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Chapter 29: Problem 20

Neutral Particle A neutral particle is at rest in a uniform magnetic field \(\vec{B}\). At time \(t=0\) it decays into two charged particles, each of mass \(m\). (a) If the charge of one of the particles is \(+q\), what is the charge of the other? (b) The two particles move off in separate paths, both of which lie in the plane perpendicular to \(\vec{B}\). At a later time the particles collide. Express the time from decay until collision in terms of \(m,|\vec{B}|\), and \(|q|\)

Short Answer

Expert verified
The charge of the other particle is \(-q\). The time from decay until collision is \(\frac{\pi m}{|q| B}\).

Step by step solution

01

Understanding the Charge Conservation

The initial neutral particle has no net charge. According to the law of charge conservation, the total charge must remain zero after the decay. Since one of the decay particles has a charge of +q, the other must have a charge of -q to keep the total charge zero.
02

Analyzing the Motion of Charged Particles

Since the magnetic field, \(\textbf{B}\), is uniform and the particles are moving perpendicularly to \(\textbf{B}\), each particle will travel in a circular path. Let's denote the radius of the circular path as \(r\). The centripetal force for a charged particle in a magnetic field is given by: \[ r = \frac{mv}{|q|B} \] where \(v\) is the velocity of the particle, \(m\) is its mass, and \(q\) is its charge.
03

Time Period of Circular Motion

The time period \(T\) for one full revolution of a charged particle in a magnetic field is: \[ T = \frac{2 \pi r}{v} \] Using the expression for \(r\) from Step 2, we get: \[ T = \frac{2 \pi }{|q|B} \times \frac{mv}{v} = \frac{2 \pi m}{|q|B} \] Thus, the time period \(T\) is \(\frac{2 \pi m}{|q| B}\).
04

Collision Time Calculation

Since the particles start from the same point, they will collide when each has completed exactly half a revolution in opposite directions. This means they will collide at half the time period. Therefore, the collision time \(t_{collision}\) is half of \(T\): \[ t_{collision} = \frac{T}{2} = \frac{2 \pi m}{|q| B} \times \frac{1}{2} = \frac{\pi m}{|q| B} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Charge Conservation
The law of charge conservation is crucial in understanding the behavior of charged particles in any physical process. When a neutral particle decays, the total charge before and after the event must remain the same. In the given exercise, the initial neutral particle has no net charge.

According to charge conservation, if one of the resultant particles from the decay carries a charge of \(+q\), the other must carry a charge of \-q\ to ensure the total charge remains zero. This principle guarantees that the net charge doesn't change, maintaining equilibrium in the system.

Think of charge conservation like balancing a see-saw: if you add a weight on one side, you must add an equal weight on the other to keep it balanced.
Circular Motion
When charged particles move in a uniform magnetic field, they experience a force perpendicular to both their velocity and the magnetic field. This force makes them travel in circular paths. For a particle of mass \(m\) and charge \(q\), moving with velocity \(v\) perpendicular to the magnetic field \(B\), the centripetal force is provided by the magnetic force, given by:

\[ r = \frac{mv}{|q|B} \]

This radius \(r\) represents the distance from the center of the circular path to the moving particle. The key insight here is that the magnetic force is always perpendicular to the velocity of the particle, causing it to move in a circular trajectory.

Imagine it like swinging a ball on a string in a circle; the tension in the string (analogous to the magnetic force) pulls the ball inward, keeping it in its circular path.
Collision Time Calculation
Understanding the time it takes for the particles to collide is essential. Both particles move in circles due to the magnetic field, and their time period \(T\) for one complete revolution is:

\[ T = \frac{2 \pi m}{|q|B} \]

This formula tells us how long it takes a charged particle to go around its circular path once. For the particles to collide, each must travel half a revolution from their starting points in opposite directions. Thus, the collision time \((t_{collision})\) will be exactly half of the time period:

\[ t_{collision} = \frac{T}{2} = \frac{2 \pi m}{|q|B} \times \frac{1}{2} = \frac{\pi m}{|q| B} \]

Think of it like two runners starting at the same point on a circular track but running in opposite directions; they will meet halfway around the circle, meaning it takes them half the total revolution time to collide.

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Most popular questions from this chapter

Horizontal Motion An electron with kinetic energy \(2.5 \mathrm{keV}\) moves horizontally into a region of space in which there is a downward-directed uniform electric field of magnitude \(10 \mathrm{kV} / \mathrm{m}\). (a) What are the magnitude and direction of the (smallest) uniform magnetic field that will cause the electron to continue to move horizontally? Ignore the gravitational force, which is small. (b) Is it possible for a proton to pass through the combination of fields undeflected? If so, under what circumstances?

