COBFP: nonuniform B

2018년 4월 10일 화요일

[플라즈마 물리][Plasma Physics]CH2 - Single Particle Motion - Nonuniform B-field Curvature Drift

Consider a magnetic field that has curvature.

\(\bullet\) Such \(B\)-field will have gradient which result in the grad-B drift.

\(\bullet\) Once particle has a velocity along the magnetic field, it experiences a centrifugal force due to the field curvature and this force gives a drift.

The centrifugal force on a charged particle with \(\vec{v_{\parallel}}\) along \(\vec{B}\) is

\[\begin{aligned} \vec{F_{cf}}= m\frac{v_{\parallel}^2}{R_c}\hat{r} = \frac{ m v_{\parallel}^2}{R_c}\frac{\vec{R_c}}{R_c}\end{aligned}\]

where \(\vec{R_c}\) is the radius of curvature. The drift velocity due to \(\vec{F_{cf}}\) (the curvature drift) is then

\[\begin{aligned} \boxed{\vec{v_R} = \frac{mv_{\parallel}^2}{qB^2}\frac{\vec{R_c}\times \vec{B}}{R_c^2}}\end{aligned}\]

Curvature drift is \(\vec{v_R}\). Now, total drift in a curved magnetic field is the combination of the grad-B drift and the drift due to the centrifugal force. From Ampere’s Law, \(\nabla \times \vec{B} = \mu_0 \vec{J}\), we have \(\nabla \times \vec{B}=0\) in a vacuum. In the curvature, B-field is defined as \(\vec{B} = B(r) \hat{\phi}\)

\[\begin{aligned} \nabla \times \vec{B} &= \left[ \frac{1}{r} \frac{\partial B_z}{\partial \phi} - \frac{\partial B_{\phi}}{\partial z } \right]\hat{r} + \left[\frac{\partial B_r}{\partial z} - \frac{\partial B_z}{\partial r} \right] \hat{\phi} + \frac{1}{r}\left[ \frac{\partial}{\partial r}(r B_{\phi}) - \frac{\partial B_r}{\partial \phi} \right]\hat{z} \\ &= \frac{1}{r}\frac{d}{dr}(rB) =0\end{aligned}\]

Hence,

\[\begin{aligned} B(r) \propto \frac{1}{r} = \frac{1}{R_c}\end{aligned}\]

Thus,

\[\begin{aligned} \frac{\nabla B}{B} &= \frac{1}{B}\frac{\partial B}{\partial r} \hat{R_c} = -\frac{\vec{R_c}}{R_c^2}\end{aligned}\]

Then the grad-B drift velocity may be written as

\[\begin{aligned} \vec{\nabla B} = \frac{\frac{1}{2}mv_{\perp}^2}{qB} \frac{\vec{B} \times \nabla B}{B^2} = \frac{\frac{1}{2}mv_{\perp}^2}{qB}\frac{\vec{B}}{B} \times \left( \frac{-\vec{R_c}}{R_c^2} \right) = \frac{\frac{1}{2}mv_{\perp}^2}{qB^2}\frac{\vec{R_c}\times \vec{B}}{R_c^2}\end{aligned}\]

Finally, the total drift in a curved magnetic field is sum of grad-B drift and centrifugal force driven drift.

\[\begin{aligned} \boxed{\vec{v_R} + \vec{v_{\nabla B}} = \frac{m}{q}\frac{\vec{R_c}\times \vec{B}}{R_c^2 B^2} \left( v_{\parallel}^2 + \frac{1}{2}v_{\perp}^2 \right)}\end{aligned}\]

As you see, since the equation depends on the charge, we will have charge separation and leads to the rise of current. Notice that \(\frac{1}{2}v_{\perp}^2\) comes from the grad-B drift and \(v_{\parallel}^2\) comes from the centrifugal driven drift; the curvature drift.

