogous to the case of electric dipole moment.so is it an infinitely small dipole (made of magnetic hypothetical monopoles) or an infinitely small current loop?

edit:its a loop

 

[granpa]

http://scienceworld.wolfram.com/physics/BoundCurrentDensity.html

http://scienceworld.wolfram.com/physics/BoundSurfaceCurrent.html

 

[insynC]

gabbagabbahey - upon reflection I can see that since R is defined as the radius of the cylinder in the question, it was poor notation for me to also use it in my expression for A. I'll refer to the R in the expression for A as r''. Clearly r'' is the magnitude of the distance from either a differential volume or surface element to the point of interest, where A is measured.

To be honest I'm not exactly sure how to express r''. Initially I considered a 2-dimensional cross section of the cylinder looking at the distance from point A, say (r, $$\phi$$, z) to a point dS or dV, say (r', $$\phi$$', z'). From that I could quite simply see:

|r''|^2=(r-r')^2 + (z-z')^2

But I'm pretty sure this is incorrect as I think r'' should also depend on $$\phi$$. I'm really struggling in picturing how to set up this integral.

Granpa - thanks for the expressions, though I didn't have any difficulty determining the bound and volume currents from the expression for the magnetisation.

I found the surface current density to be: kr$$\hat{}\phi$$ and the volume current density to be -k$$\hat{}\phi$$.

I'm pretty sure because of this precursor to the question, the way to find B is by using the expression for A I referred to in my original post. Furthermore it's an expression I've found in nearly all textbooks on electromagnetism - e.g. Jackson 3rd ed, bottom of page 197.

What is causing me grief is using my knowledge of the geometry of the situation to convert this expression into an integral that can be determined without diverging.

 

[granpa]

you know the field outside the cylinder will be zero, right?

 

[gabbagabbahey]

insync, let's start with the volume terms; your expressions for r'' and $\vec{J}_b$ look good to me. However, your volume integral shouldn't diverge; are you treating $\hat{\phi}$ as a constant?

 

[insynC]

Inasmuch as you can approximate the infinite cylinder as an infinite solenoid I was certainly expecting B=0 for r>R. Nonetheless I was more hoping that by setting up the integral correctly, results like this would fall out of the mathematics rather than trying to justify them heuristically.

I'm pretty sure that if I can work out r'' (distance from infinitesimal volume/surface element to point where A is measured) the problem will be resolved. Trying to determine it in terms of (r, phi, z) though is proving difficult.

 

[insynC]

gabbagabbahey - initially I was treating $$\vec{}\phi$$ as constant and said |r''|^2=(r-r')^2 + (z-z')^2. But I now think this is wrong, as $$\vec{}\phi$$ changes for different values inside the volume/on surface, so will change over integral. I'm currently trying to determine how I can resolve this, but no luck yet.

 

[gabbagabbahey]

Express it in terms of the Cartesian unit vectors, which are always constant (cylindrical and spherical unit vectors vary with position and shouldn't be treated as constants when integrating)

 

[granpa]

am I correct in believing that the current flowing around the center is constant at every distance from the center and that the surface current is in the opposite direction and equal to all the other currents combined?

 

[gabbagabbahey]

granpa, yes, you are; but that only tells you that the field outside is zero, it doesn't give you the field inside the cylinder.

 

[insynC]

Thanks for you help so far gabbagabbahey, I've certainly made progress. I've hit another snag though:

I firstly considered the cylinder in the r/phi plane (so cross section is a circle) with an arbitrary point for A and an arbitrary point for dV. Then I set up Cartesian coordinates so that A lay on the y-axis and the x-axis was orthogonal. I measured phi from the y-axis around to the radius from the center to dV. I could then see that:

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

If |O to dV| = r', and |OA| = r, I could also see:

|dV to A|^2 = r^2 + (r')^2 - 2rr'cos($$\phi$$) , by the cosine rule.

Then considering the cylinder in the r/z plane I could see the full distance from dV to A, r'' is given by:

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

I'm just trying to work out the volume integral at the moment, so I subbed in all these values into the integral. I ran into the integral:

integral(cos($$\phi$$)/SQRT(a-2bcos($$\phi$$))d$$\phi$$) , for constants a & b (they involve z and r).

I couldn't solve this integral, I might try integrating wrt to r first.

Am I on the right track? At least now I'm integrating with vector x & y, so it makes more sense in that way.

 

[gabbagabbahey]

Your original expression for r'' was correct; a simple vector subtraction gives

$$\vec{r}''=\vec{r}-\vec{r'}=(r\hat{r}+z\hat{z})-(r'\hat{r}+z'\hat{z})=(r-r')\hat{r}+(z-z')\hat{z}$$

the cylindrical unit vector $\hat{r}$ is perpendicular to $\hat{z}$, so

$$r''=\sqrt{\vec{r} \cdot \vec{r}}=\sqrt{(r-r')^2+(z-z')^2}$$

Your expression for $$\hat{\phi}$$ is correct; but what happens when you integrate $sin(\phi)$ or $cos(\phi)$ from 0 to 2pi?

 

[insynC]

If I take:

r'' = SQRT( (z-z')^2 + (r-r')^2 )

Then when I integrate the sin and cos components from 0 to 2pi, as you suggest, I will get 0. I feel it hard to believe that the answer is in fact A = 0 everywhere, so there must then be another phi dependence somewhere. If it is not in r'', then I'm not really sure where it is coming in from.

 

[gabbagabbahey]

Actually, I'm pretty sure that it is zero everywhere for the specified magnetization. Of course, you can check this using Ampere's Law in part (b).

