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Chapter 28: Problem 50

An electron moves in a circle of radius \(r=5.29 \times 10^{-11} \mathrm{~m}\) with speed \(2.19 \times 10^{6} \mathrm{~m} / \mathrm{s}\). Treat the circular path as a current loop with a constant current equal to the ratio of the electron's charge magnitude to the period of the motion. If the circle lies in a uniform magnetic field of magnitude \(B=7.10 \mathrm{mT}\), what is the maximum possible magnitude of the torque produced on the loop by the field?

Short Answer

Expert verified
The maximum torque is approximately \(6.55 \times 10^{-26} \mathrm{~N \cdot m}\).

Step by step solution

01

Determine the Electron's Period of Motion

First, calculate the period of motion of the electron. The period \(T\) is the time it takes to complete one full circle. Given the speed \(v = 2.19 \times 10^{6} \mathrm{~m/s}\) and the radius \(r = 5.29 \times 10^{-11} \mathrm{~m}\), use the formula for the period: \[ T = \frac{2\pi r}{v} \]Substitute the values:\[ T = \frac{2 \pi \times 5.29 \times 10^{-11}}{2.19 \times 10^{6}} \approx 1.52 \times 10^{-16} \mathrm{~s} \]
02

Calculate the Current in the Loop

The current \(I\) in the loop is defined as the charge passing through a point per unit time. The electron's charge magnitude is \(e = 1.60 \times 10^{-19} \mathrm{~C}\). The current is the electron's charge divided by the period:\[ I = \frac{e}{T} = \frac{1.60 \times 10^{-19}}{1.52 \times 10^{-16}} \approx 1.05 \times 10^{-3} \mathrm{~A} \]
03

Find the Magnetic Moment of the Loop

The magnetic moment \(\mu\) of the loop is given by the product of current and the area of the loop. The area \(A\) of the circle is \(\pi r^2\):\[ A = \pi (5.29 \times 10^{-11})^2 \approx 8.79 \times 10^{-21} \mathrm{~m^2} \]Thus, the magnetic moment is:\[ \mu = I \times A = 1.05 \times 10^{-3} \times 8.79 \times 10^{-21} \approx 9.23 \times 10^{-24} \mathrm{~A \cdot m^2} \]
04

Calculate the Maximum Torque

The torque \(\tau\) on the loop is given by the relationship:\[ \tau = \mu B \sin\theta \]where \(\theta\) is the angle between the magnetic moment \(\mu\) and the magnetic field \(B\). The maximum torque occurs when \(\sin\theta = 1\), when \(\theta = 90^\circ\).The magnitude of the magnetic field is \(7.10 \mathrm{~mT} = 7.10 \times 10^{-3} \mathrm{~T}\).Thus, the maximum torque is:\[ \tau_{\text{max}} = 9.23 \times 10^{-24} \times 7.10 \times 10^{-3} \approx 6.55 \times 10^{-26} \mathrm{~N \cdot m} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Electron Motion
Electrons are tiny particles with a negative charge. They move around the nucleus of an atom in specific paths. In this exercise, the electron travels in a circular path. This continuous circular motion is a bit like a train on a track. The path of the electron forms a circle because of the magnetic forces acting on it. When an electron moves like this, it has a specific speed and a path radius. The speed, in this case, is very fast: approximately 2.19 million meters per second. The radius is the distance from the center of this circular path to any point on its edge, measured here as approximately 0.0000000000529 meters. To calculate how long it takes for the electron to complete a full circle, we use the concept of the period of motion. The period is essential as it helps us determine a related concept: the electron behaves like a tiny current loop.
Current Loop
A current loop is formed when charges move in a closed path. Think of a loop as a simple circle of wire where electricity flows. Here, the moving electron in its circular path acts like a current loop. The idea comes from how we define current: it is the charge passing through a point in the loop per unit time. Even though there's just one electron, its continuous motion creates a loop of charge, similar to how water circulates in a fountain. The charge of one electron is known, and from the period of motion, we calculate the current. The relationship between charge, current, and time is crucial. It gives us a sense of how electric currents behave on a very small scale, like when dealing with single electrons in a path.
Magnetic Moment
The magnetic moment helps describe the magnetic strength of a loop or magnet. For a current loop, this is like associating the loop with a magnetic nail, representing its north and south poles. This concept is crucial when considering how loops interact with magnetic fields. The magnetic moment is determined by multiplying the current through the loop by the loop's area. For our electron moving in a circle, it generates a small magnetic moment. This moment is useful because it helps determine how the loop will behave when exposed to a magnetic field. Just like a compass needle aligns with Earth's magnetic field, our current loop will experience forces due to its magnetic moment.
Uniform Magnetic Field
A uniform magnetic field is one where the magnetic force is the same in strength and direction at all points. Imagine a sheet of paper with lines drawn evenly across it 鈥� that's a picture of a uniform field. When our current loop, created by the electron's motion, is placed in such a field, various interactions occur. One key interaction is the torque. Torque is like a twist or spin that acts on objects. When a magnetic field interacts with the magnetic moment of our loop, it causes the loop to experience a torque. The amount of torque depends on the angle between the field and the magnetic moment. The maximum torque happens when the magnetic moment is perpendicular to the magnetic field lines, providing a clear demonstration of these fundamental physical interactions.

