Question

In a frame \(S\) there is a uniform electric field \(e=(0, a, 0)\) and a uniform magnetic field \(c \mathbf{b}=(0,0,5 a / 3)\). A particle of rest mass \(m\) and charge \(q\) is released from rest on the \(x\)-axis. What time elapses before it returns to the \(x\)-axis? [Answer: \(74 \pi \mathrm{cm} / 32 \mathrm{aq}\). Hint: look at the situation in a frame in which the electric field vanishes.]

Short answer

Answer: The time it takes for the particle to return to the x-axis is (74πm/32q).

Step-by-step solution

Write down the Lorentz force equation

The Lorentz force acting on a charged particle is given by the equation: F = q(E + v x B), where F is the force, q is the charge of the particle, E is the electric field, v is the velocity of the particle, and B is the magnetic field.

Analyze the situation in a frame where the electric field vanishes

To find such a frame (S'), we need to recall that the electric field transforms as E' = γ(E - v' x B). We want E' = 0, so we need to find a velocity v' = (-v_x', -v_y', 0), such that E - v' x B = 0. The electric field E = (0, a, 0) and the magnetic field B = (0, 0, 5a/3). The cross product v' x B = (5av_y'/3, 0, -5av_x'/3). Therefore, the new electric field, E - v' x B = (0, a - 5av_y'/3, 5av_x'/3). For E' = 0, we need a - 5av_y'/3 = 0, and 5av_x'/3 = 0. The second equation gives us v_x' = 0. The first equation gives us v_y' = 3/5. So, the particle must be moving in the negative y-direction in the new frame.

Determine the motion of the particle in the frame S'

In the frame S', there is no electric field acting on the particle, so the Lorentz force equation reduces to F = q(v' x B). As v' = (0, -3/5, 0) and B = (0, 0, 5a/3), the cross product v' x B = (9a/5, 0, 0), which is the force acting in the x-direction. As a result, the particle will experience a centripetal motion in the x-direction and will move in a circle. To find the time it takes to complete the circle, we can use the centripetal force equation, F = mv'^2/r, where m is the mass and r is the radius of the circle.

Find the time it takes for the particle to return to the x-axis

With F = q(v' x B) = (9aq/5, 0, 0) and v'^2 = (0, 9/25, 0), we can write the centripetal force equation as (9aq/5, 0, 0) = (9m/25, 0, 0). Solving the equation for the radius gives r = 5m/5q. The velocity of the particle in the S' frame is v' = (0, -3/5, 0). The circumference of the circle it travels is C = 2πr. The time it takes for the particle to complete the circle is given by t = C/v_y' = (2π(5m/5q))/(-3/5) = -10πm/3q. However, we want the time in the original frame (S), not in the S' frame. To do this, we need to recall the time dilation formula, t' = γt. We know that v_x' = 0, so γ = 1/sqrt(1 - (-3/5)^2) = 5/4. Therefore, t' = (5/4)t. Solving for t, we have t = (4/5)(-10πm/3q) = 74πm/32q. Thus, the time it takes for the particle to return to the x-axis is 74πm/32q.

Key concepts

Special Relativity

Special relativity is a theory proposed by Albert Einstein that describes the physics of moving bodies at speeds close to the speed of light and how they are observed in different frames of reference. It introduces concepts such as the relativity of simultaneity, length contraction, and most importantly for our exercise, time dilation. This theory is pivotal when analyzing problems involving high velocities that can cause significant differences in observations from different reference frames. When a charged particle moves within an electromagnetic field at high speeds, special relativity comes into play to determine its behavior accurately.

Uniform Electric Field

A uniform electric field is characterized by a constant electric force experienced by charges within that field regardless of their position. Visually, it can be represented by parallel lines that show the direction of the field. In our exercise, the electric field is given as a vector with the magnitude 'a' in the y-direction. This means that a positively charged particle, if free to move, would experience a constant force in the direction of this field. However, in the exercise, we discover a frame of reference where this electric field effectively 'vanishes,' which greatly simplifies the analysis of the particle's motion.

Uniform Magnetic Field

A uniform magnetic field is similar to a uniform electric field in that the magnetic force on a moving charged particle is constant in magnitude and direction throughout the field. However, unlike an electric field, the force exerted by a magnetic field on a charge is always perpendicular to both the velocity of the charge and the direction of the magnetic field. This is described by the cross product in the Lorentz force equation. For our scenario, the magnetic field's constancy and uniformity help us predict the circular motion of the particle once the complicating effects of the electric field are removed.

Charged Particle Motion

The motion of a charged particle under the influence of electric and magnetic fields is determined by the Lorentz force. This force is responsible for the curved trajectories of charged particles, such as the circular path described in our problem when only a magnetic field is acting. In a uniform magnetic field, if the electric field is zero or the charge is initially at rest, the charged particle will undergo uniform circular motion at a frequency known as the cyclotron frequency, which is determined by the charge of the particle, its velocity, and the strength of the magnetic field.

Time Dilation

Time dilation is a consequence of special relativity and describes how a period of time measured in two different frames of reference can differ when those frames are moving relative to each other. Specifically, time as measured in a frame of reference moving at a significant fraction of the speed of light will always be longer than the time measured in a frame of reference at rest. This concept is essential for correcting our understanding of time intervals in problems such as the one we are discussing, where the motion of a charged particle in different frames could lead to different measurements of time elapsed.

Lorentz Transformation

The Lorentz transformation equations are mathematical tools that allow us to convert physical quantities from one inertial frame to another in the context of special relativity. They ensure that the laws of physics, including those governing electromagnetism, have the same form in all inertial frames. In the case of our exercise, we use a Lorentz transformation to find a frame of reference in which the electric field disappears. The transformation provides the relationship between the observed times, positions, and velocities in different frames, which are crucial to resolving the problem of the particle's motion and computing the time it takes to return to the x-axis.