In a vacuum chamber, a particle of charge 1 uC and mass 1 mg is projected with a velocity (i + 2j) m

Q: Why doesn't the magnetic field affect the vertical motion?

The magnetic force is given by the cross product $\vec{F} = q(\vec{v} \times \vec{B})$. Since the magnetic field $\vec{B}$ is strictly along the y-axis, the resulting force will always be perpendicular to the y-axis (i.e., in the XZ plane). Therefore, it cannot influence the vertical motion.

Q: Why is the initial y-velocity zero in the second phase?

We calculated it using the first equation of motion for the first phase: $v_y = u_y + a_y t = 2 - 10(0.2) = 0$. The particle exactly reached its maximum height at $t = 0.2\ \text{s}$.

Q: Is the trajectory in the second phase a perfect circle?

No, it's a helix. While the projection of the path on the XZ plane is a perfect circle due to the magnetic force, the particle is simultaneously accelerating downwards due to gravity, stretching the circle into a helical shape with an increasing pitch.

Q: What does 'radius of trajectory' mean in Option C?

In the context of helical motion in a uniform magnetic field, 'radius of trajectory' conventionally refers to the radius of the circular projection of the path on the plane perpendicular to the magnetic field. Here, it's the radius of the circle in the XZ plane.

Q: Why did we use $\tau = t - 0.2$ for the second phase?

Using $\tau$ simply resets our time clock to zero at the beginning of the second phase. This allows us to use the standard kinematic equations ($s = ut + \frac{1}{2}at^2$) directly without having to carry over complex time offsets.

Question

In a vacuum chamber, a particle of charge $1\ \mu	ext{C}$ and mass $1\ 	ext{mg}$ is projected with a velocity $(\hat{i} + 2\hat{j})\ 	ext{ms}^{-1}$ from the $XZ$ plane at time $t = 0$ in an electric field of $1\hat{i}\ 	ext{Vm}^{-1}$. At $t = 0.2\ 	ext{s}$, the electric field is switched off and a magnetic field of $6\hat{j}\ 	ext{T}$ is switched on. The acceleration due to gravity is $-10\hat{j}\ 	ext{ms}^{-2}$. Correct option(s) is/are:

Options:

(A) The vertical distance of the particle from the $XZ$ plane at $t = 0.3\ 	ext{s}$ is $15\ 	ext{cm}$.

(B) The vertical distance of the particle from the $XZ$ plane at $t = 0.4\ 	ext{s}$ is $10\ 	ext{cm}$.

(C) The radius of the trajectory of the particle for $t > 0.2\ 	ext{s}$ is $20\ 	ext{cm}$.

(D) The particle will be in the $XZ$ plane at $t = 0.35\ 	ext{s}$.

Sigma Percentile · Accepted Answer

### The Setup: A Tale of Two Phases

Imagine a vacuum chamber where a tiny charged particle is fired from the $XZ$ plane. We are given its mass $m = 10^{-6}\ 	ext{kg}$, charge $q = 10^{-6}\ 	ext{C}$, and an initial velocity vector $\vec{v}_0 = \hat{i} + 2\hat{j}\ 	ext{m/s}$.

This problem is a beautiful symphony of kinematics and electromagnetism. To solve it elegantly, we must break the particle's journey into two distinct phases based on the fields acting upon it.

### Phase 1: The Reign of the Electric Field ($t = 0$ to $0.2\ 	ext{s}$)

For the first $0.2\ 	ext{s}$, an electric field $\vec{E} = 1\hat{i}\ 	ext{V/m}$ is active, and gravity $\vec{g} = -10\hat{j}\ 	ext{m/s}^2$ is constantly pulling the particle down.

