Question

Two particles of charges + Q and − Q are projected from the same point with a velocity v in a region of unifrom magnetic filed B such that the velocity vector makes an angle 0 with the magnetic filed Their masses are M and 2 M respectively Then, they will meet again for the first time at a point whose distance from the point of projection is .

• A. $\frac{2 \pi M v \cos \theta}{Q B}$
• B. $\frac{4 \pi M v \cos \theta}{Q B}$
• C. $\frac{8 \pi M v \cos \theta}{Q B}$
• D. $\frac{16 \pi M v \cos \theta}{Q B}$

Answer 1

Answer: (B)

$\frac{4 \pi M v \cos \theta}{Q B}$

Answer 2

When a charged particle moves in a uniform magnetic field, its motion can be decomposed into two independent components:

1. Motion parallel to the magnetic field ($v_{\parallel}$): This component of velocity remains constant, $v_{\parallel} = v \cos \theta$. The particle moves with uniform velocity along the direction of the magnetic field.

2. Motion perpendicular to the magnetic field ($v_{\perp}$): This component of velocity causes the particle to move in a circle, $v_{\perp} = v \sin \theta$.

The combination of these two motions results in a helical path.

Key parameters for the helical path:

• Radius of the circular path (r): The magnetic force ($qv_{\perp}B$) provides the centripetal force ($\frac{mv_{\perp}^2}{r}$). $r = \frac{m v_{\perp}}{q B} = \frac{m v \sin \theta}{q B}$

• Time period of the circular motion (T): $T = \frac{2 \pi r}{v_{\perp}} = \frac{2 \pi (m v \sin \theta / q B)}{v \sin \theta} = \frac{2 \pi m}{q B}$ Note that the time period is independent of the velocity component perpendicular to the field and the angle $\theta$.

• Pitch of the helix (p): This is the distance moved along the magnetic field direction in one time period. $p = v_{\parallel} T = (v \cos \theta) \left( \frac{2 \pi m}{q B} \right)$

Let's analyze the motion of the two particles:

Particle 1:

• Charge $q_1 = +Q$
• Mass $m_1 = M$
• Velocity component parallel to B: $v_{\parallel 1} = v \cos \theta$
• Time period of circular motion: $T_1 = \frac{2 \pi M}{Q B}$

Particle 2:

• Charge $q_2 = -Q$
• Mass $m_2 = 2M$
• Velocity component parallel to B: $v_{\parallel 2} = v \cos \theta$ (same as particle 1)
• Time period of circular motion: $T_2 = \frac{2 \pi (2M)}{|-Q| B} = \frac{4 \pi M}{Q B}$

Meeting condition:

Both particles are projected from the same point. For them to meet again for the first time, they must be at the same position in space at the same time.

1. Motion along the magnetic field (z-axis, assuming B is along z): The displacement along the magnetic field at time $t$ is $z(t) = v_{\parallel} t$. Since $v_{\parallel 1} = v_{\parallel 2} = v \cos \theta$, their z-coordinates will always be identical at any given time $t$, assuming they start from the same z-position (e.g., $z=0$). So, $z_1(t) = z_2(t) = (v \cos \theta) t$.

2. Motion in the plane perpendicular to the magnetic field (x-y plane): Both particles start from the same point (origin) in the x-y plane. For them to meet again, their (x,y) coordinates must be identical at time $t$. Since their charges are opposite, they will trace circular paths in opposite directions. However, for them to meet at the same point in the x-y plane, they must both return to the origin of their circular motion. This happens when each particle completes an integer number of full revolutions. Let $t_{meet}$ be the time they meet for the first time. For particle 1 to return to the origin in the x-y plane, $t_{meet}$ must be an integer multiple of $T_1$: $t_{meet} = n_1 T_1$. For particle 2 to return to the origin in the x-y plane, $t_{meet}$ must be an integer multiple of $T_2$: $t_{meet} = n_2 T_2$. The first time this occurs is when $t_{meet}$ is the least common multiple (LCM) of $T_1$ and $T_2$.

   We have $T_1 = \frac{2 \pi M}{Q B}$ and $T_2 = \frac{4 \pi M}{Q B}$. Notice that $T_2 = 2 T_1$. The LCM of $T_1$ and $2T_1$ is $2T_1$. Therefore, the time when they meet again for the first time is $t_{meet} = 2T_1 = T_2 = \frac{4 \pi M}{Q B}$. At this time, particle 1 has completed 2 revolutions, and particle 2 has completed 1 revolution, bringing both back to the origin in the x-y plane.

Distance from the point of projection:

The distance from the point of projection is the displacement along the magnetic field direction at time $t_{meet}$. Distance $D = z(t_{meet}) = v_{\parallel} \cdot t_{meet}$ $D = (v \cos \theta) \left( \frac{4 \pi M}{Q B} \right)$ $D = \frac{4 \pi M v \cos \theta}{Q B}$