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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

Expert verified
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 particle with a charge of \(+10.0 \mu \mathrm{C}\) is moving at \(300 \cdot \mathrm{m} / \mathrm{s}\) in the positive \(z\) -direction. a) Find the minimum magnetic field required to keep it moving in a straight line at constant speed if there is a uniform electric field of magnitude \(100 . \mathrm{V} / \mathrm{m}\) pointing in the positive \(y\) -direction. b) Find the minimum magnetic field required to keep the particle moving in a straight line at constant speed if there is a uniform electric field of magnitude \(100 . \mathrm{V} / \mathrm{m}\) pointing in the positive \(z\) -direction.

A proton moving at speed \(v=1.00 \cdot 10^{6} \mathrm{~m} / \mathrm{s}\) enters a region in space where a magnetic field given by \(\vec{B}=\) \((-0.500 \mathrm{~T}) \hat{z}\) exists. The velocity vector of the proton is at an angle \(\theta=60.0^{\circ}\) with respect to the positive \(z\) -axis. a) Analyze the motion of the proton and describe its trajectory (in qualitative terms only). b) Calculate the radius, \(r\), of the trajectory projected onto a plane perpendicular to the magnetic field (in the \(x y\) -plane). c) Calculate the period, \(T,\) and frequency, \(f\), of the motion in that plane. d) Calculate the pitch of the motion (the distance traveled by the proton in the direction of the magnetic field in 1 period).

The work done by the magnetic field on a charged particle in motion in a cyclotron is zero. How, then, can a cyclotron be used as a particle accelerator, and what essential feature of the particle's motion makes it possible? A blue problem number indicates a worked-out solution is available in the Student Solutions Manual. One \(\bullet\) and two \(\bullet\) indicate increasing level of problem difficulty.

A circular coil with a radius of \(10.0 \mathrm{~cm}\) has 100 turns of wire and carries a current, \(i=100 . \mathrm{mA} .\) It is free to rotate in a region with a constant horizontal magnetic field given by \(\vec{B}=(0.0100 \mathrm{~T}) \hat{x}\). If the unit normal vector to the plane of the coil makes an angle of \(30.0^{\circ}\) with the horizontal, what is the magnitude of the net magnetic torque acting on the coil? 27.61 At \(t=0\) an electron crosses the positive \(y\) -axis (so \(x=0\) ) at \(60.0 \mathrm{~cm}\) from the origin with velocity \(2.00 \cdot 10^{5} \mathrm{~m} / \mathrm{s}\) in the positive \(x\) -direction. It is in a uniform magnetic field. a) Find the magnitude and the direction of the magnetic field that will cause the electron to cross the \(x\) -axis at \(x=60.0 \mathrm{~cm}\). b) What work is done on the electron during this motion? c) How long will the trip take from \(y\) -axis to \(x\) -axis?

A rail gun accelerates a projectile from rest by using the magnetic force on a current-carrying wire. The wire has radius \(r=5.1 \cdot 10^{-4} \mathrm{~m}\) and is made of copper having a density of \(\rho=8960 \mathrm{~kg} / \mathrm{m}^{3}\). The gun consists of rails of length \(L=1.0 \mathrm{~m}\) in a constant magnetic field of magnitude \(B=2.0 \mathrm{~T}\) oriented perpendicular to the plane defined by the rails. The wire forms an electrical connection across the rails at one end of the rails. When triggered, a current of \(1.00 \cdot 10^{4}\) A flows through the wire, which accelerates the wire along the rails. Calculate the final speed of the wire as it leaves the rails. (Neglect friction.)

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