To solve the problem of the particle moving under the influence of both electric and magnetic fields, we need to analyze the forces acting on it and how they affect its motion. The particle is subjected to an electric field \( \mathbf{E} = E_0 \hat{j} \) and a magnetic field \( \mathbf{B} = B_0 \hat{k} \). Let's break down the steps to find the speed of the particle when it reaches its maximum y-coordinate \( y_0 \).
Understanding the Forces at Play
The particle experiences two main forces due to the fields:
- Electric Force: The force due to the electric field is given by \( \mathbf{F}_E = q \mathbf{E} \), where \( q \) is the charge of the particle. Since the electric field is directed along the y-axis, this force will act in the positive y-direction.
- Magnetic Force: The magnetic force is given by \( \mathbf{F}_B = q (\mathbf{v} \times \mathbf{B}) \). The direction of this force depends on the velocity of the particle and the magnetic field direction.
Analyzing the Motion
Initially, the particle is released from rest, so it starts with zero velocity. As it moves, the electric force will accelerate it in the y-direction, while the magnetic force will influence its trajectory. The magnetic force acts perpendicular to both the velocity and the magnetic field, causing the particle to move in a curved path.
Finding the Maximum y-Coordinate
At the maximum y-coordinate \( y_0 \), the particle's velocity in the y-direction becomes zero momentarily, and it has reached its peak height. At this point, the forces acting on the particle can be analyzed:
- The electric force \( F_E = qE_0 \) acts upward.
- The magnetic force \( F_B \) acts perpendicular to the velocity, which is directed horizontally at this maximum point.
Applying Newton's Second Law
At the maximum height, the net force in the y-direction must equal the mass times the acceleration. Since the particle is momentarily at rest in the y-direction, we can write:
Net force in y-direction: \( F_E - F_B = 0 \)
Thus, we have:
\( qE_0 = qv_x B_0 \)
Here, \( v_x \) is the horizontal component of the velocity at the maximum height.
Radius of Curvature and Velocity Relationship
The radius of curvature \( R \) at the maximum height is related to the velocity and the magnetic field. The centripetal force required to keep the particle moving in a circular path is provided by the magnetic force:
\( \frac{mv^2}{R} = qvB_0 \)
From this, we can express the speed \( v \) in terms of \( R \) and \( B_0 \):
\( v = \frac{qBR}{m} \)
Combining the Equations
Now, we can substitute the expression for \( v \) back into the equation for the electric force:
From \( qE_0 = qv_x B_0 \), we can express \( v_x \) as:
\( v_x = \frac{E_0}{B_0} \)
At the maximum height, the total speed \( v \) can be expressed as:
\( v = v_x + v_y \) where \( v_y = 0 \) at maximum height.
Final Speed Calculation
Using the relationship between \( R \), \( y_0 \), and the forces, we can derive the final expression for the speed of the particle:
\( v = \frac{E_0}{B_0} \left(1 + \frac{2y_0}{R}\right) \)
Thus, the speed of the particle when its y-coordinate is \( y_0 \) is given by:
\( v = \frac{E_0}{B_0} \left(1 + \frac{2y_0}{R}\right) \)
This result shows how the interplay between electric and magnetic forces determines the motion of charged particles in fields, illustrating fundamental principles of electromagnetism.
