Edits to “Solution”, “Answer”

Anonymous edited
revision #13067 parent #13063 ← older newer →
@@ -28,10 +28,10 @@Solution
$$ m\frac{d{(v^{0}_{y}+\delta v)}}{dt}=-ev^{0}_{x}B_{0}(1+\alpha x) $$
$=>$ 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} $ , $<v^{0}_{y}>=v_{D} = \alpha \frac{mv_{0}^{2}}{4eB_{0}}$ , which is the correct way to derive the drift velocity for small gradient of B.
+$e<v^{0}_{y}x>B_{0}\alpha+eB_{0}<v^{0}_{y}>=0$, 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{2+sin2\omega T}{4} = \frac{v_{0}r_{L}}{2} $ , $<v^{0}_{y}>=v_{D} = \alpha \frac{mv_{0}^{2}}{2eB_{0}}$ , which is the correct way to derive the drift velocity for small gradient of B.
#### Answer
−$v_{D} = \alpha \frac{mv_{0}^{2}}{4eB_{0}}$
+$v_{D} = \alpha \frac{mv_{0}^{2}}{2eB_{0}}$