A particle of charge +q and mass m moving under the influence of a uniform electric field E i^ and u

Q: Why doesn't the magnetic field do any work?

Because the magnetic force $\mathbf{F}_m = q(\mathbf{v} \times \mathbf{B})$ is always perpendicular to the velocity $\mathbf{v}$. The work done is $\int \mathbf{F}_m \cdot d\mathbf{r} = \int (\mathbf{F}_m \cdot \mathbf{v}) dt = 0$.

Q: Why do we only consider the x-displacement for the electric field's work?

The electric field is uniform and points only in the x-direction ($\mathbf{E} = E\hat{i}$). The dot product $\mathbf{E} \cdot d\mathbf{r} = (E\hat{i}) \cdot (dx\hat{i} + dy\hat{j}) = E dx$. Thus, the y-displacement doesn't contribute to the work done by this electric field.

Q: How did we find the change in kinetic energy?

Kinetic energy is a scalar quantity given by $\frac{1}{2}mv^2$, where $v$ is the speed (magnitude of velocity). At $P$, speed is $v$. At $Q$, speed is $|-2v\hat{j}| = 2v$. So $\Delta K = \frac{1}{2}m(2v)^2 - \frac{1}{2}mv^2$.

Q: Why is the rate of work done at Q zero?

At point Q, the particle is moving vertically downwards ($-y$ direction), but the electric force is still pulling it horizontally ($+x$ direction). Since the force and velocity are at a $90^\circ$ angle, their dot product (power) is zero.

Q: What role does the magnetic field play if it does no work?

The magnetic field is responsible for bending the particle's path. Without it, the particle would just accelerate in a straight line along the x-axis due to the electric field. The magnetic field provides the necessary perpendicular force to curve the trajectory from P to Q.

Question

A particle of charge $+q$ and mass $m$ moving under the influence of a uniform electric field $E\hat{i}$ and uniform magnetic field $B\hat{k}$ follows a trajectory from $P$ to $Q$ as shown in figure. The velocities at $P$ and $Q$ are $v\hat{i}$ and $-2v\hat{j}$. Which of the following statement(s) is/are correct ?

Options:

(A) $E = \frac{3}{4}\left[\frac{mv^2}{qa}\right]$

(B) Rate of work done by the electric field at $P$ is $\frac{3}{4}\left[\frac{mv^3}{a}\right]$

(C) Rate of work done by the electric field at $P$ is zero

(D) Rate of work done by both the fields at $Q$ is zero

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are tracking the journey of a charged particle moving through a region filled with both an electric and a magnetic field. The particle, carrying a charge $+q$ and mass $m$, starts its journey at point $P(0, a)$ with an initial velocity $\mathbf{v}_P = v\hat{i}$. It eventually reaches point $Q(2a, 0)$ with a final velocity $\mathbf{v}_Q = -2v\hat{j}$.

We are given that the electric field is uniform and points in the positive x-direction, $\mathbf{E} = E\hat{i}$, while the magnetic field points out of the page, $\mathbf{B} = B\hat{k}$. Our goal is to determine the magnitude of the electric field and analyze the rate of work done (power) at different points along the trajectory.

### The Master Equation: Work-Energy Theorem

To find the electric field, we can rely on a fundamental principle of mechanics: the **Work-Energy Theorem**. This theorem states that the net work done on a particle equals its change in kinetic energy:

$$W_{	ext{net}} = \Delta K$$

There are two forces acting on our particle: the electric force $\mathbf{F}_e = q\mathbf{E}$ and the magnetic force $\mathbf{F}_m = q(\mathbf{v} 	imes \mathbf{B})$.

Here is a crucial insight: the magnetic force is always perpendicular to the particle's velocity. Because work is the dot product of force and displacement, a perpendicular force does absolutely **zero work**. Therefore, the entire change in the particle's kinetic energy is solely due to the work done by the electric field.

### Calculating the Electric Field

Let's set up our equations. The work done by the constant electric field is the force multiplied by the displacement in the direction of the field. Since the electric field is strictly along the x-axis, we only care about the displacement in the x-direction, which is $2a - 0 = 2a$.

$$W_e = \int \mathbf{F}_e \cdot d\mathbf{r} = \int_{0}^{2a} (qE\hat{i}) \cdot (dx\hat{i} + dy\hat{j}) = qE(2a)$$

On the other side of the equation, the change in kinetic energy is the final kinetic energy at $Q$ minus the initial kinetic energy at $P$. The speed at $Q$ is $|-2v\hat{j}| = 2v$, and the speed at $P$ is $v$.

$$\Delta K = K_Q - K_P = \frac{1}{2}m(2v)^2 - \frac{1}{2}mv^2 = \frac{1}{2}m(4v^2) - \frac{1}{2}mv^2 = \frac{3}{2}mv^2$$

Equating the work done to the change in kinetic energy:

$$qE(2a) = \frac{3}{2}mv^2$$

Rearranging this to solve for $E$, we get:

$$E = \frac{3mv^2}{4qa}$$

This perfectly matches option (a)!

### Power at Point P

Next, let's check the rate of work done, also known as power, at point $P$. Power is defined as the dot product of force and velocity: $P = \mathbf{F} \cdot \mathbf{v}$.

At point $P$, both the electric force $\mathbf{F}_e = qE\hat{i}$ and the velocity $\mathbf{v}_P = v\hat{i}$ are pointing in the positive x-direction. Their dot product is simply the product of their magnitudes:

$$P_P = (qE\hat{i}) \cdot (v\hat{i}) = qEv$$

Substituting the value of $E$ we just derived:

$$P_P = q \left(\frac{3mv^2}{4qa}ight) v = \frac{3mv^3}{4a}$$

This confirms that option (b) is correct, which automatically means option (c) is incorrect.

### Power at Point Q

Finally, what about the rate of work done at point $Q$?

At this location, the particle's velocity is purely in the negative y-direction ($\mathbf{v}_Q = -2v\hat{j}$), while the electric force remains steadfast in the positive x-direction ($\mathbf{F}_e = qE\hat{i}$).

Because the force and velocity vectors are perpendicular to each other, their dot product is zero:

$$P_{e,Q} = (qE\hat{i}) \cdot (-2v\hat{j}) = 0$$

And since the magnetic force's power is always zero regardless of the position, the total rate of work done by both fields at $Q$ is indeed zero. This makes option (d) correct as well.

By systematically applying the Work-Energy Theorem and the definition of power, we have successfully verified that options (a), (b), and (d) are the correct statements.

Select Answer:

Visualized Solution

Visualizing the Setup

Work-Energy Theorem

Setting up the Equations

Calculating Electric Field

Power at Point P

Power at Point Q

Conclusion

The Sigma Insight: Motion of a Charge in Magnetic Fields

Analyzing the Setup

The Master Equation

Calculating the Electric Field

Power at Point P

Power at Point Q

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Select Answer:

Visualized Solution

Visualizing the Setup

Work-Energy Theorem

Setting up the Equations

Calculating Electric Field

Power at Point P

Power at Point Q

Conclusion

The Sigma Insight: Motion of a Charge in Magnetic Fields

Analyzing the Setup

The Master Equation

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