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Chapter 27: Problem 12

A proton, moving in negative \(y\) -direction in a magnetic field, experiences a force of magnitude \(F\), acting in the negative \(x\) -direction. a) What is the direction of the magnetic field producing this force? b) Does your answer change if the word "proton" in the statement is replaced by “electron"?

Short Answer

Expert verified
Answer: The direction of the magnetic field is in the negative \(z\)-direction. b) Would the result change if we considered an electron moving the same way as a proton? Answer: No, the result would not change. The direction of the magnetic field would still be in the negative \(z\)-direction for an electron.

Step by step solution

01

Identify given information

We are given that the force has a magnitude \(F\) and acts in the negative \(x\)-direction. Also, the proton is moving in the negative \(y\)-direction.
02

Apply the right-hand rule

From the formula mentioned earlier, \(\vec{F} = q(\vec{v} \times \vec{B})\), we can use the right-hand rule to determine the direction of the magnetic field. Point your right hand's thumb in the direction of the proton's velocity, which is the negative \(y\)-direction. Since the proton is positively charged, curl your fingers in the direction of the force, which is the negative \(x\)-direction. The direction your palm faces now will be the direction of the magnetic field.
03

Find the direction of the magnetic field

In this case, your palm will be facing downwards, which means the magnetic field is in the negative \(z\)-direction. a) So, the direction of the magnetic field is in the negative \(z\)-direction.
04

Consider the case of an electron

Now, let's see if the result changes if we replace the proton with an electron. The only difference is that an electron has a negative charge, so using the left-hand rule instead of the right-hand rule for negatively charged particles, we follow the same process as in steps 2 and 3. b) In this case, you will find that the magnetic field is still in the negative \(z\)-direction. So, the answer does not change whether we consider a proton or an electron. The direction of the magnetic field in both cases is in the negative \(z\)-direction.

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Most popular questions from this chapter

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.

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 straight wire carrying a current of 3.41 A is placed at an angle of \(10.0^{\circ}\) to the horizontal between the pole tips of a magnet producing a field of \(0.220 \mathrm{~T}\) upward. The poles tips each have a \(10.0 \mathrm{~cm}\) diameter. The magnetic force causes the wire to move out of the space between the poles. What is the magnitude of that force?

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.

A particle with mass \(m\) and charge \(q\) is moving within both an electric field and a magnetic field, \(\vec{E}\) and \(\vec{B}\). The particle has velocity \(\vec{v},\) momentum \(\vec{p}\), and kinetic energy, \(K\). Find general expressions for \(d \vec{p} / d t\) and \(d K / d t\), in terms of these seven quantities.
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