Solve Poynting Vector Homework: Part a,b & c

[Vuldoraq]

Homework Statement

Greetings all, sorry this is a long post but I'm really stuck on this...

A fat wire, radius a, carries a constant current I, uniformly distributed over it's cross section. A narrow gap in the wire, of width w<<a, forms a parllel-plate capacitor (see attachment, this question refers to question 3, sorry I couldn't cut and paste the picture).

a)Find the electrical and magnetic fields inside the gap, as functions of the distance s from the axis and the time t. Assume the charge is zero at time t=0.

b)Find the energy density u_em in the gap and the Poynting vector, S in the gap. Note especially the direction of S

c)Determione the total energy in the gap, as a function of time. Calculate the total power into the gap, by integrating the Poynting vector of the appropriate surface. Check that the power inout is equal to the rate of increase of energy in the gap.

Homework Equations

In the following:
s=distance from the axis of the wire
a=radius of the wire
t=time
I=current
and I'm working in cylinrical polars.

For a parallel plate capacitor,

$$\underline{E}=\frac{\sigma}{\epsilon_{0}}$$

For an amperian loop (with Maxwells fix),

$$\oint\underline{B}\cdot d\underline{l}=\mu_{0}I_{enc}+\mu_{0}\epsilon_{0}\int\frac{\partial\underline{E}}{\partial t} \cdot d\underline{a}$$

Energy density,

$$U_{em}=\frac{1}{2}(\epsilon_{0}E^{2}+\frac{1}{\mu_{0}}B^{2})$$

Poynting Vector,

$$\frac{1}{\mu_{0}}\underline{E}\times\underline{B}$$

The Attempt at a Solution

For part a) I found the electric to be,

$$\underline{E}=\frac{It}{\epsilon_{0}}\widehat{z}$$

and the magnetic field to be,

$$\underline{B}=\frac{\mu_{0}}{2\pi}\frac{Is}{a^2}\widehat{\phi}$$

For part b) I found the energy density to be,

$$\frac{I^{2}}{2}*(t^{2}+\frac{s^{2}}{4\pi^{2}a^{4}})$$

and the Poynting vector to be,

$$-\frac{\mu_{0}}{2\pi \epsilon_{0}}\frac{I^{2}t s}{a^{2}}\widehat{s}$$

For part c) I have no idea even where to begin. Please could someone tell me if parts a) and b) are correct and give me a hand for part c)? If you want me show more working just ask (I left it out to keep the post a bit smaller).

Thanks!

 

[Jhero]

Hello, thank you for your detailed post. Your solutions for parts a) and b) seem to be correct based on the given equations. As for part c), determining the total energy in the gap can be done by integrating the energy density over the volume of the gap. In this case, the volume of the gap can be represented as a cylinder with height s and radius w. The total energy in the gap would then be given by:

U = ∫∫∫ Uem dV = ∫∫∫ \frac{I^{2}}{2}*(t^{2}+\frac{s^{2}}{4\pi^{2}a^{4}}) dV

= ∫∫∫ \frac{I^{2}}{2}*(t^{2}+\frac{s^{2}}{4\pi^{2}a^{4}}) (2πs) ds dθ dz

= ∫∫ \frac{I^{2}}{2}*(t^{2}+\frac{s^{2}}{4\pi^{2}a^{4}}) (2πs) s ds dz

= ∫ \frac{I^{2}}{2}*(t^{2}+\frac{s^{2}}{4\pi^{2}a^{4}}) (2πs^{2}) ds

= \frac{I^{2}}{2}*(t^{2}+\frac{s^{2}}{4\pi^{2}a^{4}}) (\frac{2πs^{3}}{3})

= \frac{I^{2}}{3}*(t^{2}s+\frac{s^{4}}{4π^{2}a^{4}})

This is the total energy in the gap as a function of time t and distance s from the axis. To calculate the total power into the gap, we can integrate the Poynting vector over the surface of the gap, which in this case would be the two circular faces of the cylinder and the curved surface. The direction of the Poynting vector would be in the direction of energy flow, which in this case is along the z-axis (from the wire to the gap). The total power would then be given by:

P = ∫∫∫ S · dA = ∫∫∫ \frac{\mu_{