0
$\begingroup$

An infinite cylinder of radius R has a uniforme current distribution through its surface $\vec \kappa (t)=\kappa(t)\vec \phi$. Find a) the magnetic field generated by the distribuition, b) the induced electric field, and c) the Poyinting vector for this configuration.

So far, I have used Ampere's Law to determine the magnetic field, which for $r \le R$ (and ignoring Maxwell's correction) is: $$ \vec B=\mu_0 \kappa(t) \hat z$$ and nule for the remaining space. Faraday's Law in integral form: ${\displaystyle \oint _{\partial \Sigma }\mathbf {E} \cdot \mathrm {d} \mathbf {l} =-{\int _\Sigma }\mathbf {\frac {\mathrm {d} {B} }{\mathrm {d} t}} \cdot \mathrm {d} \mathbf {A} }$ leads to (for $r\le R$): $$\vec E(t)=-\frac{\mu_0}{2} \frac{d \kappa(t)}{d \kappa}r \hat\phi$$

So the Poynting vector $\vec S =\frac{\vec E \times \vec B}{\mu_0}$ is nule for $R \le r$ and, for $r \le R$: $$\vec S=-\frac{\mu_0}{2}\kappa(t)\frac{d \kappa(t)}{d \kappa}r \hat r$$

I understand all the steps of the problem, but what I can't understand is why the vector $\vec S$ is discountinuous on the surface of the cylinder.

Comparing it to the case of a solenoid connected to a tension source, it makes some sense that the Poynting vector, which represents a flux of energy transported by the EM field per unit of time, is pointing (pun unintended) in the direction that energy is being "given" to the system, but I'm not sure if that's the case. Plus, eventhough there isn't a magnetic field outside the cylinder, there's still an electric field, which (at least, in my head) should still contribute to the energy flow.

Can anyone help make sense of this? What am I missing here?

$\endgroup$

2 Answers 2

3
$\begingroup$

The simple answer is that there is no $B$ field outside the cylinder, so $S$ is also zero.

What makes this setup go against intuition is the infinite cylinder part. In every realistic setup there would be a comparatively small $B$ field outside. The longer the cylinder gets, the more is it thinned out. The field lines are still there, so to speak, but get spaced out infinitely.

The same applies to the pointing vector. It also approaches zero. But the total flux, i.e. an integral of the pointing vector over an infinite surface outside would/could still have a non-zero result.

The sharp discontinuity at $R$ is also the effect of idealization of an infinitely thin current. But even for a realistic current/coil there would be a big difference between inside and outside.

$\endgroup$
1
$\begingroup$

It is worth appreciating how come the energy flux just seems to appear out of nothing at the cylinder surface.

The current distribution is that of a solenoid. When the current in the solenoid windings is increasing an EMF opposes the current so work has to be done by an external power source to increase the field. It is this external power source that is supplying the necessary energy to each bit of the current carrying wire, and from each piece of wire the delivered energy flows radially into the magnetic field in the solenoid bore ($r<R$) where it is stored in the $|B|^2/2{\mu_0}$ energy density.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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