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A charged particle is moving in a constant magnetic field. State whether each of the following statements concerning the magnetic force exerted on the particle is true or false? (Assume that the magnetic field is not parallel or antiparallel to the velocity.) a) It does no work on the particle. b) It may increase the speed of the particle. c) It may change the velocity of the particle. d) It can act only on the particle while the particle is in motion. e) It does not change the kinetic energy of the particle.

Short Answer

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a) It does no work on the particle. b) It may increase the speed of the particle. c) It may change the velocity of the particle. d) It can act only on the particle while the particle is in motion. e) It does not change the kinetic energy of the particle. Answer: a, c, d, e

Step by step solution

01

Statement a: It does no work on the particle.

Since the magnetic force is always perpendicular to the velocity, the angle between the force and the displacement is 90 degrees. The work done W is calculated as W = F*d*cos(θ). In this case, cos(90°) = 0, so, W=0. Thus, this statement is true.
02

Statement b: It may increase the speed of the particle.

As mentioned in the analysis, the magnetic force cannot change the speed of the particle, only its direction. Therefore, this statement is false.
03

Statement c: It may change the velocity of the particle.

The magnetic force can change the direction of the particle's motion, and since velocity is a vector that includes both speed and direction, the magnetic force can change the velocity of the particle. This statement is true.
04

Statement d: It can act only on the particle while the particle is in motion.

The magnetic force acts on a charged particle when it has a non-zero velocity. If the particle is stationary (i.e., its velocity is zero), the magnetic force acting on it will also be zero. So this statement is true.
05

Statement e: It does not change the kinetic energy of the particle.

Since the magnetic force does no work on the particle (as determined in statement a), the particle's kinetic energy remains unchanged. Hence, this statement is true. To summarize: a) True b) False c) True d) True e) True

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Key Concepts

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

Work Done by Magnetic Force
Understanding how magnetic forces interact with charged particles is fundamental in physics. One key principle is the concept of work done by a force. Work is defined as the product of the force applied to an object and the displacement of that object in the direction of the force. Mathematically, it's represented as \( W = F \times d \times \text{cos}(\theta) \), where \( W \) is work, \( F \) is force, \( d \) is displacement, and \( \theta \) is the angle between the force and displacement vectors.

For a charged particle moving in a magnetic field, the magnetic force is always perpendicular to its velocity, which in turn is tangent to its path. As such, the angle \( \theta \) is 90 degrees, and since \( \text{cos}(90^\circ) = 0 \), the work done by the magnetic force on the particle is zero. This is why no energy is transferred to the particle in the form of work, and why its speed remains constant when only a magnetic force is applied.

It's crucial to note that even though the magnetic force changes the direction of the particle's velocity, this change in direction doesn't constitute work, because work is only done when there is a component of force in the direction of movement. This fundamental concept is key in understanding magnetic fields' influence in various applications, from particle accelerators to the Earth's magnetic shield.
Magnetic Force and Particle Velocity
When discussing magnetism, it's important to recognize how magnetic forces affect the velocity of charged particles. Velocity is a vector quantity, which means it has both magnitude (speed) and direction. While magnetic fields can't do work on a particle, they can indeed influence its velocity.

This behavior is because the magnetic force acts perpendicular to the velocity of the particle, thus altering its direction without changing its speed. According to the equation for magnetic force \( F_m = q(v \times B) \), where \( q \) is the charge of the particle, \( v \) is its velocity, and \( B \) is the magnetic field, we see that the force is dependent on the cross product of velocity and magnetic field, emphasizing the perpendicular relationship.

The result is a circular or helical motion of the particle, where its path curves but its speed remains constant, assuming no other forces are present. This concept is widely used in designing cyclotrons and other devices where particles are steered using magnetic fields. Understanding the interplay between velocity and magnetic force is crucial for studying the trajectories of charged particles in fields, such as those in space or in laboratory experiments.
Kinetic Energy in Magnetic Field
Kinetic energy, which is the energy due to motion, is another concept closely tied to the behavior of charged particles in magnetic fields. The equation for kinetic energy is given by: \( KE = \frac{1}{2} m v^2 \), where \( m \) is mass and \( v \) is the speed of the particle.

