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Chapter 27: Problem 10

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?

Short Answer

Expert verified
Answer: The work done by a magnetic field on a charged particle moving with a specific velocity is zero. The impact of this work on the particle's motion is that there is no change in the particle's kinetic energy, and its speed remains constant. However, the magnetic field exerts a force that changes the direction of the particle's velocity, causing the particle to move along a circular path.

Step by step solution

01

Use the formula for work done

The work done on an object by a force can be calculated using the formula: \(W = Fd\cos(\theta)\). Where \(W\) represents the work done, \(F\) represents the magnitude of the force exerted, \(d\) is the distance traveled by the particle, and \(\theta\) is the angle between the force and the displacement.
02

Calculate the force exerted by the magnetic field

We can find the force exerted by the magnetic field on the charged particle using the Lorentz force formula: \(F = q(vB\sin(\alpha))\). Where \(F\) represents the magnetic force, \(q\) represents the charge of the particle, \(v\) represents the particle's velocity, \(B\) represents the magnetic field strength, and \(\alpha\) is the angle between the velocity and magnetic field direction. Since the velocity is perpendicular to the magnetic field, the angle \(\alpha = 90^{\circ}\) and \(\sin(\alpha) = 1\). Therefore, the force exerted by the magnetic field on the particle is: \(F = qvB\).
03

Find the angle between the displacement and the force exerted

The magnetic force acts perpendicular to the velocity of the particle. Therefore, the angle between the displacement (which is in the direction of the velocity) and the force is \(\theta = 90^{\circ}\).
04

Calculate the work done by the magnetic field

Using the formula for work done, we find that: \(W = Fd\cos(\theta) = qvBd\cos(90^{\circ}) = qvBd\cdot0 = 0\). Since the cosine of 90 degrees is zero, the work done by the magnetic field on the charged particle is zero.
05

Analyze the effect on the particle's motion

Since the work done by the magnetic field on the particle is zero, there is no change in the particle's kinetic energy. Therefore, the speed of the particle remains constant in the magnetic field. However, it's important to note that the magnetic field exerts a force which changes the direction of the particle's velocity, causing the particle to move along a circular path.

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Most popular questions from this chapter

A square loop of wire of side length \(\ell\) lies in the \(x y\) -plane, with its center at the origin and its sides parallel to the \(x\) - and \(y\) -axes. It carries a current, \(i\), in the counterclockwise direction, as viewed looking down the \(z\) -axis from the positive direction. The loop is in a magnetic field given by \(\vec{B}=\left(B_{0} / a\right)(z \hat{x}+x \hat{z}),\) where \(B_{0}\) is a constant field strength, \(a\) is a constant with the dimension of length, and \(\hat{x}\) and \(\hat{z}\) are unit vectors in the positive \(x\) -direction and positive \(z\) -direction. Calculate the net force on the loop.

A magnetic field is oriented in a certain direction in a horizontal plane. An electron moves in a certain direction in the horizontal plane. For this situation, there a) is one possible direction for the magnetic force on the electron. b) are two possible directions for the magnetic force on the electron. c) are infinite possible directions for the magnetic force on the electron.

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.)

An electron with a speed of \(4.0 \cdot 10^{5} \mathrm{~m} / \mathrm{s}\) enters a uniform magnetic field of magnitude \(0.040 \mathrm{~T}\) at an angle of \(35^{\circ}\) to the magnetic field lines. The electron will follow a helical path. a) Determine the radius of the helical path. b) How far forward will the electron have moved after completing one circle?

A straight wire of length \(2.00 \mathrm{~m}\) carries a current of \(24.0 \mathrm{~A} .\) It is placed on a horizontal tabletop in a uniform horizontal magnetic field. The wire makes an angle of \(30.0^{\circ}\) with the magnetic field lines. If the magnitude of the force on the wire is \(0.500 \mathrm{~N}\), what is the magnitude of the magnetic field?
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