The region between y=0 and y=d contains a magnetic field B = B k^. A particle of mass m and charge q

Q: Why does the particle move in a circular path?

The magnetic force $\mathbf{F} = q(\mathbf{v} \times \mathbf{B})$ is always perpendicular to the velocity vector. A force that is always perpendicular to motion changes the direction of the velocity but not its magnitude, which is the exact condition for uniform circular motion.

Q: How do we know the particle enters at $y=d$ and not $y=0$?

If the particle entered at $y=0$ with velocity $v\hat{\mathbf{i}}$, the magnetic force $-qvB\hat{\mathbf{j}}$ would immediately push it into the region $y<0$, where there is no magnetic field. For it to travel through the field region $0 \le y \le d$, it must enter at $y=d$.

Q: What is the difference between average and instantaneous acceleration here?

Average acceleration is the total change in velocity divided by the total time taken ($\Delta \mathbf{v} / \Delta t$). Instantaneous acceleration is the acceleration at a specific moment, given by Newton's second law $\mathbf{a} = \mathbf{F}/m$. The question asks for the acceleration *at the point of emergence*, which implies instantaneous acceleration.

Q: Why doesn't the calculated answer match any of the options?

This appears to be an error in the original exam question. The options might have been derived using an incorrect assumption or a sign error during the problem's creation. The mathematically rigorous derivation leads to the answer we found.

Q: Could the question be asking for the magnitude of acceleration?

The options are given as vectors (with $\hat{\mathbf{i}}$ and $\hat{\mathbf{j}}$ components), so it is definitely asking for the acceleration vector, not just its magnitude. The magnitude would simply be $qvB/m$ throughout the motion.

Question

The region between $y=0$ and $y=d$ contains a magnetic field $\mathbf{B} = B\hat{\mathbf{k}}$. A particle of mass $m$ and charge $q$ enters the region with a velocity $\mathbf{v} = v\hat{\mathbf{i}}$. If $d = \frac{mv}{2qB}$, then the acceleration of the charged particle at the point of its emergence at the other side is

Options:

(A) $\frac{qvB}{m}\left(\frac{\sqrt{3}}{2}\hat{\mathbf{i}} + \frac{1}{2}\hat{\mathbf{j}}\right)$

(B) $\frac{qvB}{m}\left(\frac{1}{2}\hat{\mathbf{i}} - \frac{\sqrt{3}}{2}\hat{\mathbf{j}}\right)$

(C) $\frac{qvB}{m}\left(\frac{-\hat{\mathbf{j}} + \hat{\mathbf{i}}}{\sqrt{2}}\right)$

(D) $\frac{qvB}{m}\left(\frac{\hat{\mathbf{i}} + \hat{\mathbf{j}}}{\sqrt{2}}\right)$

Sigma Percentile · Accepted Answer

### The Curious Case of the Emerging Particle

Imagine a charged particle, full of kinetic energy, suddenly entering a region permeated by a uniform magnetic field. What happens next is a beautiful dance dictated by the laws of electromagnetism. The magnetic field exerts a force that is always perpendicular to the particle's velocity, acting as a perfect centripetal force. This forces the particle into a circular trajectory.

### Analyzing the Setup

In our specific problem, the particle enters a magnetic field $\mathbf{B} = B\hat{\mathbf{k}}$ that exists only between $y=0$ and $y=d$. It enters with an initial velocity $\mathbf{v} = v\hat{\mathbf{i}}$. Using the right-hand rule for the Lorentz force, $\mathbf{F} = q(\mathbf{v} 	imes \mathbf{B})$, we find that the initial force is directed along $-\hat{\mathbf{j}}$. This means the particle will curve downwards into the region.

We are given a crucial piece of information: the width of the region is $d = \frac{mv}{2qB}$. We also know that the radius of the circular path is $r = \frac{mv}{qB}$.

**This immediately tells us that $d = \frac{r}{2}$.**

Because the particle curves downwards from $y=d$, the center of its circular path must lie at a distance $r$ below the entry point, placing the center at $(0, d-r) = (0, -d)$.

### The Master Equation for Deflection

The particle will emerge from the magnetic field when it crosses the boundary at $y=0$. Let's determine the angle $	heta$ by which its velocity vector has rotated. From the geometry of the circular path, the vertical distance the particle travels before emerging is $r(1 - \cos	heta)$.

Equating this to the width of the region $d$, we get:

$$d = r(1 - \cos	heta)$$

Substituting $d = \frac{r}{2}$:

$$\frac{r}{2} = r(1 - \cos	heta)$$

$$\cos	heta = \frac{1}{2}$$

**This reveals that the angle of deflection is exactly $	heta = 60^\circ$.**

### Final Calculation: Velocity and Acceleration

Since the velocity vector has rotated $60^\circ$ clockwise, the final velocity $\mathbf{v}_f$ at the point of emergence is:

$$\mathbf{v}_f = v\cos(60^\circ)\hat{\mathbf{i}} - v\sin(60^\circ)\hat{\mathbf{j}} = \frac{v}{2}\hat{\mathbf{i}} - \frac{v\sqrt{3}}{2}\hat{\mathbf{j}}$$

The question asks for the instantaneous acceleration at this exact point. According to Newton's second law and the Lorentz force:

$$\mathbf{a} = \frac{q}{m}(\mathbf{v}_f 	imes \mathbf{B})$$

Let's carefully compute the cross product:

$$\mathbf{a} = \frac{q}{m} \left( \left(\frac{v}{2}\hat{\mathbf{i}} - \frac{v\sqrt{3}}{2}\hat{\mathbf{j}}ight) 	imes B\hat{\mathbf{k}} ight)$$

$$\mathbf{a} = \frac{qB}{m} \left( -\frac{v}{2}\hat{\mathbf{j}} - \frac{v\sqrt{3}}{2}\hat{\mathbf{i}} ight)$$

**$$\mathbf{a} = -\frac{qvB}{m} \left( \frac{\sqrt{3}}{2}\hat{\mathbf{i}} + \frac{1}{2}\hat{\mathbf{j}} ight)$$**

Interestingly, if you look closely at the options provided in the original exam, none of them perfectly match this mathematically rigorous result. This is a classic example of a flawed question where trusting your fundamental derivations is more important than trying to force-fit an answer into incorrect options.

Select Answer:

Visualized Solution

$R a d i u s o f C i r c u l a r P a t h$

$A na l y z in g t h e G eo m e t r y$

$F in d in g t h e A n g l e o f D e f l ec t i o n$

$V e l oc i t y a t E m er g e n ce$

$C a l c u l a t in g I n s t an t an eo u s A cce l er a t i o n$

The Sigma Insight: Motion of a Charge in Magnetic Fields

The Curious Case of the Emerging Particle

Analyzing the Setup

The Master Equation for Deflection

Final Calculation

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