Since the magnetic force does no work on the particle, as earlier explained, and work is the mechanism by which energy is transferred to an object, the kinetic energy of a charged particle moving in a magnetic field does not change as a result of the magnetic force alone. This is consistent with the principle of conservation of energy, which states that energy in a closed system remains constant if there are no external work inputs or outputs.

The constant kinetic energy implies constant speed for the particle as it moves around in a circular or helical path within the magnetic field. This invariance of kinetic energy is a foundational concept that has practical applications in the operation of spectrographs and magnetic confinement in fusion devices. Appreciating the stability of kinetic energy in a magnetic field helps one to understand how charged particles can be manipulated without impacting their energy states, enabling advances in various scientific and technological fields.

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Most popular questions from this chapter

A proton moving with a speed of \(4.0 \cdot 10^{5} \mathrm{~m} / \mathrm{s}\) in the positive \(y\) -direction enters a uniform magnetic field of \(0.40 \mathrm{~T}\) pointing in the positive \(x\) -direction. Calculate the magnitude of the force on the proton.

A current-carrying wire is positioned within a large, uniform magnetic field, \(\vec{B}\). However, the wire experiences no force. Explain how this might be possible.

A copper wire with density \(\rho=8960 \mathrm{~kg} / \mathrm{m}^{3}\) is formed into a circular loop of radius \(50.0 \mathrm{~cm} .\) The cross-sectional area of the wire is \(1.00 \cdot 10^{-5} \mathrm{~m}^{2},\) and a potential difference of \(0.012 \mathrm{~V}\) is applied to the wire. What is the maximum angular acceleration of the loop when it is placed in a magnetic field of magnitude \(0.25 \mathrm{~T}\) ? The loop rotates about an axis through a diameter.

The figure shows a schematic diagram of a simple mass spectrometer, consisting of a velocity selector and a particle detector and being used to separate singly ionized atoms \(\left(q=+e=1.60 \cdot 10^{-19} \mathrm{C}\right)\) of gold \((\mathrm{Au})\) and molybdenum (Mo). The electric field inside the velocity selector has magnitude \(E=1.789 \cdot 10^{4} \mathrm{~V} / \mathrm{m}\) and points toward the top of the page, and the magnetic field has magnitude \(B_{1}=1.00 \mathrm{~T}\) and points out of the page. a) Draw the electric force vector, \(\vec{F}_{E},\) and the magnetic force vector, \(\vec{F}_{B},\) acting on the ions inside the velocity selector. b) Calculate the velocity, \(v_{0}\), of the ions that make it through the velocity selector (those that travel in a straight line). Does \(v_{0}\) depend on the type of ion (gold versus molybdenum), or is it the same for both types of ions? c) Write the equation for the radius of the semicircular path of an ion in the particle detector: \(R=R\left(m, v_{0}, q, B_{2}\right)\). d) The gold ions (represented by the black circles) exit the particle detector at a distance \(d_{2}=40.00 \mathrm{~cm}\) from the entrance slit, while the molybdenum ions (represented by the gray circles) exit the particle detector at a distance \(d_{1}=19.81 \mathrm{~cm}\) from the entrance slit. The mass of a gold ion is \(m_{\text {gold }}=\) \(3.27 \cdot 10^{-25}\) kg. Calculate the mass of a molybdenum ion.

The magnetic field in a region in space (where \(x>0\) and \(y>0\) ) is given by \(B=(x-a z) \hat{y}+(x y-b) \hat{z},\) where \(a\) and \(b\) are positive constants. An electron moving with a constant velocity, \(\vec{v}=v_{0} \hat{x},\) enters this region. What are the coordinates of the points at which the net force acting on the electron is zero?

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