Using Newton's second law, the net acceleration is:

$$\vec{a}_1 = \frac{q\vec{E}}{m} + \vec{g} = \frac{10^{-6}(1\hat{i})}{10^{-6}} - 10\hat{j} = \hat{i} - 10\hat{j}\ 	ext{m/s}^2$$

Notice how the $x$ and $y$ motions are completely independent. Let's find out how fast it's moving right when the electric field is switched off at $t = 0.2\ 	ext{s}$. Using the first equation of motion:

$$\vec{v}_{(0.2)} = \vec{v}_0 + \vec{a}_1 t = (\hat{i} + 2\hat{j}) + (\hat{i} - 10\hat{j})(0.2)$$

$$\vec{v}_{(0.2)} = \hat{i} + 2\hat{j} + 0.2\hat{i} - 2\hat{j} = 1.2\hat{i}\ 	ext{m/s}$$

The $y$-component of velocity perfectly cancels out! At this exact moment, the particle has reached its peak and is moving purely along the $x$-axis. But how high is this peak? Let's use the second equation of motion for the $y$-axis:

$$y_{(0.2)} = u_y t + \frac{1}{2} a_y t^2 = 2(0.2) - \frac{1}{2}(10)(0.2)^2 = 0.4 - 0.2 = 0.2\ 	ext{m}$$

So, at $t = 0.2\ 	ext{s}$, the particle is $20\ 	ext{cm}$ above the $XZ$ plane.

### Phase 2: The Magnetic Twist ($t > 0.2\ 	ext{s}$)

Now comes the twist. The electric field vanishes, and a magnetic field $\vec{B} = 6\hat{j}\ 	ext{T}$ turns on. The particle is moving along the $x$-axis, so the magnetic force $\vec{F}_B = q(\vec{v} 	imes \vec{B})$ pushes it along the $z$-axis.

Crucially, the magnetic force acts only in the $XZ$ plane. It has **absolutely no effect** on the vertical $y$-motion! The particle simply falls freely under gravity from its peak.

Let's define a new time variable $	au = t - 0.2$ to make our lives easier. The vertical position is given by:

$$y(	au) = y_0 + u_y 	au + \frac{1}{2} a_y 	au^2 = 0.2 + 0 - 5	au^2$$

Let's check the options. At $t = 0.3\ 	ext{s}$ (which means $	au = 0.1\ 	ext{s}$):

$$y = 0.2 - 5(0.1)^2 = 0.2 - 0.05 = 0.15\ 	ext{m} = 15\ 	ext{cm}$$

This makes **Option A absolutely correct!**

What about $t = 0.4\ 	ext{s}$ ($	au = 0.2\ 	ext{s}$)?

$$y = 0.2 - 5(0.2)^2 = 0.2 - 0.2 = 0\ 	ext{m}$$

The particle hits the $XZ$ plane exactly at $t = 0.4\ 	ext{s}$. This means Option B is incorrect, and Option D is also incorrect because it hits at $0.4\ 	ext{s}$, not $0.35\ 	ext{s}$.

### The Helical Trajectory

Finally, let's look at the circular motion in the $XZ$ plane caused by the magnetic field. The radius of this circular projection is given by the classic formula:

$$R = \frac{mv_{\perp}}{qB} = \frac{10^{-6} 	imes 1.2}{10^{-6} 	imes 6} = 0.2\ 	ext{m} = 20\ 	ext{cm}$$

So, **Option C is also correct!** The particle traces a beautiful downward helix, spiraling with a radius of $20\ 	ext{cm}$ while accelerating towards the floor.

Select Answer:

Visualized Solution

\text{Initial Setup}

\text{Phase 1: } t \in [0, 0.2]\ \text{s}

\text{Velocity at } t = 0.2\ \text{s}

\text{Position at } t = 0.2\ \text{s}

\text{Phase 2: } t > 0.2\ \text{s}

\text{Vertical Motion for } t > 0.2\ \text{s}

\text{Checking Options B and D}

\text{Radius of Trajectory}

The Sigma Insight: Motion of a Charge in Magnetic Fields

The Setup

Phase 1

Phase 2

The Helical Trajectory

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