This is the second part of a problem in Griffith's "Introduction to Electrodynamics 4th Edition" (problem 6.16).

The first part was to find the magnetic field inside a coaxial cable (2 concentric cylindrical shells with radii $a$ and $b$, $b>a$, and with a linearly magnetizable media of $X_{m}$ between them). I was able to accomplish this using Ampere's Law for $\textbf{H}$.

The second part is to check this solution by finding the bound currents from their definitions (in terms of magnetization $\textbf{M}$, and then finding the field produced by those bound currents.

I found that $\textbf{J}_{b}=0$ while $$\textbf{K}_{b}=\frac{X_{m}I}{2\pi s}\hat{I}$$ Where $s$ is the radius of the cylindrical shell and $\hat{I}$ is the direction of the current on that particular shell. The problem then should reduce to finding the magnetic field produced by the surface current.

I tried doing just that by calculating the vector potential $\textbf{A}$ of each cylinder and then via the superposition principle, get the total $\textbf{A}$ from which I get obtain the total $\textbf{B}$ via $\textbf{B}=\nabla\times\textbf{A}$.

The problem I ran into is when I tried integrating $$ \textbf{A}=\frac{\mu_{0}}{4\pi}\int\frac{\textbf{K}}{R}da' $$ over the entire area of the cylinder, I get a non converging integral. In cylindrical coordinates, $$R=\sqrt{s^{2}+a^{2}-2as*cos(\phi'-\phi)+z^{2}-2zz'+z'^{2}}$$ $$da'=ad\phi'dz'$$ $$\textbf{K}=K\hat{z}$$ with limits of integration: $\phi': (0,2\pi)$, $z': (-\infty,\infty)$.

What I did was rewrote $R$ by completing the square for $z'$ and making the substitution $z'-\alpha=\beta tan\theta$, which changes the limits of integration in $\theta$ to $(-\frac{\pi}{2},\frac{\pi}{2})$. But this also reduces the integrand to $sec\theta d\theta$ and the integral diverges.

Did I make a calculation error? Assuming the superposition principle applies for $\textbf{A}$, then the only way I can get a finite $\textbf{B}$ is if the sum of the infinite $\textbf{A}$'s is somehow finite.


I believe you are misinterpreting what the question is asking - I don't think there is any reason to solve for $\mathbf{A}$.

Your $\mathbf{K}_{b}$ is wrong I think also - it should be running up the cylinder on the outside, and the running down on the inside of the cylinder.

You have $\mathbf{J}_{b}$. Once you compute $\mathbf{K}_{b}$ correctly, I would encourage you to try and find the current enclosed by an Amperian loop; then use the integral form of Ampere's law to find the overall magnetic field $\mathbf{B}$ (I believe this is what part (b) is actually asking for). Then make sure this $\mathbf{B}$ matches with the one you would compute using $\mathbf{M}$ and $\mathbf{H}$.

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    $\begingroup$ I see, so after finding Kb, we can treat this situation as equivalent to one without an insulator but where the bound current is added to the free current. Doing so, I do get the same B I got with M and H. Using Kb = M x n, I got Kb running in the same direction as the free current, and this Kb allowed me to get the correct B. $\endgroup$ – user279043 Mar 9 '15 at 13:14

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