 

[insynC]

Fair enough, I guess I just need to trust the maths is right - which it seems to be. I'll have a look at (b) to see if I can confirm this.

Thankyou for all your help.

 

[gabbagabbahey]

Yes, for a finite cylinder, it wouldn't necessarily be zero because you would have to add the surface terms at the ends of the cylinder. But for the infinite cylinder, I think it is zero.

 

[granpa]

normally you think of the field as decreasing as a function of distance from the wire but in the case of an infinite solenoid, due to the symmetry of the situation, the external field due to the current on each side would be inversely proportional to distance from the center. and of course equal and opposite to each other.

thats weird

 

[insynC]

Ampere's law also gives B = 0 both inside and outside the cylinder. Certainly not what I was expecting.

Thanks again for the help.

 

[insynC]

I'm still not 100% sure about the derivation of this result. Firstly I had some concerns about the way I used Ampere's law to show B=0, so I'm now not sure about that. When I looked over the derivation using plain the magnetic potential the following step concerns me:

In finding r'' = SQRT( (z-z')^2 + (r-r')^2 ), we used r'' = r - r'.

Does simple vector subtraction apply for vectors in cylindrical coordinates? I'm not convinced it does at least on geometrical grounds.

 

[gabbagabbahey]

Why wouldn't it? $\hat{r}$ and $\hat{z}$ are perpendicular.

How did you apply Ampere's law? Why are you concerned?

 

[insynC]

This is why I'm concerned:

In Cartesian coordinates vector subtraction is as follows: $$\vec{}a$$ - $$\vec{}b$$ = (x1,y1,z1)-(x2,y2,z2) = (x1-x2) $$\hat{}x$$ + (y1-y2) $$\hat{}y$$ + (z1-z2) $$\hat{}z$$ =(x1-x2,y1-y2,z1-z1). This works because both vectors a & b have the same direction for unit vectors x y & z.

But in cylindrical coordinates, a general vector a = (r1,phi1,z1) = r1 $$\hat{}r1$$ + phi $$\hat{}phi1$$ + z1 $$\hat{}z1$$ , and the direction of $$\hat{}r1$$ is not necessarily the same direction as for some other vector b. Consequently you can't simply do vector subtraction of the form (r1-r2) because the unit vectors are different.

I did Ampere's law initially with a loop in the r/phi plane (so cross section is a circle). I though this demonstrated that B = 0, but in truth all this demonstrated was that there could be no B directed radially or in phi direction, it didn't rule out having a B directed in the z direction (which obviously if B was directed anywhere this is where it would be directed).

So you need to consider a loop in the z/r plane. In this case the integrals don't go to zero, and although I haven't yet worked out what they give B as, it is not clearly 0.

Btw the version of Ampere's law I'm using is:

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

 

[granpa]

doesnt the surface current by itself produce a uniform magnetic field within the cylinder?

 

[insynC]

I think it might, but if this is the case I think you need to use the cosine rule to determine r'', and then the integrals become horrendous.

 

[gabbagabbahey]

When working in a single coordinate system, the basis unit vectors are always unique. It makes no sense to say that $\hat{r}_1$ is different from $\hat{r}_2$. The direction of to vectors $\vec{a}$ and $\vec{b}$ may not be the same, but that direction is determined by their projection onto each basis vector.

As an example, say that $\vec{a}$ has magnitude 4 and points along the direction of the x-axis, then $\vec{a}=4\hat{x}=4(cos(\phi)\hat{r}-sin(\phi)\hat{\phi})=4(cos(0)\hat{r}-sin(0)\hat{\phi})=4\hat{r}$.

Now suppose the vector $\vec{b}$ has magnitude 4 and points along the negative x-axis. Clearly, $\phi=\pi$ so you get $\vec{b}=4(cos(\pi)\hat{r}-sin(\pi)\hat{\phi})=-4\hat{r}$

The vectors point in the opposite direction, as expected. The direction of $\hat{r}$ in terms of Cartesian unit vector is determined by $\phi$ and vice versa. So while $\hat{r}$ points in different directions for different vectors in terms of cartesian unit vectors, in Cylindrical coordinates $\hat{r}$ is $\hat{r}$ for all vectors.

 

[insynC]

To give an example of why I'm confused, say:

x = (1, 0, 0) and y = (1, pi/2, 0)

Geometrically it is clear that |x-y| = Sqrt(2).

But to do the approach used earlier to find r'' would give: x-y = (1-1,0-pi/2,0) = (0,-pi/2,0), and would give |x-y|=0

So does this mean I need to use the cosine rule to work out r''? If this is the case r'' involves at phi dependence, and so there is no guarantee the integrals go to 0.

In fact get the integral: integral(cos(phi)/SQRT(a-b cos(phi)) dphi), which I couldn't determine.

 

[insynC]

I'll have another look at the Ampere's Law bit. I thought it only applied to free current and so do 'bound' currents that arise from the magnetization actually count as currents as far as Ampere's Law is concerned?

But in terms of the magnetic potential approach, do I need to use the cosine rule to find r'' and if so how do I determine that integral?

 

[gabbagabbahey]

Hmm. now that I think about it , you may be right.

Ampere's law only applies to free current so H is zero both inside and outside, however M is only zero outside, so inside there should be a magnetic field $\vec{B}=\mu_0 \vec{M}=\mu_0 kr \hat{z}$.

Yes, it appears that you do need to use the law of cosines to find the distance r''...arrg , my apologies for leading you astray!

 

[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)?