In the xy-plane, the region y 0 has a uniform magnetic field B1 k^ and the region y

Q: Why does the particle move in a semicircle and not a full circle?

The magnetic field exists only in specific regions ($y>0$ and $y<0$). When the particle completes a semicircle, it crosses the boundary (X-axis) into a region with a different magnetic field, which changes its path.

Q: Why is the force in the second region directed towards the negative X-axis?

When the particle crosses the X-axis, its velocity is directed along the negative Y-axis ($-v_0 \hat{\mathbf{j}}$). The magnetic field is still along the positive Z-axis ($B_2 \hat{\mathbf{k}}$). The cross product $(-\hat{\mathbf{j}}) \times \hat{\mathbf{k}}$ results in $-\hat{\mathbf{i}}$, meaning the force is in the negative X direction.

Q: Why do we add the distances $2R_1$ and $2R_2$ instead of subtracting them?

The question asks for the average *speed* along the X-axis, which is a scalar quantity defined as total distance over total time. Even though the particle moves left in the second region, the distance covered is always positive. If it asked for average *velocity*, we would subtract them.

Q: Does the speed of the particle change when it enters the second magnetic field?

No, magnetic forces do no work because they always act perpendicular to the velocity. Therefore, the kinetic energy and the speed of the particle remain constant throughout the motion.

Q: What would happen if $B_2$ was in the $-\hat{\mathbf{k}}$ direction?

If $B_2$ was in the $-\hat{\mathbf{k}}$ direction, the force in the second region would be in the positive X direction. The particle would continue to drift to the right, and both its distance and displacement along the X-axis would be $2R_1 + 2R_2$.

Question

In the $xy$-plane, the region $y > 0$ has a uniform magnetic field $B_1 \hat{\mathbf{k}}$ and the region $y < 0$ has another uniform magnetic field $B_2 \hat{\mathbf{k}}$. A positively charged particle is projected from the origin along the positive $Y$-axis with speed $v_0 = \pi 	ext{ ms}^{-1}$ at $t = 0$, as shown in figure. Neglect gravity in this problem. Let $t = T$ be the time when the particle crosses the $X$-axis from below for the first time. If $B_2 = 4B_1$, the average speed of the particle, in $	ext{ms}^{-1}$, along the $X$-axis in the time interval $T$ is ......... .

Sigma Percentile · Accepted Answer

The problem of a charged particle moving through multiple magnetic fields is a classic test of both your conceptual clarity and your mathematical precision. In this scenario, we are dealing with a positively charged particle navigating two distinct regions of uniform magnetic fields. Let's break down the journey step-by-step.

### Analyzing the Setup

The particle is projected from the origin along the positive Y-axis with an initial speed $v_0 = \pi 	ext{ ms}^{-1}$. The $xy$-plane is divided into two regions:
1.  **Region 1 ($y > 0$):** The magnetic field is $\mathbf{B}_1 = B_1 \hat{\mathbf{k}}$.
2.  **Region 2 ($y < 0$):** The magnetic field is $\mathbf{B}_2 = B_2 \hat{\mathbf{k}}$, where $B_2 = 4B_1$.

Our goal is to find the **average speed** of the particle along the X-axis from the moment it is projected until it crosses the X-axis from below for the first time.

### The First Semicircle: A Rightward Arc

As the particle enters Region 1, its velocity is $\mathbf{v}_0 = v_0 \hat{\mathbf{j}}$. The magnetic force acting on it is given by the Lorentz force law:
$$ \mathbf{F}_1 = q(\mathbf{v}_0 	imes \mathbf{B}_1) = q(v_0 \hat{\mathbf{j}} 	imes B_1 \hat{\mathbf{k}}) = qv_0 B_1 \hat{\mathbf{i}} $$
Since the force is directed along the positive X-axis, the particle will trace a semicircular path to the right. The radius of this path is:
$$ R_1 = \frac{mv_0}{qB_1} $$
The time taken to complete this semicircle is half of the full time period:
$$ t_1 = \frac{\pi m}{qB_1} $$
Upon completing this semicircle, the particle crosses the X-axis at a distance of $2R_1$ from the origin.

### The Second Semicircle: A Leftward Hook

At the instant the particle crosses the X-axis into Region 2, its velocity is directed downwards, so $\mathbf{v} = -v_0 \hat{\mathbf{j}}$. The magnetic field here is $\mathbf{B}_2 = B_2 \hat{\mathbf{k}}$. The new magnetic force is:
$$ \mathbf{F}_2 = q(-v_0 \hat{\mathbf{j}} 	imes B_2 \hat{\mathbf{k}}) = -qv_0 B_2 \hat{\mathbf{i}} $$
This force is directed along the negative X-axis! Consequently, the particle traces a new semicircle, this time hooking back towards the left. The radius of this second semicircle is:
$$ R_2 = \frac{mv_0}{qB_2} $$
And the time taken is:
$$ t_2 = \frac{\pi m}{qB_2} $$

### The Crucial Distinction: Speed vs Velocity

Here lies the trap that catches many students. The question asks for the **average speed** along the X-axis, not the average velocity. 
Average speed is defined as the total distance traveled divided by the total time. 
- In the first region, the distance traveled along the X-axis is $2R_1$.
- In the second region, the distance traveled along the X-axis is $2R_2$.
Therefore, the total distance traveled along the X-axis is $d = 2R_1 + 2R_2$. The total time elapsed is $T = t_1 + t_2$.

### The Final Calculation

We are given that $B_2 = 4B_1$. This relationship allows us to express $R_2$ and $t_2$ in terms of $R_1$ and $t_1$:
$$ R_2 = \frac{mv_0}{q(4B_1)} = \frac{R_1}{4} $$
$$ t_2 = \frac{\pi m}{q(4B_1)} = \frac{t_1}{4} $$
Now, let's substitute these into our expressions for total distance and total time:
$$ d = 2R_1 + 2\left(\frac{R_1}{4}ight) = \frac{5}{2} R_1 = \frac{5mv_0}{2qB_1} $$
$$ T = t_1 + \frac{t_1}{4} = \frac{5}{4} t_1 = \frac{5\pi m}{4qB_1} $$
Finally, we calculate the average speed:
$$ 	ext{Average Speed} = \frac{d}{T} = \frac{\frac{5mv_0}{2qB_1}}{\frac{5\pi m}{4qB_1}} = \frac{2v_0}{\pi} $$
Given that the initial speed $v_0 = \pi 	ext{ ms}^{-1}$, we find:
$$ 	ext{Average Speed} = \frac{2(\pi)}{\pi} = \mathbf{2 	ext{ ms}^{-1}} $$

Enter Numerical Value:

Visualized Solution

\text{Initial Setup}

\text{Motion in Region 1 } (y > 0)

\text{Motion in Region 2 } (y < 0)

\text{Distance and Time}

\text{Substitution}

\text{Final Calculation}

The Sigma Insight: Motion of a Charge in Magnetic Fields

Analyzing the Setup

The First Semicircle

The Second Semicircle

The Crucial Distinction

The Final Calculation

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