Generals 2004 II 1

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[edit] Part a

The ion-ion collision frequency is the same as the electron-electron collision frequency:

\frac{\nu_{ii}}{\nu_{ee}}\sim Z_{\mathrm{eff}}


The electron-ion energy exchange rate is that of the slowest component, ions slowing on electrons, νie. Therefore:

\frac{\nu_{\epsilon}}{\nu_{ee}}\sim\frac{\nu_{ie}}{\nu_{ee}}\sim\frac{m_{e}}{m_{i}}


[edit] Part b

The drift velocity is v_{D}\sim\frac{1}{\Omega}\mu\nabla B\sim v_{t}^{2}/R_{0}\Omega. The bounce frequency is vtε1 / 2 / qR. The banana width is thus Λj˜vD / ωb˜vtq / ε1 / 2Ω˜ρq / ε1 / 2. Since q = εBt / Bp, this is Λj˜ρε1 / 2BT / Bp. ε is the aspect ration r / R0.


[edit] Part c

The magnetic field must increase to B_{1}\sim B_{0}a_{1}^{2}/a_{0}^{2}, because of the frozen-in condition. The total energy density is proportional to B_{0}^{2}+p, and the total energy is this times the area. Since the total energy must be conserved:

\frac{B_{0}^{2}a_{0}^{2}}{4\pi}+\frac{p_{0}a_{0}^{2}}{\gamma-1}=\frac{p_{1}a_{1}^{2}}{\gamma-1}+\frac{B_{0}^{2}a_{1}^{4}/a_{0}^{2}}{4\pi}


This means that the new energy density is:

p_{1}=\left(\frac{a_{0}^{2}}{a_{1}^{2}}-\frac{a_{1}^{2}}{a_{0}^{2}}\right)B_{0}^{2}\frac{\left(\gamma-1\right)}{4\pi}+\frac{a_{0}^{2}}{a_{1}^{2}}p_{0}
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