To solve the problem of a charged particle moving in a uniform magnetic field, we need to apply the principles of electromagnetism, particularly the Lorentz force. The particle is released from the origin with an initial velocity in the positive x-direction, and it experiences a magnetic field directed in the negative z-direction. Let's break down the problem step by step.

Understanding the Motion of the Particle

The Lorentz force acting on a charged particle in a magnetic field is given by the equation:

F = q(v × B)

Where:

• F is the force on the particle.
• q is the charge of the particle.
• v is the velocity vector of the particle.
• B is the magnetic field vector.

In this case:

• The velocity vector is v = v₀ i, where i is the unit vector in the x-direction.
• The magnetic field vector is B = -B₀ k, where k is the unit vector in the z-direction.

Calculating the Force

Now, we can calculate the cross product v × B:

v × B = (v₀ i) × (-B₀ k)

Using the right-hand rule for the cross product, we find:

v × B = -v₀B₀ (i × k) = -v₀B₀ j

Thus, the force acting on the particle is:

F = q(-v₀B₀ j)

Equations of Motion

The force results in an acceleration given by Newton's second law:

F = m a

Where m is the mass of the particle and a is its acceleration. Therefore, we have:

ma = -qv₀B₀ j

From this, we can express the acceleration:

a = -\frac{qv₀B₀}{m} j

Finding the Motion in the y-Direction

The acceleration in the y-direction is constant, which means we can use the kinematic equations to find the position of the particle as it moves. The initial velocity in the y-direction is zero, and the particle starts from the origin:

The equation for position in the y-direction is:

y(t) = y₀ + v_{y₀}t + \frac{1}{2} a_y t²

Substituting the known values:

y(t) = 0 + 0 + \frac{1}{2} \left(-\frac{qv₀B₀}{m}\right) t²

This simplifies to:

y(t) = -\frac{qv₀B₀}{2m} t²

Determining the Time of Flight

Next, we need to find the time t when the particle reaches the point (0, y, 0). The motion in the x-direction is uniform since there is no force acting in that direction:

x(t) = v₀ t

At the moment the particle passes through (0, y, 0), we have:

x(t) = 0

This means that the particle must have completed a half-cycle of circular motion in the xy-plane, which occurs when:

t = \frac{2\pi m}{qB₀}

Calculating the y-Position

Substituting this time back into the equation for y(t) gives:

y\left(\frac{2\pi m}{qB₀}\right) = -\frac{qv₀B₀}{2m} \left(\frac{2\pi m}{qB₀}\right)²

After simplifying, we find:

y = -\frac{qv₀}{2B₀} \cdot \frac{2\pi m}{qB₀} = -\frac{v₀\pi m}{B₀²}

Thus, the value of y when the particle passes through (0, y, 0) is:

y = -\frac{v₀\pi m}{B₀²}

In summary, the particle's trajectory in the magnetic field results in a circular motion in the xy-plane, and the y-coordinate can be determined using the derived equations. If you have any further questions or need clarification on any part of the solution, feel free to ask!