Generals 2005 II 2

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[edit] Part a

Pressure balance gives:

\frac{1}{c}\mathbf{J}\times\mathbf{B}=\nabla p


Where:

\mathbf{J}=nq\mathbf{v}


So:

\frac{nq}{c}\mathbf{v}_{\perp}\times\mathbf{B}=\nabla p


Crossing with \mathbf{B}:

\mathbf{v}_{\perp}=\frac{c}{nqB_{0}^{2}}\mathbf{B}\times\nabla p


This is called the diamagnetic velocity because it only seems to exist, since with a pressure gradient there are more or faster ions spinning on one side than the other.


[edit] Part b

The acceleration of the particle is, to zero order:

\frac{d\mathbf{v}}{dt}=\frac{q}{m}\frac{1}{c}\mathbf{v}\times\mathbf{B}


We can write the velocity as:

\mathbf{v}=v_{\|}\hat{n}+\mathbf{v}_{\perp}


Over a gyro orbit, there is no net perpendicular velocity, so that:

\left\langle \mathbf{v}_{\perp}\right\rangle =0


Going to next order, we write the velocity:

\mathbf{v}=v_{\|}\hat{n}+\mathbf{v}_{\perp}+\mathbf{v}_{E}


Then the cyclotron averaged acceleration is:

0=\left\langle \frac{q}{m}\left(\mathbf{E}+\frac{1}{c}\left(\mathbf{v}_{\perp}+\mathbf{v}_{E}\right)\times\mathbf{B}\right)\right\rangle


So:

\mathbf{E}=-\frac{1}{c}\mathbf{v}_{E}\times\mathbf{B}


And:

\mathbf{v}_{E}=-\frac{c}{B}\hat{n}\times\mathbf{E}_{\perp}


Going to next order:

\mathbf{v}=v_{\|}\hat{n}+\mathbf{v}_{\perp}+\mathbf{v}_{E}+\mathbf{v}_{pol}


The averaged acceleration becomes:

\left\langle \frac{d\mathbf{v}_{E}}{dt}\right\rangle =\left\langle \frac{q}{m}\left(\mathbf{E}+\frac{1}{c}\left(\mathbf{v}_{\perp}+\mathbf{v}_{E}+\mathbf{v}_{pol}\right)\times\mathbf{B}\right)\right\rangle


Solving for \mathbf{v}_{pol}:

\mathbf{v}_{pol}=\frac{mc}{qB}\hat{n}\times\frac{d\mathbf{v}_{E}}{dt}=\frac{1}{\Omega}\hat{n}\times\frac{d}{dt}\left(-\frac{c}{B}\hat{n}\times\mathbf{E}_{\perp}\right)=-\frac{c}{\Omega B}\hat{n}\times\left(\hat{n}\times\frac{d\mathbf{E}_{\perp}}{dt}\right)
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