2005 II 3A

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We have the dispersion relation:

n^{2}=\frac{RL}{S}

The wave frequency is:

\omega=2\pi f=1.9\times10^{8}\,\mathrm{s}^{-1}

Other parameters:

\omega_{pi}=1.32\times10^{3}Z\mu^{-1/2}n_{i}^{1/2}=5.1\times10^{9}\,\mathrm{s}^{-1}
\omega_{pe}=5.64\times10^{4}n_{e}^{1/2}=3.1\times10^{11}\mathrm{s}^{-1}
\Omega_{i}=9.58\times10^{3}Z\mu^{-1}B=2.7\times10^{7}\mathrm{s}^{-1}
\Omega_{e}=1.76\times10^{7}B=7.0\times10^{10}\mathrm{s}^{-1}

Then:

R\approx1-\frac{\omega_{pe}^{2}}{\omega\Omega_{e}}-\frac{\omega_{pi}^{2}}{\omega^{2}}\approx7510
L\approx1-\frac{\omega_{pe}^{2}}{\omega\Omega_{e}}+\frac{\omega_{pi}^{2}}{\omega^{2}}\approx-6941
S=\frac{1}{2}\left(R+L\right)\approx1-\frac{\omega_{pi}^{2}}{\omega^{2}}=-285

So:

n^{2}=\frac{c^{2}k^{2}}{\omega^{2}}=\frac{7510\cdot\left(-6941\right)}{-285}=1.8\times10^{5}

And:

\lambda_{\perp}^{-2}=\frac{f^{2}}{c^{2}}\left(1.8\times10^{5}\right)=0.18\,\mathrm{cm}^{-2}

Or the wavelength is:

\lambda_{\perp}=2.4\,\mathrm{cm}

Since \lambda_{\perp}\ll a, ray tracing methods will work.

This page was recovered in October 2009 from the Plasmagicians page on Generals_2005_II_3A dated 20:27, 15 May 2007.

Retrieved from "https://qed.princeton.edu/main/PlasmaWiki/Generals/2005_II_3A"
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