/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 70 A plastic circular loop of radiu... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 27: Problem 70

A plastic circular loop of radius \(R\) and a positive charge \(q\) is distributed uniformly around the circumference of the loop. The loop is then rotated around its central axis, perpendicular to the plane of the loop, with angular speed \(\omega\) . If the loop is in a region where there is a uniform magnetic field \(\overrightarrow{\boldsymbol{B}}\) directed parallel to the plane of the loop, calculate the magnitude of the magnetic torque on the loop.

Short Answer

Expert verified
The magnetic torque magnitude is \(\frac{q R^2 \cdot \omega \cdot B}{2}\).

Step by step solution

01

Understand the Problem

We are given a circular loop with a radius \(R\) and a uniformly distributed charge \(q\). The loop rotates with an angular speed \(\omega\) around its central axis which is perpendicular to its plane. A magnetic field \(\overrightarrow{\boldsymbol{B}}\) is present parallel to the loop's plane, and we want to determine the magnetic torque on the loop.
02

Magnetic Dipole Moment Calculation

The magnetic dipole moment \(\mu\) of a charged rotating loop is given by \(\mu = q \cdot A \cdot \omega / (2\pi)\), where \(A\) is the area of the loop. Since the loop is circular, \(A = \pi R^2\). Plugging in, \(\mu = \frac{q \cdot (\pi R^2) \cdot \omega}{2\pi} = \frac{q R^2 \cdot \omega}{2}\).
03

Torque on a Magnetic Dipole

The torque \(\tau\) on a magnetic dipole in a magnetic field is given by \(\tau = \mu B \sin(\theta)\), where \(\theta\) is the angle between the magnetic dipole moment and the magnetic field. Since the field is parallel to the loop's plane, \(\theta = 90^\circ\), giving \(\sin(90^\circ) = 1\). Therefore, \(\tau = \mu B = \frac{q R^2 \cdot \omega}{2} B\).
04

Final Expression for Torque

Substitute the expression for \(\mu\) into the torque expression: \[\tau = \frac{q R^2 \cdot \omega}{2} \cdot B = \frac{q R^2 \cdot \omega \cdot B}{2}\]. This is the magnitude of the magnetic torque on the loop.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Magnetic Dipole Moment
The magnetic dipole moment is a fundamental concept in electromagnetism, especially in the context of loops and coils. It's essentially a measure of the strength and direction of a magnetic source that can be thought of as a magnet or a coil with a current running through it.
For a loop of radius \(R\) carrying a charge \(q\), when this loop rotates, it creates a kind of magnetic field similar to that of a small magnet. This is quantified by the magnetic dipole moment.
To calculate it, you use the formula \( \mu = \frac{q \cdot (\pi R^2) \cdot \omega}{2\pi} \). Here:
  • \( \pi R^2 \) represents the area of the loop
  • \(\omega\) stands for the angular speed, indicating how fast the loop is rotating
  • \(q\) is the charge uniformly distributed
Simplifying this gives \( \mu = \frac{q R^2 \cdot \omega}{2} \), showing us how the rotation and the charge contribute to the magnetic moment.
Angular Speed
Angular speed, not to be confused with linear speed, is the measure of how fast something rotates or revolves relative to another point. It is denoted by \( \omega \) and usually measured in radians per second (rad/s).
In the context of rotational motion, angular speed is key to understanding phenomena such as rotational kinetic energy and torque. For our loop, it determines how frequently the loop completes a circle along its axis of rotation.
This links directly to the magnetic dipole moment because:
  • Higher angular speeds (\(\omega\): many complete cycles in a short time) result in a larger magnetic dipole moment.
  • As \(\omega \) increases, the strength of the loop's equivalent magnetic field also increases.
Thus, the faster the loop spins, the more significant its magnetic effects become.
Uniform Magnetic Field
A uniform magnetic field is one where the magnetic force experienced is constant in both magnitude and direction over a defined space. This consistency is important for predicting the interactions of objects with the field, like our charged, rotating loop.
In this exercise, the magnetic field \(\overrightarrow{B} \) is described as being parallel to the plane of the loop.
The uniformity here implies:
  • The field lines are straight and evenly spaced.
  • Any point within the designated area experiences the same magnetic force due to \(\overrightarrow{B} \).
When the magnetic dipole moment of the loop interacts with this field:
  • The torque (turning effect) on the loop can be calculated as \(\tau = \mu B \,\sin(\theta)\), with \(\theta = 90^\circ\).
  • This results in maximum torque since \(\sin(90^\circ) = 1\).
Hence, the uniform magnetic field and its orientation play a crucial role in determining the torque experienced by the loop.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Determining the Mass of an Isotope. The electric field between the plates of the velocity selector in a Bainbridge mass spectrometer (see Fig. 27.22) is 1.12 \(\times 10^{5} \mathrm{V} / \mathrm{m}\) , and the magnetic field in both regions is 0.540 T. A stream of singly charged selenium ions moves in a circular path with a radius of 31.0 \(\mathrm{cm}\) in the magnetic field. Determine the mass of one selenium ion and the mass number of this selenium isotope. (The mass number is equal to the mass of the isotope in atomic mass units, rounded to the nearest integer. One atomic mass unit \(=1 \mathbf{u}=1.66 \times 10^{-27} \mathrm{kg} .\) .

