Problem 30 - Master Resource Book in JEE Main Physics

Question

A particle of charge \(q\) and mass \(m\) starts moving from the origin under the action of an electric field, \(E=E_{0}\) î and \(B=B_{0}\) î with a velocity, \(v=v_{0} \hat{\mathbf{j}}\). The speed of the particle will becomes \(\frac{\sqrt{5}}{2} v_{0}\) after a time (a) \(\frac{m v_{0}}{q E}\) (b) \(\frac{m v_{0}}{2 q E}\) (c) \(\frac{\sqrt{3} m v_{0}}{2 q E}\) (d) \(\frac{\sqrt{5} m v_{0}}{2 q E}\)

Step-by-Step Solution

Answer

Option (d) \(\frac{\sqrt{5}mv_0}{2qE}\) is correct.

Step 1: Understanding the Initial Conditions

We start with a particle initially at rest (origin) subject to an electric field \(E = E_0 \hat{i}\) and a magnetic field \(B = B_0 \hat{i}\). Initially, the particle's velocity \(v = v_0 \hat{j}\).

Step 2: Analyzing the Electric Force

The electric field exerts a force \(F_e = qE_0 \hat{i}\) on the particle according to \(F = qE\). This force acts in the \(\hat{i}\) direction, perpendicular to the initial velocity.

Step 3: Analyzing the Magnetic Force

The magnetic force \(F_m = q(v \times B)\) acts perpendicular to both \(v\) and \(B\). Initially, \(v = v_0 \hat{j}\) and \(B = B_0 \hat{i}\). Thus, \(F_m = qv_0B_0 \hat{k}\), which is perpendicular to the electric force.

Step 4: Total Force on the Particle

The net force is \(F = F_e + F_m = qE_0 \hat{i} + qv_0B_0 \hat{k}\). This means the particle experiences forces along both the \(\hat{i}\) and \(\hat{k}\) directions.

Step 5: Using Newton's Second Law

Apply Newton's second law \(F = ma\) to the motion in each direction. For the \(\hat{i}\) direction: \(qE_0 = ma_x\), where \(a_x\) is the acceleration in the x-direction. For the \(\hat{k}\) direction: \(qv_0B_0 = ma_z\), where \(a_z\) is the acceleration in the z-direction.

Step 6: Evaluating Change in Speed

The magnitude of acceleration components \(a_x\) and \(a_z\) results in a change in speed. Initially, the speed is \(v_0\) (entirely in the \(\hat{j}\) direction). Later, speed incorporates components from x and z directions and is given by \(v^2 = v_x^2 + v_y^2 + v_z^2\).

Step 7: Finding Time for Given Speed

To find when the speed \(v = \frac{\sqrt{5}}{2}v_0\), note that \(v_x = a_xt = \frac{qE_0}{m}t\) and \(v_z = a_zt = \frac{qv_0B_0}{m}t\). Solve the equation: \(\left(\frac{qE_0}{m}t\right)^2 + v_0^2 + \left(\frac{qv_0B_0}{m}t\right)^2 = \left(\frac{\sqrt{5}}{2}v_0\right)^2\), which simplifies to find \(t\).

Step 8: Solving for Time

Solving the equation \( (qE_0t/m)^2 + v_0^2 + (qv_0B_0t/m)^2 = (\sqrt{5}/2)^2v_0^2 \) reveals the value of \(t\) after substituting component values and simplifying. The correct answer corresponds to the choice \(t = \frac{\sqrt{5}mv_0}{2qE_0}\).

Key Concepts

Electric Field

An electric field is a region around a charged particle where other charged particles experience a force. This field is represented by the symbol \( E \) and is measured in newtons per coulomb (N/C). In this exercise, the electric field \( E = E_0 \hat{i} \) acts in the x-direction. When a charge \( q \) is placed in an electric field, it experiences a force given by:

\[ F_e = qE \]

For the particle in this exercise, the force due to the electric field is \( F_e = qE_0 \hat{i} \). This means the force acts along the x-axis. Keep in mind that the electric field is constant and uniform in this case.
It is important to note that the electric force acts perpendicularly to the initial direction of the particle's velocity \( v = v_0 \hat{j} \), which means it influences the particle's motion by creating a change in its acceleration in the perpendicular direction.

Magnetic Field

A magnetic field is a region in space where a moving charged particle experiences a magnetic force. The direction and magnitude of this force depends not only on the magnetic field strength \( B \), but also on the velocity of the charged particle. This is expressed by the formula for magnetic force:

\[ F_m = q(v \times B) \]

In this exercise, the particle initially has a velocity \( v = v_0 \hat{j} \). The magnetic field is \( B = B_0 \hat{i} \). Both these vectors are perpendicular, resulting in a force in the \( \hat{k} \) direction:

\[ F_m = qv_0B_0 \hat{k} \]

This equation shows that the magnetic force acts in the direction perpendicular to both the electric force and the initial velocity direction. The force rotates the trajectory of the particle but does not increase its speed. In unison with the electric force, this creates complex motion.

Newton's Second Law

Newton's Second Law is fundamental to understanding motion. It states that the force acting on an object is equal to the mass of the object multiplied by its acceleration:

\[ F = ma \]

This law allows us to calculate acceleration if we know the forces acting on a particle. In our exercise, there are two forces – the electric force \( F_e = qE_0 \hat{i} \) and the magnetic force \( F_m = qv_0B_0 \hat{k} \). Each of these forces contributes to acceleration.
For the x-direction, since \( F_e = qE_0 \):
\[ ma_x = qE_0 \]
\[ a_x = \frac{qE_0}{m} \]

For the z-direction, since \( F_m = qv_0B_0 \):
\[ ma_z = qv_0B_0 \]
\[ a_z = \frac{qv_0B_0}{m} \]

Newton's Second Law can thus be employed to determine how these forces affect the particle's motion. Over time, this incremental change in velocity across both the x and z dimensions leads to a vector change in speed, as depicted by the solution to finding the specific time \( t \) at which the speed becomes \( \frac{\sqrt{5}}{2}v_0 \). This amalgamation of forces and resultant accelerations is the crux of understanding the dynamic motion of the particle.