A charged particle of mass m and charge q moving under the influence of uniform electric field E i-c

Q: Why does the magnetic field do zero work?

The magnetic force is always perpendicular to the velocity of the charged particle ($\mathbf{F}_B = q(\mathbf{v} \times \mathbf{B})$). Since work is the dot product of force and displacement (or velocity for power), and the dot product of perpendicular vectors is zero, the magnetic field does no work.

Q: How did we know the work done by the electric field is $2qaE$?

The electric field is uniform and points along the positive x-axis ($\mathbf{E} = E\hat{i}$). The particle moves from $x=0$ to $x=2a$. The work done is the integral of $q\mathbf{E} \cdot d\mathbf{r}$, which simplifies to $qE \int_0^{2a} dx = 2qaE$.

Q: Why is the power at point Q zero?

At point Q, the particle is moving purely in the negative y-direction. The electric force is in the x-direction, and the magnetic force is in the x-direction. Since both forces are perpendicular to the velocity, their dot products with velocity are zero, resulting in zero power.

Q: How do you calculate the angular momentum at point P?

Angular momentum is $\mathbf{L} = \mathbf{r} \times \mathbf{p}$. At point P, the position vector is $\mathbf{r} = a\hat{j}$ and the momentum is $\mathbf{p} = mv\hat{i}$. The cross product $a\hat{j} \times mv\hat{i}$ gives $-mva\hat{k}$.

Q: What is the difference between the magnitude of angular momentum and the magnitude of the difference?

The magnitude of angular momentum at P is $mva$, and at Q is $4mva$. The difference between these magnitudes is $4mva - mva = 3mva$. The magnitude of the difference would be $|\mathbf{L}_Q - \mathbf{L}_P| = |-4mva\hat{k} - (-mva\hat{k})| = |-3mva\hat{k}| = 3mva$. In this specific case, they yield the same result, but it's important to read the question carefully!

Question

A charged particle of mass $m$ and charge $q$ moving under the influence of uniform electric field $E\hat{i}$ and a uniform magnetic field $B\hat{k}$ follows a trajectory from point $P$ to $Q$ as shown in the figure. The velocities at $P$ and $Q$ are respectively, $v\hat{i}$ and $-2v\hat{j}$. Then, which of the following statements (1, 2, 3, 4) are the correct?
(Trajectory shown is schematic and not to scale)
1. $E = \frac{3}{4}\left(\frac{mv^2}{qa}\right)$
2. Rate of work done by the electric field at $P$ is $\frac{3}{4}\left(\frac{mv^3}{a}\right)$.
3. Rate of work done by both the fields at $Q$ is zero.
4. The difference between the magnitude of angular momentum of the particle at $P$ and $Q$ is $2mav$.

Options:

(A) (1), (3), (4)

(B) (1), (2), (3)

(C) (1), (2), (3), (4)

(D) (2), (3), (4)

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine a charged particle embarking on a journey through a region filled with both an electric and a magnetic field. The particle starts at point $P(0, a)$ with an initial velocity $\mathbf{v}_P = v\hat{i}$. It curves through space and eventually reaches point $Q(2a, 0)$ with a new velocity $\mathbf{v}_Q = -2v\hat{j}$.

The environment is governed by a uniform electric field pointing to the right ($\mathbf{E} = E\hat{i}$) and a uniform magnetic field pointing out of the page ($\mathbf{B} = B\hat{k}$). Our goal is to evaluate four distinct physical statements about this journey.

### The Master Equation: Work-Energy Theorem

To determine the magnitude of the electric field $E$, we need a bridge between the forces acting on the particle and its change in speed. The Work-Energy Theorem is the perfect tool: $W_{net} = \Delta K$.

Crucially, the magnetic force $\mathbf{F}_B = q(\mathbf{v} 	imes \mathbf{B})$ is always perpendicular to the velocity vector. Because work is the dot product of force and displacement, a perpendicular force does **zero work**. Therefore, the entire change in kinetic energy is solely due to the electric field.

The work done by the uniform electric field as the particle moves from $x = 0$ to $x = 2a$ is:
$$W_E = \int_P^Q \mathbf{F}_E \cdot d\mathbf{r} = \int_0^{2a} qE \, dx = 2qaE$$

The change in kinetic energy from $P$ to $Q$ is:
$$\Delta K = \frac{1}{2}m(2v)^2 - \frac{1}{2}mv^2 = \frac{3}{2}mv^2$$

Equating the two yields the electric field:
$$2qaE = \frac{3}{2}mv^2 \implies E = \frac{3}{4}\left(\frac{mv^2}{qa}ight)$$
This confirms that **Statement 1 is correct**.

### Power and Rate of Work Done

The "rate of work done" is simply the mechanical power, defined as $P = \mathbf{F}_{net} \cdot \mathbf{v}$. Let's evaluate this at both points.

**At Point P:**
The velocity is $\mathbf{v}_P = v\hat{i}$. The net force is $\mathbf{F}_{net} = qE\hat{i} + q(v\hat{i} 	imes B\hat{k}) = qE\hat{i} - qvB\hat{j}$.
Taking the dot product:
$$P_P = (qE\hat{i} - qvB\hat{j}) \cdot (v\hat{i}) = qEv$$
Substituting our expression for $E$:
$$P_P = q \left(\frac{3}{4}\frac{mv^2}{qa}ight) v = \frac{3}{4}\left(\frac{mv^3}{a}ight)$$
This confirms that **Statement 2 is correct**.

**At Point Q:**
The velocity is $\mathbf{v}_Q = -2v\hat{j}$. The electric force is $qE\hat{i}$, which is perpendicular to $\mathbf{v}_Q$, so its power contribution is zero. The magnetic force is always perpendicular to velocity, so its power contribution is also zero. Thus, the total rate of work done at $Q$ is exactly zero. 
This confirms that **Statement 3 is correct**.

### The Angular Momentum Check

Finally, let's check the angular momentum, defined as $\mathbf{L} = \mathbf{r} 	imes \mathbf{p} = \mathbf{r} 	imes m\mathbf{v}$.

**At Point P:**
$$\mathbf{L}_P = (a\hat{j}) 	imes (mv\hat{i}) = -mva\hat{k}$$
The magnitude is $|\mathbf{L}_P| = mva$.

**At Point Q:**
$$\mathbf{L}_Q = (2a\hat{i}) 	imes (-2mv\hat{j}) = -4mva\hat{k}$$
The magnitude is $|\mathbf{L}_Q| = 4mva$.

The difference between these magnitudes is:
$$\Delta L = 4mva - mva = 3mva$$

Statement 4 claims the difference is $2mav$, which is mathematically false. Therefore, **Statement 4 is incorrect**.

Since statements 1, 2, and 3 are correct, the right choice is option (b).

Select Answer:

Visualized Solution

Initial Setup of $E$ and $B$ Fields

Work-Energy Theorem

Work Done by Electric Field

Evaluating Statement 1

Power at Point $P$

Evaluating Statement 2

Evaluating Statement 3

Angular Momentum Calculation

Evaluating Statement 4

The Sigma Insight: Magnetic Force on Charged Particle in Magnetic Field

Analyzing the Setup

The Master Equation

Power and Rate of Work Done

The Angular Momentum Check