ABCDEFA Figure \(\quad 29-36\) \(\begin{array}{l}\text { shows } a & \text { current } \text { loop }\end{array}\) \(A B C D E F A\) carrying a current \(i\) \(=5.00 \mathrm{~A}\). The sides of the loop are parallel to the coordinate axes, with \(A B=20.0 \mathrm{~cm}, B C=\) \(30.0 \mathrm{~cm}\), and \(F A=10.0 \mathrm{~cm}\). Calculate the magnitude and direction of the magnetic dipole moment of this loop. (Hint: Imagine equal and opposite currents \(i\) in the line segment \(A D\); then treat the two rectangular loops \(A B C D A\) and \(A D E F A .)\)

A Velocity Selector A group of physicists at Argonne National Laboratory in Illinois wants to bombard metals with monoenergetic beams of alpha particles to study radiation damage. (Alpha particles are helium nuclei, which consist of two neutrons and two protons and thus have a net charge of \(+2 e\) where \(e\) is the amount of the charge on the electron.) They have managed to create a beam of alpha particles from the decay of radioactive elements, but some of the alpha particles lose energy as they collide with other atoms in the source. As a new physicist assigned to the group you have been asked to use a velocity selector to select only the alpha particles in the beam that are close to one velocity and get rid of the others. The velocity selector consists of: (1) a power supply capable of de- livering large potential differences between capacitor plates and (2) a large permanent magnet that has a uniform magnetic field perpendicular to the beam. The setup for the velocity selector is shown in Fig. 29-38. The direction of the \(\mathrm{B}\) -field is out of the paper. Your magnet has a field of \(0.22 \mathrm{~T}\) and the capacitor plates have a spacing of \(2.5 \mathrm{~cm}\). You are asked to figure out how the velocity selector works and then tell your group what voltage to put across the capacitor plates to select a velocity of \(4.2 \times 10^{6} \mathrm{~m} / \mathrm{s} .\) This is your first job and you feel overwhelmed by the assignment, but you calm down and begin to analyze the situation one step at a time. You come up with the following: (a) The magnet is oriented so its magnetic field is out of the paper in the diagram you are given, so you use the right-hand rule to determine the direction of the magnetic force on an alpha particle passing from left to right into the magnetic field. What direction did you come up with for the force? (b) You realize that by using \(\vec{F}^{\prime} \mathrm{mag}=q \vec{v} \times \vec{B}\), you can calculate the magnitude of force on an alpha particle moving at speed \(v\) just as it enters the uniform magnetic field as a function of the charge on the alpha particle and the magnitude of the magnetic field \(B\). What is the expression for the magnitude of the force in terms of \(e\), \(v\), and \(B ?\) (c) You realize that you might be able to put just the right voltage across the two capacitor plates so that the electrical force on a given alpha particle will be equal in magnitude and opposite in direction to the Lorentz magnetic force. Then any alpha particles with just the right velocity will pass straight through the poles of the magnet without being deflected. First you think about whether the voltage on the upper capacitor plate should be positive or negative to give a canceling force. What do you decide? (d) Next you realize that if you know the electric field between the plates and the charge on the alpha particle then you can compute the electrical force on it. What is the relationship between the electrical force \(\vec{F}^{\text {elec }}\), charge, \(q\), and electric field \(\vec{E} ?\) (e) Finally, you use the fact that the magnitude of the electric field between capacitor plates is given by \(E=|\Delta V| / d\) where \(d\) is the spacing between the plates. Show that the voltage needed to have the electrical force and the magnetic force be "equal and opposite" can be calculated using the equation \(|\Delta V|=v B d\). Calculate the voltage needed.

Heavy Ions Physicist S. A. Goudsmit devised a method for measuring the masses of heavy ions by timing their periods of revolution in a known magnetic field. A singly charged ion of iodine makes \(7.00\) rev in a field of \(45.0 \mathrm{mT}\) in \(1.29 \mathrm{~ms}\). Calculate its mass, in atomic mass units. (Actually, the method allows mass measurements to be carried out to much greater accuracy than these approximate data suggest.)

Electric and Magnetic Field An electric field of magnitude \(1.50 \mathrm{kV} / \mathrm{m}\) and a magnetic field of \(0.400 \mathrm{~T}\) act on a moving electron to produce no net force. (a) Calculate the minimum speed \(|\vec{v}|\) of the electron. (b) Draw a set of vectors \(\vec{E}, \vec{B}\), and \(\vec{v}\) that could yield the net force.
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