Magnetic Field, Field Intensity and Magnetisation

[Mr_Allod]

Homework Statement

A conducting wire is placed along the z axis, carries a current I in the +z direction and
is embedded in a nonconducting magnetic material with permeability $\mu$. Find the
magnitude and direction of $\vec H$, $\vec B$, $\vec M$ and the equivalent current $\vec J_b$ at the point (x,0,0).

Relevant Equations

Ampere's Law: $\oint \vec B \cdot d\vec l$
$\vec H$ and $\vec B$ Relation: $\vec H = \frac 1 \mu_0 \vec B - \vec M$
Permeability: $\mu = \mu_0 (1+ \chi_m)$
Magnetisation: $\vec M = \chi_m \vec H$

Hello there, I've worked through this problem and I would just like to check whether I've understood it correctly. I found $\vec H$, $\vec B$ and $\vec M$ using Ampere's Law and the above relations as I would for any thin current carrying wire and these were my answers:
$$\vec H = \frac I {2\pi s} \hat \phi$$ $$\vec B = \frac {\mu I} {2\pi s} \hat \phi$$ $$\vec M = \chi_m \frac I{2\pi s} \hat \phi$$

For the point (x,0,0) I would simply swap $x$ for $s$.

Then using $\vec J_b = \nabla \times \vec M$ I tried to calculate $\vec J_b$. All but one of the terms of the cross product evaluate to 0 leaving:
$$\vec J_b = \frac 1 s \frac \partial {\partial s}(s M_\phi) \hat z = \frac 1 s \frac \partial {\partial s}(s \chi_m \frac {\mu I}{2\pi s}) \hat z$$

Which as it turns out also evaluates to 0. This leads me to believe I must have misunderstood something about the question since I don't expect I would be asked to find $\vec J_b$ if it was simply 0. If somebody could help me figure out what I've done wrong I'd appreciate it.

 

[andrewkirk]

Without comparing any of the equations to what we know about electromagnetic theory, we can see there must be an error in the first part, as substituting your formula for $\vec{\mathit H}$ into your magnetisation equation gives
$$\vec M = \chi_m \frac {I}{2\pi s} \hat \phi$$
whereas you have written
$$\vec M = \chi_m \frac {\mu I}{2\pi s} \hat \phi$$

Since the problem did not specify that $\mu=1$, at least one of those two must be wrong.

 

[Mr_Allod]

Without comparing any of the equations to what we know about electromagnetic theory, we can see there must be an error in the first part, as substituting your formula for $\vec{\mathit H}$ into your magnetisation equation gives
$$\vec M = \chi_m \frac {I}{2\pi s} \hat \phi$$
whereas you have written
$$\vec M = \chi_m \frac {\mu I}{2\pi s} \hat \phi$$

Since the problem did not specify that $\mu=1$, at least one of those two must be wrong.

Yes you are quite right, I accidentally put in the result for $\vec B$ instead of $\vec H$ when I wrote the expression for $\vec M$. Thank you for pointing that out, I have edited it to make it correct.

 

[andrewkirk]

It still contains algebraic errors.

Add your formulas for $\vec H$ and $\vec M$ together, then multiply by $\mu_0$ to get a formula for $\vec B$. That formula doesn't match the formula for $\vec B$ you have written above, and again both can't be correct unless $\mu = 1$, which is not given.

 

[Mr_Allod]

I'm sorry but I don't see what you mean. Applying your correction and doing as you say:
$$\vec H + \vec M = \frac I {2\pi s} \hat \phi + \chi_m \frac I{2\pi s} \hat \phi = (1 + \chi_m) \frac I{2\pi s} \hat \phi$$
Multiplying by $\mu_0$:
$$\mu_0(\vec H + \vec M) = \mu_0 (1 + \chi_m) \frac I{2\pi s} \hat \phi$$

Which gives the expression for $\vec B$:
$$\vec B = \frac {\mu I} {2\pi s} \hat \phi = \mu_0 (1 + \chi_m) \frac I{2\pi s} \hat \phi$$

 

[Delta2]

I think you correctly calculate the curl of $\vec{M}$ as zero for all $s\neq 0$.

My only concern is what happens when $s=0$. Is it $|\vec{M}|=\infty$and $|\vec{J_b}|=\infty$ for $s=0$?

 

[Mr_Allod]

I think you correctly calculate the curl of $\vec{M}$ as zero for all $s\neq 0$.

My only concern is what happens when $s=0$. Is it $|\vec{M}|=\infty$and $|\vec{J_b}|=\infty$ for $s=0$?

Yes you are right that is a problem in this case. I think it's because the wire is taken to be infinitely thin in this question. A different but very similar problem I have done before involves a cylindrical wire of radius $a$ and magnetic susceptibility $\chi_m$ where the current density inside the wire is proportional to the radius. For that problem I calculated:
$$\vec J_b = \chi_m I \frac {3s}{2\pi a^3} \hat \phi$$
Which would have $\left|\vec J_b \right|\rightarrow 0$ as $s \rightarrow 0$, avoiding that issue.