0
$\begingroup$

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.

$\endgroup$

1 Answer 1

0
$\begingroup$

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 $\mathbf{H}$.

$\endgroup$
1
  • 1
    $\begingroup$ 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
    Commented Mar 9, 2015 at 13:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.