10.2.2∗. Estimate the electron drift velocity across an inhomogeneous magnetic field, the induction components of which are $B_{x}=0$,$B_{y}=0$, $B_{z}=B_{0}(1+\alpha x)$.Electron velocity :$v <<\frac{e B_{0}}{\alpha m_{e}}$.
Solution
Example Solution: The equation of motion for electron in magnetic field is :
Calculating the cross product yields two equations :
$$O\hat{i} : m\frac{d{v_{x}}}{dt}=ev_{y}B_{0}(1+\alpha x)$$ $$O\hat{j} : m\frac{d{v_{y}}}{dt}=-ev_{x}B_{0}(1+\alpha x)$$ The drift velocity should be perpendicular to the direction in which the field varies . Using the fact that $\alpha$ is small , and using the solutions for the case in which $\alpha=0$ : $$x^{(0)}(t)=r_{L}cos(\omega t)$$ $$y^{(0)}(t)=r_{L}sin(\omega t)$$ , $r_{L} = \frac{mv_{0}}{eB_{0}}$, by perturbation theory $v \approx v^{0}+\delta v$,
$$m\frac{d{v^{0}_{x}}}{dt}=e(v^{0}_{y}+\delta v)B_{0}(1+\alpha x)$$ $$m\frac{d{(v^{0}_{y}+\delta v)}}{dt}=-ev^{0}_{x}B_{0}(1+\alpha x)$$ $=>$ , Averaging over time yields yields: left part equal to zero and : $e<v^{0}_{y}x>B_{0}\alpha+eB_{0}<v^{0}_{y}>=0$, time average over one period of $<v^{0}_{y}x> = \frac{1}{T}\int_{0}^{T} \omega r_{L}^{2} cos^{2}\omega t \,dt = \frac{\omega^{2} T }{2\pi} r_{L}^{2} \frac{1+sin2\omega T}{4} = \frac{v_{0}r_{L}}{4}$ , $v_{D} = \alpha \frac{mv_{0}^{2}}{4eB_{0}}$
