Solution
Problem Analysis:
A particle starts from the origin with velocity $u$ at an angle $\theta$ to the magnetic field (x-direction). It travels a distance $d$ along x, hits a rigid plane elastically, loses half its charge (becomes $q/2$), and returns to the origin. On the return path, it crosses the x-axis $n$ times.
Step 1: Collision Dynamics
The collision is elastic with a rigid plane perpendicular to the x-axis. The component of velocity parallel to the x-axis ($u_x = u \cos\theta$) reverses direction. The perpendicular component ($u_{\perp} = u \sin\theta$) remains unchanged.
After collision:
- Velocity along x: $-u \cos\theta$ (moves back towards origin).
- New Charge: $q’ = q/2$.
Step 2: Return Journey Analysis
The particle travels back distance $d$. The condition “crossing the x-axis $n$ times” implies that the distance $d$ corresponds to $n$ full pitches of the helical path with the new charge parameters.
The new time period of revolution is:
$$ T’ = \frac{2\pi m}{q’ B} = \frac{2\pi m}{(q/2) B} = \frac{4\pi m}{qB} $$The pitch on the return journey ($p’$) is:
$$ p’ = |u_x| \cdot T’ = (u \cos\theta) \frac{4\pi m}{qB} $$Step 3: Calculating Distance d
Since the particle crosses the x-axis $n$ times during the return, the total distance $d$ is $n$ times the new pitch:
$$ d = n \times p’ $$ $$ d = n \left[ \frac{4\pi m u \cos\theta}{qB} \right] $$