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Chapter 27: Problem 15

A charged particle moves under the influence of an electric field only. Is it possible for the particle to move with a constant speed? What if the electric field is replaced with a magnetic field?

Short Answer

Expert verified
Can it do so in a magnetic field? Explain your answer. Answer: A charged particle cannot move with a constant speed in an electric field, as the force it experiences due to the electric field causes it to either accelerate or decelerate. However, in a magnetic field, the force experienced by the charged particle is always perpendicular to its velocity vector, meaning that it does not change the particle's speed, only its direction. Therefore, it is possible for a charged particle to move with a constant speed in a magnetic field.

Step by step solution

01

Understanding the force on a charged particle in an electric field

A charged particle in an electric field experiences a force given by the formula: \(F_e = qE\), where \(q\) is the charge of the particle and \(E\) is the strength of the electric field. This force will cause the particle to accelerate or decelerate, which depends on the direction of the electric field and the charge of the particle.
02

Determining if a particle can move with a constant speed in an electric field

For a particle to move with a constant speed, the net force acting on the particle should be zero. In the case of an electric field, the charged particle experiences a force due to the field (\(F_e = qE\)). If there are no other forces acting on the particle, then it will always either accelerate or decelerate, making it impossible for the particle to move at a constant speed within an electric field.
03

Understanding the force on a charged particle in a magnetic field

A charged particle in a magnetic field experiences a force given by the formula: \(F_m = q(\mathbf{v} \times \mathbf{B})\), where \(q\) is the charge of the particle, \(\mathbf{v}\) is the velocity vector of the particle, and \(\mathbf{B}\) is the magnetic field vector. The force is always perpendicular to the velocity vector, which means that the magnetic force will not change the speed of the particle, only its direction.
04

Determining if a particle can move with a constant speed in a magnetic field

In the case of a magnetic field, the charged particle experiences a force that is always perpendicular to its velocity vector (\(F_m = q(\mathbf{v} \times \mathbf{B})\)). Since this force does not cause any change in speed, it is possible for the particle to move at a constant speed within a magnetic field.

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Most popular questions from this chapter

A particle with a charge of \(20.0 \mu \mathrm{C}\) moves along the \(x\) -axis with a speed of \(50.0 \mathrm{~m} / \mathrm{s}\). It enters a magnetic field given by \(\vec{B}=0.300 \hat{y}+0.700 \hat{z},\) in teslas. Determine the magnitude and the direction of the magnetic force on the particle.

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.

A particle with mass \(m\), charge \(q\), and velocity \(v\) enters a magnetic field of magnitude \(B\) and with direction perpendicular to the initial velocity of the particle. What is the work done by the magnetic field on the particle? How does this affect the particle's motion?

An electron is moving with a constant velocity. When it enters an electric field that is perpendicular to its velocity, the electron will follow a ________ trajectory. When the electron enters a magnetic field that is perpendicular to its velocity, it will follow a ________ trajectory.

The magnitude of the magnetic force on a particle with charge \(-2 e\) moving with speed \(v=1.0 \cdot 10^{5} \mathrm{~m} / \mathrm{s}\) is \(3.0 \cdot 10^{-18} \mathrm{~N}\). What is the magnitude of the magnetic field component perpendicular to the direction of motion of the particle?
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