Solution: Average Velocity in Regions of Different B-Field
Analysis: The particle drifts along the x-axis by performing alternating arcs in the two regions. Assume $B_2 > B_1$, so radius $R_1 > R_2$. The particle moves forward in region 1 and loops back partially in region 2.
1. Displacement per cycle ($\Delta x$):
The chord length of an arc leaving at angle $\theta$ and returning is $2R \sin\theta$.
Displacement in Region 1 (Forward): $x_1 = 2 R_1 \sin\theta$
Displacement in Region 2 (Backward/Looping): $x_2 = -2 R_2 \sin\theta$
Net Displacement:
$$ D = 2(R_1 – R_2)\sin\theta = 2 \frac{mv}{q} \sin\theta \left( \frac{1}{B_1} – \frac{1}{B_2} \right) $$
$$ D = \frac{2mv \sin\theta}{q} \frac{B_2 – B_1}{B_1 B_2} $$
2. Time per cycle ($\Delta t$): In Region 1, the velocity vector rotates by $2\theta$. $$ t_1 = \frac{2\theta}{\omega_1} = \frac{2\theta m}{qB_1} $$ In Region 2, the particle completes the loop, turning by $2\pi – 2\theta$. $$ t_2 = \frac{2\pi – 2\theta}{\omega_2} = \frac{(2\pi – 2\theta) m}{qB_2} $$ Total Time $T = t_1 + t_2 = \frac{2m}{q} \left[ \frac{\theta}{B_1} + \frac{\pi – \theta}{B_2} \right]$ $$ T = \frac{2m}{q} \left[ \frac{\theta B_2 + (\pi – \theta) B_1}{B_1 B_2} \right] $$
3. Average Velocity ($\langle v \rangle$): $$ \langle v \rangle = \frac{D}{T} = \frac{\frac{2mv \sin\theta (B_2 – B_1)}{q B_1 B_2}}{\frac{2m (\theta B_2 + \pi B_1 – \theta B_1)}{q B_1 B_2}} $$ Canceling common terms ($2m/q$, $B_1 B_2$):