Generals 2003 I 4

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[edit] Part a

The ion cyclotron frequency is:

f_{ci}=1.52\times10^{3}Z\mu^{-1}B\,\mathrm{Hz}=38\,\mathrm{MHz}

The electron cyclotron frequency is:

f_{ce}=2.80\times10^{6}B\,\mathrm{Hz}=140\,\mathrm{GHz}

The plasma frequency is:

f_{pe}=8.98\times10^{3}n_{e}^{1/2}\mathrm{Hz}=110\,\mathrm{GHz}

The ion plasma frequency is:

f_{pi}=2.10\times10^{2}Z\mu^{-1/2}n_{i}^{1/2}\mathrm{Hz}=1.8\,\mathrm{GHz}

The source frequency is:

f_{s}=4.6\,\mathrm{GHz}

So:

f_{ci}\ll f_{pi}<f_{s}\ll f_{pe}<f_{ce}

[edit] Part b

The cold, electrostatic dispersion relation is:

\mathbf{k}\cdot\mathbf{\epsilon}\cdot\mathbf{k}=0

Or:

Sk_{\perp}^{2}+Pk_{\|}^{2}=0

Since \gamma\gg1, and \Omega_{i}\ll\omega\ll\Omega_{e}:

R=1-\sum\frac{\omega_{ps}^{2}}{\omega\left(\omega+\Omega_{s}\right)}\approx1-\frac{\omega_{pi}^{2}}{\omega^{2}}-\frac{\omega_{pe}^{2}}{\omega\Omega_{e}}
L=1-\sum\frac{\omega_{ps}^{2}}{\omega\left(\omega-\Omega_{s}\right)}\approx1-\frac{\omega_{pi}^{2}}{\omega^{2}}+\frac{\omega_{pe}^{2}}{\omega\Omega_{e}}

So:

S=\frac{1}{2}\left(R+L\right)\approx1-\frac{f_{pi}^{2}}{f^{2}}=0.85

And:

P\approx-\frac{f_{pe}^{2}}{f^{2}}=-571

So:

\frac{1}{\lambda_{\perp}^{2}}=\frac{-Pn_{\|}^{2}f^{2}}{Sc^{2}}=84\,\mathrm{cm}^{-2}

So \lambda_{\perp}=0.1\,\mathrm{cm}.

[edit] Part c

We can calculate n_{\perp}:

n_{\perp}=\frac{c}{f\lambda_{\perp}}=65

Since n_{\perp}\gg n_{\|} the phase velocity \mathbf{v}_{p}\sim c\mathbf{n}/n^{2} is approximately perpendicular. Since for cold electrostatic waves the phase velocity is perpendicular to the group velocity, \mathbf{v}_{g} is approximately parallel. Therefore the power will flow in the toroidal direction.

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