Find B and H everywhere for a magnetized infinite cylinder

[Potatochip911]

Homework Statement

An infinitely long cylinder of radius a has its axis along the z-direction. It has Magnetization $M=M_0(s/a)^2\hat{\phi}$ in cylindrical coordinates where $M_0$ is a constant and s is the perpendicular distance from the axis. Find the values of $\vec{B}$ and $\vec{H}$ everywhere.

Homework Equations

$\vec{k_b} = \vec{M}\times \hat{n}$
$\vec{J_b}=\vec{\nabla}\times \vec{M}$
$\vec{H} = \frac{1}{\mu_0}\vec{B}-\vec{M}$

The Attempt at a Solution

[/B]
$\vec{k_b}$ and $\vec{J_b}$ are straightforward to calculate so I will just list the results.

$$ \vec{k_b}|_{r=a}=-M_0 \\ \vec{J_b}=\frac{3M_0s}{a^2}$$

I'm pretty sure these are correct since summing up the total bound charge gives 0:
$$Q=\int_0^a J_b\cdot da + k_b * (2\pi a) = \frac{6\pi M_0}{a^2}\int_0^a s^2ds -2\pi aM_0 = 2\pi M_0a-2\pi M_0a=0$$

Since there's no free charge in the problem $\vec{H}=0$ everywhere. Then for $r < a$ we have $$B=\mu_0\vec{M}=\mu_0M_0\frac{s^2}{a^2}\hat{\phi}$$

However, if I try to calculate this using Ampere's law with an amperian loop similar to those used for a solenoid I find: (a-s)L area of cylinder inside amperian loop for r < a)

$$ B(s)\cdot L = \mu_0 I_{encl} = \mu_0(\vec{J_b}\cdot(a-s)L +\vec{k_b}L)=\mu_0M_0L\left(\frac{3s(a-s)}{a^2}-1\right)$$

and this obviously isn't going to give the same result as I found using $\vec{H}$. I'm not sure where I'm going wrong as this is very similar to problem 6.12 in griffiths electrodynamics 4th edition and I'm basically employing all of the same techniques.

 

[Charles Link]

Everything else looks good, (other than you didn't specify the direction of $\vec{J}_m$ and $\vec{K}_m$ and they are both in the $\hat{z}$ direction), until your last equation, which I think is simply incorrect. You are computing $\oint \vec{B} \cdot \, dl=\mu_o \int \vec{J}_m \, dA$. That is fairly straightforward for this problem where the current is all in the +z direction for $s<a$. Try it again=I think you will get it to work. You can also predict that $B$ needs to be zero for $s>a$, since $M=0$ for $s>a$, and I think you should be able to get that result as well. (Edit: I left out a $\mu_o$ in the equation above that I now put in there).

 

[Potatochip911]

Everything else looks good, (other than you didn't specify the direction of $\vec{J}_m$ and $\vec{K}_m$ and they are both in the $\hat{z}$ direction), until your last equation, which I think is simply incorrect. You are computing $\oint \vec{B} \cdot \, dl=\mu_o \int \vec{J}_m \, dA$.

Ok, the difference was in the book that their $\vec{J_b}$ was a constant so they could just multiply by the area to get the result of the integral however, my $\vec{J_b}$ is linear w.r.t. s although I am still running into trouble.

Defining $dA =l \,ds\Longrightarrow \int_s^a J_d\cdot dA =\frac{3M_0l}{a^2}\int_s^a s\,ds$ results in $\int J_d\cdot dA = \frac{3M_0l}{2}(1-(s/a)^2)$. I'm clearly still missing something as the 3/2 coefficient is a problem.

 

[Charles Link]

You have the geometry all confused. This is not a solenoid type problem. You have current ($\vec{J}_m$) going along a wire in the +z direction. Computing $B$ for that is fairly simple with Ampere's law= You need a circular loop for your Ampere's law path. I think with that hint, you might also now find it very easy.

 

[Potatochip911]

You have the geometry all confused. This is not a solenoid type problem. You have current ($\vec{J}_m$) going along a wire in the +z direction. Computing $B$ for that is fairly simple with Ampere's law= You need a circular loop for your Ampere's law path. I think with that hint, you might also now find it very easy.

Yes I agree that geometry doesn't make much sense to me either I was just copying what they had done. Setting up the problem as $$\vec{B}(2\pi s) = \mu_0(\int_s^a J_d\cdot 2\pi s\, ds - M_0\cdot 2\pi a) $$ results in the answer of $$\vec{B} = -\mu_0 M_0 (s/a)^2\hat{\phi}$$ which is just the negative of what I was finding before.