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Most popular questions from this chapter

A magnetic dipole with a dipole moment of magnitude \(0.020 \mathrm{~J} / \mathrm{T}\) is released from rest in a uniform magnetic field of magnitude \(52 \mathrm{mT}\). The rotation of the dipole due to the magnetic force on it is unimpeded. When the dipole rotates through the orientation where its dipole moment is aligned with the magnetic field, its kinetic energy is \(0.80 \mathrm{~mJ}\). (a) What is the initial angle between the dipole moment and the magnetic field? (b) What is the angle when the dipole is next (momentarily) at rest?

A mass spectrometer (Fig. 28-12) is used to separate uranium ions of mass \(3.92 \times 10^{-25} \mathrm{~kg}\) and charge \(3.20 \times 10^{-19} \mathrm{C}\) from related species. The ions are accelerated through a potential difference of \(100 \mathrm{kV}\) and then pass into a uniform magnetic field, where they are bent in a path of radius \(1.00 \mathrm{~m}\). After traveling through \(180^{\circ}\) and passing through a slit of width \(1.00 \mathrm{~mm}\) and height \(1.00 \mathrm{~cm}\), they are collected in a cup. (a) What is the magnitude of the (perpendicular) magnetic field in the separator? If the machine is used to separate out \(100 \mathrm{mg}\) of material per hour, calculate (b) the current of the desired ions in the machine and (c) the thermal energy produced in the cup in \(1.00 \mathrm{~h}\).

A wire of length \(25.0 \mathrm{~cm}\) carrying a current of \(4.51 \mathrm{~mA}\) is to be formed into a circular coil and placed in a uniform magnetic field \(\vec{B}\) of magnitude \(5.71 \mathrm{mT}\). If the torque on the coil from the field is maximized, what are (a) the angle between \(\vec{B}\) and the coil's magnetic dipole moment and (b) the number of turns in the coil? (c) What is the magnitude of that maximum torque?

An alpha particle travels at a velocity \(\vec{v}\) of magnitude \(550 \mathrm{~m} / \mathrm{s}\) through a uniform magnetic field \(\vec{B}\) of magnitude \(0.045 \mathrm{~T}\). (An alpha particle has a charge of \(+3.2 \times 10^{-19} \mathrm{C}\) and a mass of \(6.6 \times\) \(\left.10^{-27} \mathrm{~kg} .\right)\) The angle between \(\vec{v}\) and \(\vec{B}\) is \(52^{\circ} .\) What is the magnitude of (a) the force \(\vec{F}_{B}\) acting on the particle due to the field and (b) the acceleration of the particle due to \(\vec{F}_{B}\) ? (c) Does the speed of the particle increase, decrease, or remain the same?

A beam of electrons whose kinetic energy is \(K\) emerges from a thin-foil "window" at the end of an accelerator tube. A metal plate at distance \(d\) from this window is perpendicular to the direction of the emerging beam (Fig. 28-53). (a) Show that we can prevent the beam from hitting the plate if we apply a uniform magnetic field such that $$ B \geq \sqrt{\frac{2 m K}{e^{2} d^{2}}} $$ in which \(m\) and \(e\) are the electron mass and charge. (b) How should \(\vec{B}\) be oriented?
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