Step-by-step Derivations

1. Motion in the region y > 0

A positively charged particle with charge $q$ and mass $m$ is projected from the origin with velocity $\vec{v} = v_0 \widehat{j}$. In the region $y > 0$, the magnetic field is $\vec{B_1} = B_1 \widehat{k}$. The magnetic Lorentz force on the particle is given by $\vec{F_1} = q(\vec{v} \times \vec{B_1})$. Initially, at $t=0$, the force is: $\vec{F_1} = q(v_0 \widehat{j} \times B_1 \widehat{k}) = qv_0 B_1 (\widehat{j} \times \widehat{k}) = qv_0 B_1 \widehat{i}$ Since the force is perpendicular to the velocity, the particle moves in a circular path. The force is directed towards the center of the circle. As the initial force is in the $+\widehat{i}$ direction, the particle curves to the right, and the center of the circle must lie on the positive x-axis. The radius of this circular path, $R_1$, is determined by the balance between the magnetic force and the centripetal force: $qv_0 B_1 = \frac{mv_0^2}{R_1} \implies R_1 = \frac{mv_0}{qB_1}$ The particle completes a semicircle in the $y>0$ region, starting from the origin $(0,0)$. It will cross the x-axis again at a point whose x-coordinate is the diameter of the semicircle, i.e., at $(2R_1, 0)$. The time taken to complete this semicircle, $t_1$, is half the time period of the circular motion: $t_1 = \frac{\text{distance}}{\text{speed}} = \frac{\pi R_1}{v_0} = \frac{\pi}{v_0} \left( \frac{mv_0}{qB_1} \right) = \frac{\pi m}{qB_1}$ At time $t_1$, the particle is at $(2R_1, 0)$ and its velocity is $\vec{v} = -v_0 \widehat{j}$.

2. Motion in the region y < 0

The particle now enters the region $y < 0$ with velocity $\vec{v} = -v_0 \widehat{j}$. The magnetic field here is $\vec{B_2} = B_2 \widehat{k}$. The magnetic force is $\vec{F_2} = q(\vec{v} \times \vec{B_2})$: $\vec{F_2} = q(-v_0 \widehat{j} \times B_2 \widehat{k}) = -qv_0 B_2 (\widehat{j} \times \widehat{k}) = -qv_0 B_2 \widehat{i}$ The force is in the $-\widehat{i}$ direction, so the particle curves to the left. It will again follow a semicircular path in the $y<0$ region. The radius of this path, $R_2$, is: $R_2 = \frac{mv_0}{qB_2}$ The particle starts this semicircle at $(2R_1, 0)$ and completes it, crossing the x-axis again. This is the first time it crosses the x-axis from below. The diameter of this semicircle is $2R_2$. The particle's final x-coordinate will be $2R_1 - 2R_2$. The time taken for this second semicircle, $t_2$, is: $t_2 = \frac{\pi R_2}{v_0} = \frac{\pi m}{qB_2}$ The total time elapsed until the particle crosses the x-axis from below for the first time is $T = t_1 + t_2$.

3. Calculating the Average Speed along the x-axis

The question asks for the "average speed of the particle ... along the x-axis". This is interpreted as the total distance traveled in the x-direction divided by the total time interval $T$.

• Total distance traveled along x-axis ($d_x$): In the first semicircle, the particle's x-coordinate changes from $0$ to $2R_1$. The distance covered along x is $d_{x1} = 2R_1$. In the second semicircle, the x-coordinate changes from $2R_1$ to $2R_1 - 2R_2$. The distance covered along x is $d_{x2} = |(2R_1 - 2R_2) - 2R_1| = |-2R_2| = 2R_2$. The total distance is $d_x = d_{x1} + d_{x2} = 2R_1 + 2R_2 = 2(R_1 + R_2)$.

• Total time taken ($T$): $T = t_1 + t_2 = \frac{\pi R_1}{v_0} + \frac{\pi R_2}{v_0} = \frac{\pi}{v_0}(R_1 + R_2)$.

• Average speed along x-axis ($\langle s_x \rangle$): $\langle s_x \rangle = \frac{d_x}{T} = \frac{2(R_1 + R_2)}{\frac{\pi}{v_0}(R_1 + R_2)}$ The term $(R_1 + R_2)$ cancels out. $\langle s_x \rangle = \frac{2}{\pi/v_0} = \frac{2v_0}{\pi}$ Notice that this result is independent of the magnetic field strengths $B_1$ and $B_2$, and hence the condition $B_2 = 4B_1$ is not needed for this calculation. It would be required if we were to calculate the average velocity along the x-axis (rac{\Delta x}{T}).

4. Final Calculation

We are given the initial speed $v_0 = \pi \, m{s^{-1}}$. Substituting this value into our expression for the average speed along the x-axis: $\langle s_x \rangle = \frac{2(\pi)}{\pi} = 2 \, m{s^{-1}}$

The average speed of the particle along the x-axis is 2 m/s.