I'm attaching how they solved a similar problem in the solutions manual since they are using the flat ampere loop for a cylinder and getting the correct result for their problem. Perhaps you can explain how it's working?

 

[Charles Link]

The left side is correct. For the right side, the limits should be $0$ to $s$, the $dA=2 \pi s \, ds$, and any current outside of the circular loop of radius $s$ isn't counted. This is a simple Ampere's law calculation for the current that travels through the loop which is the circle of the $\oint \vec{B} \cdot \, dl$ integral.

 

[Charles Link]

@Potatochip911 In the attachment of your previous post, the magnetization $\vec{M}$ is in the +z direction. For that case, the surface currents have the same geometry as the currents of a solenoid. $\\$ The problem you are working here has magnetization $\vec{M}$ in the $\phi$ direction.

 

[Potatochip911]

@Potatochip911 In the attachment of your previous post, the magnetization $\vec{M}$ is in the +z direction. For that case, the surface currents have the same geometry as the currents of a solenoid. $\\$ The problem you are working here has magnetization $\vec{M}$ in the $\phi$ direction.

Thanks! This pieces everything together.

 

[Charles Link]

@Potatochip911 I have one addition/correction to your solution above: The concept is somewhat advanced and actually involves a detail that comes from the "magnetic pole model" of solving magnetostatic problems, but you may find it of interest: Your statement that $\vec{H} =0$ is correct, but your reason is incomplete. $\\$ There are actually two possible sources for $\vec{H}$ that need to be considered:$\\$ 1) Free currents in conductors (and you are correct=these are absent) $\\$ 2) Magnetic "poles" where magnetic pole (also called "magnetic charge") density $\rho_m=- \mu_o \nabla \cdot \vec{M}$, and in addition, surface magnetic pole density $\sigma_m= \mu_o \vec{M} \cdot \hat{n}$. $\\$ (I'm using your same units here where you use $\vec{B}=\mu_o \vec{ H} +\mu_o \vec{M}$. In another very common units, $\vec{B}=\mu_o \vec{ H}+ \vec{M}$, without any $\mu_o$ on the $\vec{M}$). $\\$ ........................... $\\$ In the problem at hand, $\nabla \cdot \vec{M}=0$ , and $\sigma_m=\mu_o \vec{M} \cdot \hat{n}=0$, so indeed, you can say $\vec{H}=0$. $\\$ .............................. $\\$ Meanwhile, if you consider the cylinder of "finite" length of uniform magnetization $\vec{M}$ along the +z axis as in your attachment of post 5, (it is a very typical permanent magnet), it will have magnetic "poles" on its endfaces, (with $\sigma_m=\mu_o \vec{M} \cdot \hat{n}$, with the dot product generating a $"+"$ pole on one endface, and a $"-"$ pole on the other), so that $\vec{H}$ is not zero everywhere for a magnetized cylinder of finite length. The $\vec{H}$ from these poles is found by the inverse square law, in a manner analogous to how the electric field $\vec{E}$ is computed from electric charges. The magnetic field is then $\vec{B}=\mu_o \vec{H}+\mu_o \vec{M}$. (They are really computing it for a very long cylinder, but it is worthwhile to consider the cylinder of "finite" length). $\\$ Alternatively, you could do a magnetic surface current calculation, (it's a rather difficult calculation for the magnetized cylinder of finite length=you can do it on-axis, but off-axis, it is extremely difficult), and compute the magnetic field $\vec{B}$ for this case. When you then go about computing $\vec{H}=\frac{\vec{B}}{\mu_o}-\vec{M}$, you will get the very same result for $\vec{H}$ that the pole model calculation gives.$\\$ (And they do have it correct in the attachment though that $\vec{H}=0$ and $\oint \vec{H} \cdot dl=0$ everywhere, because there are no poles in the case of a magnetized cylinder of infinite length. It is also of interest that the contribution to $\vec{H}$ that originates from magnetic poles does have $\nabla \times \vec{H}= 0$ everywhere, so that $\oint \vec{H} \cdot dl=0$ for this contribution to $\vec{H}$. However, for the case of a cylinder of finite length, $\vec{H}$ will be non-zero, even though $\oint \vec{H} \cdot dl=0$ ). $\\$ Inside a permanent magnet, $\vec{H}$ actually points opposite the direction of $\vec{M}$. And outside a permanent magnet, $\vec{H}=\frac{\vec{B}}{\mu_o}$ is also non-zero. $\\$ You might also find this previous posting of interest: See in particular post 2 of this thread: https://www.physicsforums.com/threads/magnetic-field-of-a-ferromagnetic-cylinder.863066/