GATE EE 2023 Set 1 — Question 43

The problem describes the motion of a charge in crossed electric and magnetic fields. 1. Analyze the Fields and Forces:
First, we must note a discrepancy between the text and the figure. The text states that $\vec{E}$ is in the x-direction and $\vec{B}$ is in the z-direction. However, the figure clearly shows the electric field $\vec{E}$ along the positive z-axis and the magnetic field $\vec{B}$ along the positive x-axis. In such cases, the figure is typically considered the ground truth. We will proceed with the directions from the figure:
$\vec{E} = E \hat{k}$
$\vec{B} = B \hat{i}$The force on the positive charge 'q' is given by the Lorentz force equation:$$\vec{F} = q(\vec{E} + \vec{v} \times \vec{B})$$2. Initial Motion:
The charge is released from rest, so its initial velocity $\vec{v}(0) = 0$. The initial force is purely due to the electric field:$$\vec{F}(0) = q\vec{E} = qE\hat{k}$$This force is in the positive z-direction, so the charge begins to accelerate along the z-axis.3. Subsequent Motion and Equations of Motion:
As the charge gains velocity, the magnetic force comes into play. Let the velocity at any time be $\vec{v} = v_x \hat{i} + v_y \hat{j} + v_z \hat{k}$. The force equation becomes:$$m\vec{a} = q(E\hat{k} + (v_x \hat{i} + v_y \hat{j} + v_z \hat{k}) \times B\hat{i})$$$$m\vec{a} = q(E\hat{k} + v_x B (\hat{i} \times \hat{i}) + v_y B (\hat{j} \times \hat{i}) + v_z B (\hat{k} \times \hat{i}))$$Using the cross product rules ($\hat{i} \times \hat{i} = 0$, $\hat{j} \times \hat{i} = -\hat{k}$, $\hat{k} \times \hat{i} = \hat{j}$):$$m\vec{a} = q(E\hat{k} - v_y B \hat{k} + v_z B \hat{j})$$$$m\vec{a} = qv_z B \hat{j} + q(E - v_y B)\hat{k}$$4. Evaluate the Options:

• Option (B): The charge will always move on the y-z plane only.

From the force equation, $m\vec{a} = m(a_x \hat{i} + a_y \hat{j} + a_z \hat{k}) = 0 \cdot \hat{i} + qv_z B \hat{j} + q(E - v_y B)\hat{k}$.
The x-component of the acceleration is $a_x = 0$. Since the charge starts from rest ($v_x(0) = 0$), its velocity in the x-direction will always remain zero. Therefore, the motion is confined to the y-z plane. Option (B) is correct.

• Option (D): The charge will progress in the direction of y.

The motion in crossed E and B fields results in a cycloidal path superimposed on a constant drift velocity. The drift velocity is given by:
$$\vec{v}_d = \frac{\vec{E} \times \vec{B}}{B^2}$$ Substituting the field vectors:
$$\vec{E} \times \vec{B} = (E\hat{k}) \times (B\hat{i}) = EB (\hat{k} \times \hat{i}) = EB \hat{j}$$ So, the drift velocity is:
$$\vec{v}_d = \frac{EB \hat{j}}{B^2} = \frac{E}{B} \hat{j}$$ The drift velocity is in the positive y-direction. This means that while the charge follows an oscillatory path, its average position moves, or 'progresses', in the positive y-direction. Option (D) is correct.

• Option (A): The charge will move in the direction of z with constant velocity.

This is incorrect. The force, and thus acceleration, has components in both y and z directions, and they are not constant (they depend on velocity components). The velocity will not be constant or purely in the z-direction.

• Option (C): The trajectory of the charge will be a circle.

This is incorrect. Circular motion occurs in a pure magnetic field. The presence of the crossed electric field results in a cycloidal trajectory, not a circle.Therefore, the correct statements are (B) and (D).