Generals 2003 II 2

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[edit] Part a

Drift will primarily come from the \mathbf{E}\times\mathbf{B} term:

\mathbf{v}_{D}= \frac{1}{cB_{0}}\left(\mathbf{E}\times\mathbf{B}\right)

So that the radial term is:

\left(v_{D}\right)_{r}=\frac{E_{\zeta}B_{p}}{cB_{T}}

[edit] Part b

The canonical angular momentum is:

P_{\zeta}=mRv_{\zeta}-\frac{e}{c}RA_{\zeta}

Since \mathbf{B}=\nabla\times\mathbf{A}:

B_{p}=-\frac{\partial}{\partial r}A_{\zeta}

And so:

ψ = − RAζ

Giving:

P_{\zeta}=mRv_{\zeta}+\frac{e}{c}\psi

We then solve for ψ :

\psi=\frac{c}{e}\left(P_{\zeta}-mRv_{\zeta}\right)

Taking the Lagrangian derivative:

\frac{d\psi}{dt}= \frac{c}{e}\left(\frac{dP_{\zeta}}{dt}-mRv_{\zeta}\right)

Since Pζ is conserved, the first term is zero. Bounce averaging:

\left\langle \frac{d\psi}{dt}\right\rangle =-\left\langle \frac{mc}{e}Rv_{\zeta}\right\rangle

Since the particle returns to the same position and velocity after each bounce, \left\langle Rv_{\zeta}\right\rangle =0. Then:

\left\langle \frac{d\psi}{dt}\right\rangle =\left\langle \frac{\partial\psi}{\partial t}+\mathbf{V}\cdot\nabla\psi\right\rangle =0

So that:

\frac{\partial\psi}{\partial t}=-\mathbf{V}\cdot\nabla\psi

[edit] Part c

Faraday's law is:

\nabla\times\mathbf{E}= -\frac{1}{c}\frac{\partial\mathbf{B}}{\partial t}

Writing \mathbf{B}=\nabla\times\mathbf{A}:

\nabla\times\mathbf{E}=-\frac{1}{c} \nabla\times\frac{\partial\mathbf{A}}{\partial t}

So:

\mathbf{E}=-\frac{1}{c}\frac{\partial\mathbf{A}}{\partial t}

In the parallel direction:

E_{\|}=-\frac{1}{c}\frac{\partial A_{\zeta}}{\partial t}=\frac{1}{cR}\frac{\partial\psi}{\partial t}

Plugging in:

cRE_{\|}=-\mathbf{V}\cdot\nabla\psi

using Vr and \left(\nabla\psi\right)_{r}=B_{\theta}R:

cRE_{\|}=-V_{r}B_{\theta}R

Or:

V_{r}=-\frac{cE_{\|}}{B_{\theta}}
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