/*! 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 56 A circular wire loop of radius \... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A circular wire loop of radius \(15.0 \mathrm{~cm}\) carries a current of \(2.60 \mathrm{~A}\). It is placed so that the normal to its plane makes an angle of \(41.0^{\circ}\) with a uniform magnetic field of magni- tude \(12.0\) T. (a) Calculate the magnitude of the magnetic dipole moment of the loop. (b) What is the magnitude of the torque acting on the loop?

Short Answer

Expert verified
(a) Magnitude of the magnetic dipole moment: \(0.0184 \, \mathrm{Am^2}\). (b) Magnitude of the torque: \(0.199 \, \mathrm{Nm}\).

Step by step solution

01

Define the magnetic dipole moment formula

The magnetic dipole moment \( \mu \) for a circular loop of wire carrying a current \( I \) is given by the formula \( \mu = I \cdot A \), where \( A \) is the area of the loop.
02

Calculate the area of the loop

The area \( A \) of a circle is calculated using the formula \( A = \pi r^2 \), where \( r = 15.0 \ \mathrm{cm} = 0.15 \ \mathrm{m} \).

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 91Ó°ÊÓ!

Key Concepts

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

Torque on Current Loop
When a current-carrying loop is placed in a magnetic field, it experiences a torque. Torque is essentially a twist or a rotational force that is applied on an object. For a loop, this torque results from the magnetic field interacting with the current flowing through it.

The formula for calculating the torque (\( \tau \)) on a current loop is given by \( \tau = \mu B \sin \theta \). Here, \( \mu \) is the magnetic dipole moment, \( B \) is the magnitude of the magnetic field, and \( \theta \) is the angle between the magnetic field and the normal to the plane of the loop. The sin component captures how torque is only effective when the magnetic field direction is not parallel to the normal of the loop.

Some important things to remember about torque in this setup:
  • If the loop is completely parallel to the magnetic field, \( \theta = 0 \), and hence, \( \sin \theta = 0 \). This means no torque is applied in this orientation.
  • The maximum torque occurs when \( \theta = 90^\circ \) because \( \sin(90^\circ) = 1 \).
Circular Wire Loop
A circular wire loop is often used in physics to study magnetic properties and their interactions. The loop's shape makes it mathematically convenient because its circular symmetry allows for straightforward calculations.

For a circular loop, the area \( A \) that it encloses is a critical factor in determining the magnetic dipole moment \( \mu \). The area \( A \) is found using the formula \( A = \pi r^2 \), where \( r \) is the radius of the loop. In our example, \( r = 15.0 \ \mathrm{cm} \) which can also be converted to meters as \( r = 0.15 \ \mathrm{m} \). So, substituting it gives the area as \( A = \pi \times (0.15)^2 \).

The concept of a circular loop is essential in a variety of applications such as:
  • Magnetic coil designs in electronics.
  • Antennas for wireless communication.
  • Basic solenoids in electromagnets.
Magnetic Field Interaction
Magnetic fields are regions around magnets or electric currents in which magnetic forces can be felt. When a wire loop carrying a current is immersed in a magnetic field, several interesting phenomena take place due to this interaction.

The interaction of a magnetic field with the circular loop is governed primarily by the relationship between the field and the current. The total magnetic dipole moment \( \mu \), calculated by \( \mu = I \cdot A \) (where \( I \) is the current and \( A \) is the area), interacts with this field. This results in physical effects like the exertion of torque on the loop as discussed.

These interactions allow:
  • Control of the orientation of devices by altering the angle \( \theta \).
  • Understanding how motors and generators function.
  • Development of technologies such as magnetic levitation.
The interplay between current-carrying loops and magnetic fields serves as a foundational concept in understanding electromagnetism and its practical applications.

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

An electron follows a helical path in a uniform magnetic field of magnitude \(0.300 \mathrm{~T}\). The pitch of the path is \(6.00 \mu \mathrm{m}\), and the magnitude of the magnetic force on the electron is \(2.00 \times 10^{-15} \mathrm{~N}\). What is the electron's speed?

A strip of copper \(150 \mu \mathrm{m}\) thick and \(4.5 \mathrm{~mm}\) wide is placed in a uniform magnetic field \(\vec{B}\) of magnitude \(0.65 \mathrm{~T}\), with \(\vec{B}\) perpendicular to the strip. A current \(i=23 \mathrm{~A}\) is then sent through the strip such that a Hall potential difference \(V\) appears across the width of the strip. Calculate \(V\). (The number of charge carriers per unit volume for copper is \(8.47 \times 10^{28}\) electrons \(/ \mathrm{m}^{3}\).)

A \(13.0 \mathrm{~g}\) wire of length \(L=62.0 \mathrm{~cm}\) is suspended by a pair of flexible leads in a uniform magnetic field of magnitude \(0.440 \mathrm{~T}\) (Fig. \(28-41\) ). What are the (a) magnitude and (b) direction (left or right) of the current required to remove the tension in the supporting leads?

An alpha particle can be produced in certain radioactive decays of nuclei and consists of two protons and two neutrons. The particle has a charge of \(q=+2 e\) and a mass of \(4.00 \mathrm{u}\), where \(\mathrm{u}\) is the atomic mass unit, with \(1 \mathrm{u}=1.661 \times 10^{-27} \mathrm{~kg}\). Suppose an alpha particle travels in a circular path of radius \(4.50 \mathrm{~cm}\) in a uniform magnetic field with \(B=1.20 \mathrm{~T}\). Calculate (a) its speed, (b) its period of revolution, (c) its kinetic energy, and (d) the potential difference through which it would have to be accelerated to achieve this energy.

An electric field of \(1.50 \mathrm{kV} / \mathrm{m}\) and a perpendicular magnetic field of \(0.400 \mathrm{~T}\) act on a moving electron to produce no net force. What is the electron's speed?

See all solutions

Recommended explanations on Physics Textbooks

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.