Average Speed of Charged Particle in Split Magnetic Fields

To find the average speed of the particle along the $x$-axis in the time interval $T$, we first analyze the motion of the particle in both regions.

1. Motion in the Region $y > 0$:

The particle with charge $q > 0$ and mass $m$ enters the region $y > 0$ at the origin $(0,0)$ with initial velocity $\vec{v} = v_0 \hat{j}$. The magnetic field in this region is $\vec{B}_1 = B_1 \hat{k}$.

The magnetic Lorentz force acting on the particle is: $\vec{F}_1 = q(\vec{v} \times \vec{B}_1) = q(v_0 \hat{j} \times B_1 \hat{k}) = q v_0 B_1 \hat{i}$

Since the force is perpendicular to the velocity and initially directed along the positive $x$-axis, the particle executes uniform circular motion in the $xy$-plane with radius $R_1$: $R_1 = \frac{m v_0}{q B_1}$

The particle traces a semicircle in the upper half-plane:

• It starts at $(0,0)$ and reaches the $x$-axis again at $x_1 = 2 R_1$.
• The time spent in the upper half-plane is: $t_1 = \frac{\pi R_1}{v_0} = \frac{\pi m}{q B_1}$
• At $t = t_1$, the particle is at $(2 R_1, 0)$ with velocity $\vec{v} = -v_0 \hat{j}$.

2. Motion in the Region $y < 0$:

The particle enters the region $y < 0$ at $x = 2 R_1$ with velocity $\vec{v} = -v_0 \hat{j}$. The magnetic field in this region is $\vec{B}_2 = B_2 \hat{k} = 4 B_1 \hat{k}$.

The magnetic Lorentz force acting on the particle is: $\vec{F}_2 = q(\vec{v} \times \vec{B}_2) = q(-v_0 \hat{j} \times 4 B_1 \hat{k}) = -4 q v_0 B_1 \hat{i}$

This force is directed along the negative $x$-axis, causing the particle to trace another semicircle in the lower half-plane with radius $R_2$: $R_2 = \frac{m v_0}{q B_2} = \frac{m v_0}{4 q B_1} = \frac{R_1}{4}$

• The particle curves towards the negative $x$-direction and crosses the $x$-axis from below at: $x_2 = 2 R_1 - 2 R_2 = 2 R_1 - \frac{R_1}{2} = \frac{3}{2} R_1$
• The time spent in this region is: $t_2 = \frac{\pi R_2}{v_0} = \frac{\pi R_1}{4 v_0} = \frac{t_1}{4}$

Thus, the total time until the particle crosses the $x$-axis from below for the first time is: $T = t_1 + t_2 = t_1 + \frac{t_1}{4} = \frac{5}{4} t_1 = \frac{5 \pi R_1}{4 v_0}$

3. Average Speed along the $x$-axis:

The speed along the $x$-axis at any instant is $|v_x|$. The total distance traveled by the particle along the $x$-direction is: $d_x = \int_0^T |v_x|\, dt$

• In the time interval $t \in [0, t_1]$, the $x$-coordinate increases monotonically from $0$ to $2 R_1$, so the distance traveled along the $x$-axis is: $d_{x,1} = 2 R_1$

• In the time interval $t \in [t_1, T]$, the $x$-coordinate decreases monotonically from $2 R_1$ to $2 R_1 - 2 R_2$, so the distance traveled along the $x$-axis is: $d_{x,2} = 2 R_2 = \frac{R_1}{2}$

Thus, the total distance traveled along the $x$-axis is: $d_x = 2 R_1 + 2 R_2 = 2 R_1 + \frac{R_1}{2} = \frac{5}{2} R_1$

The average speed along the $x$-axis is given by: $v_{\text{avg}, x} = \frac{d_x}{T} = \frac{\frac{5}{2} R_1}{\frac{5 \pi R_1}{4 v_0}} = \frac{2 v_0}{\pi}$

Given $v_0 = \pi \text{ m s}^{-1}$: $v_{\text{avg}, x} = \frac{2 \pi}{\pi} = 2 \text{ m s}^{-1}$