Difference between revisions of "ECE PhD QE FO3 2011 Problem1.1" - Rhea

Latest revision as of 10:35, 18 June 2017

1)

$ \nabla\times\bar{H} = \bar{J}+\frac{\partial D}{\partial t} = 0\hspace{1cm}d\bar{l} = dr\hat{r} + rd\phi\hat{\phi} + dz\hat{z} $

$ \nabla\times\bar{E} = -\frac{\partial \bar{B}}{\partial t}\hspace{1cm}dv=r dr d\phi dz $

$ \oint\bar{E}\cdot d\bar{l} = -\frac{\partial}{\partial t}\int_S \bar{B}\cdot d\bar{S} = -\frac{\partial\Phi}{\partial t} $

$ \Phi = \int_0^{2\pi}\int_0^rB_0\sin\omega t(rd\phi dr) = B_0\sin(\omega t)(\pi r^2) $

$ \int_0^{2\pi}E_{\phi}(rd\phi) = -\omega B_0\cos(\omega t)(\pi r^2) $

$ \bar{E} = \frac{-\omega B_0(\pi r^2)\cos(\omega t)}{2\pi r}\hat{\phi} $

$ \bar{E} = \frac{-\omega B_0 r\cos(\omega t)}{2}\hat{\phi} $

$ P = \frac{1}{2}\int_v\bar{J}\cdot\bar{E}dv = \frac{\sigma}{2}\int_v\bar{E}\cdot\bar{E}dv = \frac{\sigma}{2}\int_v|E|^2dv $

$ P = \frac{\sigma}{2}\int_0^h\int_0^{2\pi}\int_0^a\frac{\omega^2B_0^2r^2\cos^2(\omega t)}{4}(r dr d\phi dz) $

$ P = \frac{\sigma}{8}\bigg(\omega^2B_0\cos^2(\omega t)(2\pi h)\bigg)\bigg(\frac{a^4}{4}\bigg) $

$ P = \frac{\pi\sigma B_0^2ha^4\cos^2(\omega t)}{16}W $

-\underline{time-average}: $ \cos^2(\omega t) $ drops out (gives 1/2 factor)

2)

$ \nabla\times\bar{E} = -\frac{\partial B}{\partial t} = 0 \to\oint\bar{E}\cdot d\bar{l} = 0 $

a)

$ \oint\bar{H}\cdot d\bar{l} = \int\bar{J}\cdot ds + \int\frac{\partial D}{\partial t}\cdot d\bar{s} $

\underline{for left figure:}

$ \int_0^{2\pi}\bar{H}_\phi r d\phi = I_0 \cos\omega t +0 $

$ (2\pi r)\bar{H}_\phi = I_0\cos\omega t $

$ \bar{H} = \frac{I_0\cos(\omega t)}{2\pi r}\hat{\phi} $

\underline{for right figure:}

$ \bar{E} $ between plates :

$ \oint\bar{D}\cdot ds = \int_v \rho_v dv = Q $

$ \int_0^{2\pi} \int_0^a D_z rdrd\phi = Q = \int I_0 \cos(\omega t) dt $

$ \bar{D}_z(\pi a^2) = \frac{I_0}{\omega} \sin(\omega t) $

$ \bar{E} = \frac{I_0}{\omega \pi a^2 \epsilon_0}\sin(\omega t)\hat{z} $

$ \to (2 \pi r)\bar{H}_{\phi} = 0 + \frac{\partial}{\partial t}\int_0^{2\pi} \int_0^a \frac{I_0}{\omega \pi a^2}\sin (\omega t) rdrd\phi $

$ (2 \pi r)\bar{H}_{\phi} = \frac{\partial}{\partial t} \Big(\frac{I_0}{\omega \cancel{\pi a^2}}\sin (\omega t)\Big) (\cancel{\pi a^2}) $

$ (2 \pi r)\bar{H}_{\phi} = I_0 \cos(\omega t) $

$ \bar{H} = \frac{I_0 \cos (\omega t)}{2 \pi r} $

b) $ D = 9\epsilon_0\bar{E} $

--- $ \bar{E} $ will decrease by a factor of 9 between plates.

--- $ \bar{H} $ stays the same.

--- Surface integrals remain the same; $ \bar{E} $ field only changes.

c-i)$ W_e = \frac{1}{2} \int_v\bar{D}\cdot\bar{E}dv = \frac{1}{2}\int_v|\bar{E}|^2dv $

$ = \frac{1}{2}\int_v\frac{I_0^2\sin^2\omega t}{\omega^2(\pi a^2)^2\epsilon^2}dv $

$ =\frac{I_0^2\sin^2\omega t}{2\omega^2(\pi a^2)^2\epsilon^2}(\pi a^2d) $

$ W_e = \frac{I_0^2d\sin^2\omega t}{2\omega^2(\pi a^2)\epsilon^2}\xrightarrow{\text{time average}} W_e = \frac{I_0^2d}{4\omega^2(\pi a^2)\epsilon^2} $

c-ii) $ W_m = \frac{1}{2}\int_v\bar{B}\cdot\bar{H}dv = \frac{1}{2}\int_v|H|^2dv $

$ \bar{E} = \frac{I_0}{\omega \pi a^2\epsilon_0}\sin(\omega t)\hat{z} $

$ H_\phi(2\pi r) = \frac{\partial}{\partial t}\int_0^{2\pi}\int_0^r\frac{I_0\sin(\omega t)}{\omega \pi a^2}(r dr d\phi) $

$ H_\phi(2\pi r) = I_0\cos(\omega t)\bigg[\frac{\pi r^2}{\pi a^2}\bigg] $

$ \bar{H} = \frac{I_0\cos(\omega t)}{2\pi}\bigg(\frac{r}{a^2}\bigg) $

$ W_m = \frac{1}{2}\int_v|H|^2dv $

$ \frac{1}{2}\int_0^d\int_0^{2\pi}\int_0^a\frac{I_0^2\cos^2\omega t}{4\pi^2}\bigg(\frac{r^2}{a^4}\bigg)d dr d\phi dz $

$ = \frac{I_0^2\cos^2(\omega t)}{8\pi^2a^4}\bigg(\frac{a^4}{4}\bigg) $

$ W_m = \frac{I_0^2\cos^2(\omega t)}{32\pi^2}\xrightarrow{\text{time average}} W_m = \frac{I_0^2}{64\pi^2} $

$ \frac{\cancel{I_0^2}}{64\pi^2} < \frac{\cancel{I_0^2}d}{4\omega^2(\pi a^2)\epsilon^2} $ $ \omega^2 <\frac{16\pi d}{a^2\epsilon^2} $

$ \omega < \frac{4\sqrt{\pi d}}{a\epsilon} $

--- quasi-static assumption ignores retardation effect on coupled $ \bar{E} $ and $ \bar{H} $ fields. The assumption holds when the frequency $ \frac{c}{\lambda} $ is small ($ \lambda $ is large)

--- as d gets large, the frequency must get smaller and smaller to ignore retardation effects.