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Say we have an infinite cylinder, with constant magnetization $M(z)\hat{z}$ along its axis. This vanishes for large $|z|$, say like $e^{-z^2}$.

It seems to me that the solution to the equations $$\vec{\nabla}\times\vec{H}=0=\vec{\nabla}\times\vec{M}$$ and $$\vec{\nabla}\cdot\vec{H}=-\vec{\nabla}\cdot\vec{M}$$ is simply $$\vec{H}=-\vec{M},$$ which would lead to $\vec{B}=0$ from the equation $\vec{B}=\mu_0(\vec{H}+\vec{M})$. Is this correct?

On the other hand, since the cylinder is infinite and there is no free current, I would be tempted to conclude that $\vec{H}=0$.

Perhaps $\vec{H}$ is neither $-\vec{M}$ nor $0$, but how do I compute it?

I am confused about magnetostatics...

thedude
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2 Answers2

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Since $\overrightarrow{M}\neq\overrightarrow{0}$ inside the cylinder and $\overrightarrow{M}=\overrightarrow{0}$ outside, $\overrightarrow{\nabla}\times\overrightarrow{M}\neq\overrightarrow{0}$ at the cylinder sides. Therefore, $\overrightarrow{H}+\overrightarrow{M}\neq \overrightarrow{0}$ holds at almost all points.

[Edit #1] Work in the cylindrical coordinate $(\rho,\phi,z)$ and assuming that the radius is $\rho_0$. If the magnetization is given of the form $M_z=M_z^0-M_z^0H(\rho-\rho_0)$, where $H(\rho)$ is the Heaviside step function. $$ \vec{\nabla}\times\vec{M}|_{\hat{\phi}}=\frac{\partial M_\rho}{\partial z} -\frac{\partial M_z}{\partial \rho} =-\frac{\partial M_z}{\partial \rho} =\frac{\partial M_z^0H(\rho-\rho_0)}{\partial \rho} =+M_z^0\frac{\partial H(\rho-\rho_0)}{\partial \rho} =+M_z^0\delta(\rho-\rho_0) $$ Here, $\delta(\rho)$ is the Dirac's delta function. Thus, $\vec{\nabla}\times\vec{M}$ is not zero at the side position of cylinder.

HEMMI
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Let me try to answer my own question to see if what I am thinking makes sense.

Step 1) There are no free currents, so $\vec{\nabla}\times\vec{H}=0$ and we can define a scalar magnetic potential $U$ so that $\vec{H}=\vec{\nabla}U$. Then it must satisfy $$\nabla^2 U=-\vec{\nabla}\cdot \vec{M}=2ze^{z^2}.$$ Also, $U$ is continuous at the cylinder boundary, and its partial derivatives $\frac{\partial U}{\partial r}$ and $\frac{\partial U}{\partial z}$ are also continuous.

Step 2) Writing the laplacian in cylindrical coordinates and assuming azimuthal symmetry, we have inside the cylinder $$ \frac{1}{r}\frac{\partial}{\partial r}\left(r\frac{\partial U}{\partial r}\right)+\frac{\partial^2 U}{\partial z^2}=2ze^{z^2}.$$ With $U=R_{in}(r)Z_{in}(z)$ we can separate variables to get $$\frac{1}{r}\frac{d}{d r}\left(r\frac{dR_{in}}{d r}\right)=-k^2R_{in}$$ and $$\frac{d^2 Z_{in}}{d z^2}=k^2Z_{in}+2ze^{z^2}Z_{in}$$ for some separation constant $k$.

So the function $R_{in}(r)=J_0(kr)$ is a Bessel function and function $Z_{in}(z)$ is something complicated.

Step 3) Again writing the laplacian in cylindrical coordinates and assuming azimuthal symmetry, we have outside the cylinder $$ \frac{1}{r}\frac{\partial}{\partial r}\left(r\frac{\partial U}{\partial r}\right)+\frac{\partial^2 U}{\partial z^2}=0.$$ With $U=R_{out}(r)Z_{out}(z)$ we can separate variables to get $R_{out}(r)=J_0(qr)$ and now $$\frac{d^2 Z_{out}}{d z^2}=q^2Z_{out},$$ for some separation constant $q$.

Step 4) Continuity of $U$ and its radial derivative imply $q=k$. Continuity of derivative with respect to $z$ for every $z$ is not possible since $Z_{out}$ is very different from $Z_{in}$. I think the only way out is if $U(R,z)=0$ for all $z$, which implies $k$ must be $\gamma_n/R$ where $\gamma_n$ is a zero of $J_0$.

So $$U(r,z)=\sum_n J_0(\gamma_nr/R)Z_{in}(n;z)$$ for $r<R$ and $$U(r,z)=\sum_n J_0(\gamma_nr/R)Z_{out}(n;z)$$ for $r>R$.

Assuming I could find $Z_{in}(n;z)$, I could compute the gradient of this function and it would give me the field $\vec{H}$.

Is this calculation correct?

thedude
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