A uniform magnetic field B exists in the region between x = 0 and x = 3R/2 (region 2 in the figure)

Q: Why does the particle move in a circle?

Because the magnetic force is always perpendicular to the velocity vector, it acts as a centripetal force, changing the direction of the particle without changing its speed, resulting in uniform circular motion.

Q: How do we know the center of the circle is on the y-axis?

The particle enters region 2 at $x=0$ moving purely in the x-direction. The magnetic force at this instant is purely in the y-direction. Since the force points towards the center of the circular path, the center must lie on the y-axis.

Q: What does 'longest possible path in region 2' mean?

It means the particle completes a full semicircle within region 2, just grazing the boundary of region 3 at $x = 3R/2$. This corresponds to the maximum possible radius $r = 3R/2$ for a particle that re-enters region 1.

Q: Why is the change in momentum not just zero if the speed is constant?

Momentum is a vector quantity. Even though the speed (magnitude of velocity) is constant, the direction of velocity changes continuously. Therefore, the momentum vector changes, and the change in momentum is non-zero.

Q: Why is the distance of re-entry directly proportional to mass?

The distance of re-entry is the diameter of the circular path, $2r$. Since $r = \frac{mv}{qB}$, for a fixed velocity, charge, and magnetic field, the radius (and thus the re-entry distance) increases linearly with the mass of the particle.

Question

A uniform magnetic field $B$ exists in the region between $x = 0$ and $x = \frac{3R}{2}$ (region 2 in the figure) pointing normally into the plane of the paper. A particle with charge $+Q$ and momentum $p$ directed along $X$-axis enters region 2 from region 1 at point $P_1(y = -R)$. Which of the following option(s) is/are correct?

Options:

(A) When the particle re-enters region 1 through the longest possible path in region 2, the magnitude of the change in its linear momentum between point $P_1$ and the farthest point from $Y$-axis is $\frac{p}{\sqrt{2}}$.

(B) For $B = \frac{8}{13} \frac{p}{QR}$, the particle will enter region 3 through the point $P_2$ on $X$-axis.

(C) For $B > \frac{2}{3} \frac{p}{QR}$, the particle will re-enter region 1.

(D) For a fixed $B$, particles of same charge $Q$ and same velocity $v$, the distance between the point $P_1$ and the point of re-entry into region 1 is inversely proportional to the mass of the particle.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine a charged particle, carrying a positive charge $+Q$, zooming along the $X$-axis with a momentum $p$. It crosses the boundary at $x=0$ and enters a mysterious zone—Region 2. This region is filled with a uniform magnetic field $B$ that points directly into the page.

What happens the moment it crosses the threshold? The magnetic field grabs hold of it. According to the Lorentz force law, the magnetic force is given by the cross product:

$$ \mathbf{F} = Q(\mathbf{v} 	imes \mathbf{B}) $$

Since the velocity $\mathbf{v}$ is along the positive $x$-direction ($\hat{i}$) and the magnetic field $\mathbf{B}$ is into the page ($-\hat{k}$), the right-hand rule tells us the force is directed along the positive $y$-axis ($\hat{j}$). This perpendicular force acts as a centripetal force, causing the particle to veer upwards and trace a circular path.

### The Master Equation of the Path

The radius $r$ of this circular path is a classic result of balancing the magnetic force with the required centripetal force:

$$ r = \frac{mv}{QB} = \frac{p}{QB} $$

Because the particle starts at $P_1(0, -R)$ and curves upwards, the center of its circular orbit must lie on the $y$-axis, exactly a distance $r$ away from the starting point. Thus, the center $C$ is located at $(0, -R+r)$. The mathematical equation governing its journey is:

$$ x^2 + (y - (r-R))^2 = r^2 $$

### Evaluating the Options

**Option (a): The Longest Path**
The longest possible path the particle can take in Region 2 without spilling over into Region 3 is a perfect semicircle that just grazes the boundary at $x = \frac{3R}{2}$. This means the maximum radius is $r = \frac{3R}{2}$.

At the starting point $P_1$, the momentum is purely horizontal: $\mathbf{p}_1 = p \hat{i}$. At the farthest point from the $y$-axis (the apex of the semicircle), the particle is moving purely vertically, so its momentum is $\mathbf{p}_2 = p \hat{j}$. The change in momentum is:

$$ \Delta \mathbf{p} = \mathbf{p}_2 - \mathbf{p}_1 = p \hat{j} - p \hat{i} $$

The magnitude of this change is $|\Delta \mathbf{p}| = \sqrt{p^2 + (-p)^2} = p\sqrt{2}$. Option (a) incorrectly claims it is $\frac{p}{\sqrt{2}}$.

**Option (b): Hitting the Target $P_2$**
What if the particle perfectly threads the needle and hits point $P_2\left(\frac{3R}{2}, 0ight)$? We simply plug these coordinates into our master circle equation:

$$ \left(\frac{3R}{2}ight)^2 + (0 - (r-R))^2 = r^2 $$

Expanding this yields:

$$ \frac{9R^2}{4} + r^2 - 2rR + R^2 = r^2 $$

The $r^2$ terms beautifully cancel out, leaving us with $\frac{13R^2}{4} = 2rR$, which simplifies to $r = \frac{13R}{8}$.

Since we know $r = \frac{p}{QB}$, we can equate the two:

$$ \frac{p}{QB} = \frac{13R}{8} \implies B = \frac{8p}{13QR} $$

This matches Option (b) perfectly!

**Option (c): The U-Turn**
For the particle to execute a U-turn and re-enter Region 1, its circular path must be entirely contained within Region 2. This dictates that its radius $r$ must be strictly less than the width of the region:

$$ r < \frac{3R}{2} $$

Substituting our expression for $r$:

$$ \frac{p}{QB} < \frac{3R}{2} \implies B > \frac{2p}{3QR} $$

Option (c) is absolutely correct.

**Option (d): The Re-entry Distance**
When the particle completes its semicircle and re-enters Region 1, it emerges at a distance of $2r$ from its starting point $P_1$.

$$ 	ext{Distance} = 2r = \frac{2mv}{QB} $$

For a fixed magnetic field $B$, charge $Q$, and velocity $v$, this distance is **directly proportional** to the mass $m$. Option (d) falsely claims it is inversely proportional.

Select Answer:

Visualized Solution

\text{Initial Setup and Magnetic Force}

\text{Geometry of the Circular Path}

\text{Checking Option (a): Change in Momentum}

\text{Checking Option (b): Passing through } P_2

\text{Checking Option (c): Re-entering Region 1}

\text{Checking Option (d): Distance of Re-entry}

The Sigma Insight: Motion of a Charge in Magnetic Fields

Analyzing the Setup

The Master Equation of the Path

Evaluating the Options

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