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[플라즈마물리][Plasma Physics]CH2 - Single Particle Motion - Nonuniform B-field Grad-B drift

Case Five: Non Uniform Magnetic Field

Let us assume gradient B-field exist in the z-direction with gradient along the y-direction.

\[\begin{aligned} \vec{B}(r) = B(y)\hat{z}\end{aligned}\]

Assume \(B\) is slightly in-homogeneous.

\[r_L\left| \frac{\nabla B}{B} \right|<< 1\]

\(\vec{B}\) can be expressed by the Taylor series.

\[\begin{aligned} \vec{B} = \vec{B_0} + (\vec{r} \cdot \nabla)\vec{B} + ...\end{aligned}\]

so that we have

\[\begin{aligned} \vec{B} = \vec{B_0} + (\vec{r} \cdot \nabla)\vec{B} = \vec{B_0} + y\frac{\partial \vec{B}}{\partial y}\end{aligned}\]

From the Lorentz force, we will compute the average force and determine the guiding center drift.

\[\begin{aligned} \frac{dv_x}{dt} &=\frac{q}{m}v_y B(y) \hat{x}\\ \frac{dv_y}{dt} &=-\frac{q}{m}v_x B(y) \hat{y}\end{aligned}\]

\[\begin{aligned} v_x &= v_{\perp}\cos\omega_c t \\ v_y &= \mp v_{\perp}\sin\omega_c t\end{aligned}\]

\[\begin{aligned} x &= r_L \sin(\omega_c t) \\ y &=\pm r_L \cos \omega_c t\end{aligned}\]

Hence, using above equation, we can approximate the Lorentz force in inhomogeneous B-field.

\[\begin{aligned} F_x &= qv_y B =\mp q v_{\perp}\sin(\omega_c t) \left( B_0 \pm r_L \cos(\omega_c t)\frac{\partial B}{\partial y} \right)\\ F_y &=-qv_x B = -qv_{\perp}\cos\omega_c t \left(B_0 \pm r_L \cos(\omega_c t) \frac{\partial B}{\partial y} \right)\\\end{aligned}\]

Since

\[\begin{aligned} <\sin(\omega t)> &= \frac{1}{T} \int_{0}^{T}\sin \omega t dt = 0\\ <\cos(\omega t)> &= \frac{1}{T} \int_{0}^{T} \cos \omega t dt = 0\\ <\sin^2(\omega t)> &= \frac{1}{T} \int_{0}^{T}\sin^2 \omega t dt = \frac{1}{T} \int_{0}^{T} \frac{1-\cos(2\omega t)}{2}dt = \frac{1}{2}\\ <\cos^2(\omega t)> &= \frac{1}{T} \int_{0}^{T} \cos^2 \omega t dt = \frac{1}{T} \int_{0}^{T} \frac{1+\cos(2\omega t)}{2}dt = \frac{1}{2}\\ <\sin \omega t \cos \omega t> &=\frac{1}{T} \int_{0}^{T} \cos(\omega t) \sin(\omega t)dt = \frac{1}{T} \int_{0}^{T} \frac{1}{2}\sin(2\omega t)dt =0\end{aligned}\]

we get

\[\begin{aligned} <F_x> &=\frac{1}{T}\int_{0}^{T}F_x dt =0\\ <F_y> &=\frac{1}{T}\int_{0}^{T}F_y dt \\ &=\frac{1}{T}\int_{0}^{T}F_y dt \\ &=\mp qv_{\perp}r_L \frac{\partial B}{\partial y}\frac{1}{2}\\ &=-\mu \frac{\partial B}{\partial y}\end{aligned}\]

Notice that \(\mu = \pm \frac{1}{2}q v_{\perp} r_L\). By plugging this into the guiding center drift equation for general case,

\[\begin{aligned} \vec{v}_{\nabla B} &=\frac{\vec{F}\times \vec{B}}{qB^2} \\ &= \frac{F_y \hat{y} \times B \hat{z}}{qB^2}\\ &=\frac{F_y}{qB}\hat{x} \\ &= -\frac{\mu}{qB}\frac{\partial B}{\partial y}\hat{x}\end{aligned}\]

The expression can be generalized as

\[\begin{aligned} \vec{v}_{\nabla B} = \frac{\mu}{q}\frac{\vec{B} \times \nabla B}{B^2} \end{aligned}\]

\[\begin{aligned} \boxed{\vec{v}_{\nabla_B} = \frac{\frac{1}{2}mv_{\perp}^2}{qB}\frac{\vec{B} \times \nabla B}{B^2}} \end{aligned}\]

The grad-B drift depends of the charge \(q\), it can cause the plasma currents and charge separation.