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 $v \approx v^{0}+\delta v$,
$e<v^{0}_{y}x>B_{0}\alpha+eB_{0}<v^{0}_{y}>=0$, time average over one period of $<v^{0}_{y}x> = \frac{1}{T}\int_{0}^{T}\omega r_{L}^{2} cos^{2}\omega t \,dt = \frac{\omega^{2} T }{2\pi} r_{L}^{2}\frac{1+sin2\omega T}{4} = \frac{v_{0}r_{L}}{4}$ , $v_{D} = \alpha\frac{mv_{0}^{2}}{4eB_{0}}$
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$e<v^{0}_{y}x>B_{0}\alpha+eB_{0}<v^{0}_{y}>=0$, time average over one period of $<v^{0}_{y}x> = \frac{1}{T}\int_{0}^{T}\omega r_{L}^{2} cos^{2}\omega t \,dt = \frac{\omega^{2} T }{2\pi} r_{L}^{2}\frac{1+sin2\omega T}{4} = \frac{v_{0}r_{L}}{4}$ , $<v^{0}_{y}>=v_{D} = \alpha\frac{mv_{0}^{2}}{4eB_{0}}$ , which is the correct way to derive the drift velocity for small gradient of B.
#### Answer
$v_{D} = \alpha\frac{mv_{0}^{2}}{4eB_{0}}$
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### Statement
### Statement
$10.2.2.$ Estimate the electron drift velocity across an inhomogeneous magnetic field,
$10.2.2.$ Estimate the electron drift velocity across an inhomogeneous magnetic field,
the induction components of which are $B_{x}=0 $,$B_{y}=0 $,
the induction components of which are $B_{x}=0 $,$B_{y}=0 $,
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 $\alpha=0$ :
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 $\alpha=0$ :
$$x^{(0)}(t)=r_{L}cos(\omega t)$$
$$x^{(0)}(t)=r_{L}cos(\omega t)$$
$$y^{(0)}(t)=r_{L}sin(\omega t)$$
$$y^{(0)}(t)=r_{L}sin(\omega t)$$
, $r_{L} = \frac{mv_{0}}{eB_{0}}$, by perturbation theory $v \approx v^{0}+\delta v$,
, $r_{L} = \frac{mv_{0}}{eB_{0}}$, by perturbation theory $v \approx v^{0}+\delta v$,
$e<v^{0}_{y}x>B_{0}\alpha+eB_{0}<v^{0}_{y}>=0$, time average over one period of $<v^{0}_{y}x> = \frac{1}{T}\int_{0}^{T}\omega r_{L}^{2} cos^{2}\omega t \,dt = \frac{\omega^{2} T }{2\pi} r_{L}^{2}\frac{1+sin2\omega T}{4} = \frac{v_{0}r_{L}}{4}$ , $v_{D} = \alpha\frac{mv_{0}^{2}}{4eB_{0}}$
$e<v^{0}_{y}x>B_{0}\alpha+eB_{0}<v^{0}_{y}>=0$, time average over one period of $<v^{0}_{y}x> = \frac{1}{T}\int_{0}^{T}\omega r_{L}^{2} cos^{2}\omega t \,dt = \frac{\omega^{2} T }{2\pi} r_{L}^{2}\frac{1+sin2\omega T}{4} = \frac{v_{0}r_{L}}{4}$ , $<v^{0}_{y}>=v_{D} = \alpha\frac{mv_{0}^{2}}{4eB_{0}}$ , which is the correct way to derive the drift velocity for small gradient of B.