An Electromagnetic Rail Gun. A conducting bar with mass \(m\) and length \(L\) slides over horizontal rails that are connected to a voltage source. The voltage source maintains a constant current \(I\) in the rails and bar, and a constant, uniform, vertical magnetic field \(\overrightarrow{\boldsymbol{B}}\) fills the region between the rails (Fig. 27.63\()\) . (a) Find the magnitude and direction of the net force on the conducting bar Ignore friction, air resistance, and electrical resistance. (b) If the bar has mass \(m\) , find the distance \(d\) that the bar must move along the rails from rest to attain speed \(v\) . (c) It has been suggested that rail guns based on this principle could accelerate payloads into earth orbit or beyond. Find the distance the bar must travel along the rails if it is to reach the escape speed for the earth \((11.2 \mathrm{km} / \mathrm{s}) .\) Let \(B=0.50 \mathrm{T}, \quad I=2.0 \times 10^{3} \mathrm{A}\) \(m=25 \mathrm{kg},\) and \(L=50 \mathrm{cm} .\) For simplicity assume the net force on the object is equal to the magnetic force, as in parts \((a)\) and \((b)\) , even though gravity plays an important role in an actual launch in space.

A circular area with a radius of 6.50 \(\mathrm{cm}\) lies in the \(x y\) -plane. What is the magnitude of the magnetic flux through this circle due to a uniform magnetic field \(B=0.230 \mathrm{T}\) (a) in the \(+z\) -direction; \((b)\) at an angle of \(53.1^{\circ}\) from the \(+z\) -direction; \((\mathrm{c})\) in the \(+y\) -direction?

A circular ring with area \(4.45 \mathrm{~cm}^{2}\) is carrying a current of 12.5 A. The ring is free to rotate about a diameter. The ring, initially at rest, is immersed in a region of uniform magnetic field given by \(\vec{B}=\left(1.15 \times 10^{-2} \mathrm{~T}\right)(12 \hat{\imath}+3 \hat{\jmath}-4 \hat{k}) .\) The ring is positioned initially such that its magnetic moment is given by \(\vec{\mu}_{1}=\mu(-0.800 \hat{\imath}+0.600 \hat{\jmath}),\) where \(\mu\) is the (positive) magnitude of the magnetic moment. The ring is released and turns through an angle of \(90.0^{\circ},\) at which point its magnetic moment is given by \(\vec{\mu}_{f}=-\mu \hat{k} .\) (a) Determine the decrease in potential energy. (b) If the moment of inertia of the ring about a diameter is \(8.50 \times 10^{-7} \mathrm{~kg} \cdot \mathrm{m}^{2},\) determine the angular speed of the ring as it passes through the second position.

In a shunt-wound dc motor with the field coils and rotor connected in parallel (Fig. \(27.56 ),\) the resistance \(R_{t}\) of the field coils is \(106 \Omega,\) and the resistance \(R_{r}\) of the rotor is 5.9\(\Omega\) . When a potential difference of 120 \(\mathrm{V}\) is applied to the brushes and the motor is running at full speed delivering mechanical power, the current supplied to it is 4.82 \(\mathrm{A}\) (a) What is the current in the field coils? (b) What is the current in the rotor? (c) What is the induced emf developed by the motor? (d) How much mechanical power is developed by this motor?
See all solutions

Recommended explanations on Physics Textbooks

Astrophysics

Read Explanation

Fields in Physics

Read Explanation

Particle Model of Matter

Read Explanation

Mechanics and Materials

Read Explanation

Scientific Method Physics

Read Explanation

Wave Optics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.