Generals 2005 II 3A

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.