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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 conducting rod of length \(L\) slides freely down an inclined plane, as shown in the figure. The plane is inclined at an angle \(\theta\) from the horizontal. A uniform magnetic field of strength \(B\) acts in the positive \(y\) -direction. Determine the magnitude and the direction of the current that would have to be passed through the rod to hold it in position on the inclined plane.

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 coil consists of 120 circular loops of wire of radius \(4.8 \mathrm{~cm} .\) A current of 0.49 A runs through the coil, which is oriented vertically and is free to rotate about a vertical axis (parallel to the \(z\) -axis). It experiences a uniform horizontal magnetic field in the positive \(x\) -direction. When the coil is oriented parallel to the \(x\) -axis, a force of \(1.2 \mathrm{~N}\) applied to the edge of the coil in the positive \(y\) -direction can keep it from rotating. Calculate the strength of the magnetic field.

A small aluminum ball with a mass of \(5.00 \mathrm{~g}\) and a charge of \(15.0 \mathrm{C}\) is moving northward at \(3000 \mathrm{~m} / \mathrm{s}\). You want the ball to travel in a horizontal circle with a radius of \(2.00 \mathrm{~m}\), in a clockwise sense when viewed from above. Ignoring gravity, what is the magnitude and the direction of the magnetic field that must be applied to the aluminum ball to cause it to have this motion?

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.
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