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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

An alpha particle \(\left(m=6.6 \cdot 10^{-27} \mathrm{~kg}, q=+2 e\right)\) is accelerated by a potential difference of \(2700 \mathrm{~V}\) and moves in a plane perpendicular to a constant magnetic field of magnitude \(0.340 \mathrm{~T}\), which curves the trajectory of the alpha particle. Determine the radius of curvature and the period of revolution.

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.

A semicircular loop of wire of radius \(R\) is in the \(x y\) -plane, centered about the origin. The wire carries a current, \(i\), counterclockwise around the semicircle, from \(x=-R\) to \(x=+R\) on the \(x\) -axis. A magnetic field, \(\vec{B}\), is pointing out of the plane, in the positive \(z\) -direction. Calculate the net force on the semicircular loop.

In your laboratory, you set up an experiment with an electron gun that emits electrons with energy of \(7.50 \mathrm{keV}\) toward an atomic target. What deflection (magnitude and direction) would Earth's magnetic field \((0.300 \mathrm{G})\) produce in the beam of electrons if the beam is initially directed due east and covers a distance of \(1.00 \mathrm{~m}\) from the gun to the target? (Hint: First calculate the radius of curvature, and then determine how far away from a straight line the electron beam has deviated after \(1.00 \mathrm{~m}\).)

A 30 -turn square coil with a mass of \(0.250 \mathrm{~kg}\) and a side length of \(0.200 \mathrm{~m}\) is hinged along a horizontal side and carries a 5.00 -A current. It is placed in a magnetic field pointing vertically downward and having a magnitude of \(0.00500 \mathrm{~T}\). Determine the angle that the plane of the coil makes with the vertical when the coil is in equilibrium. Use \(g=9.81 \mathrm{~m} / \mathrm{s}^{2}\).

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