/*! 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 51 A magnet in the form of a cylind... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 32: Problem 51

A magnet in the form of a cylindrical rod has a length of \(5.00 \mathrm{~cm}\) and a diameter of \(1.00 \mathrm{~cm}\). It has a uniform magnetization of \(5.30 \times 10^{3} \mathrm{~A} / \mathrm{m}\). What is the magnitude of its magnetic dipole moment?

Short Answer

Expert verified
The magnetic dipole moment is \( 0.0208 \; A\cdot m^2 \).

Step by step solution

01

Determine the Volume of the Cylindrical Rod

First, calculate the volume of the cylindrical rod. The volume of a cylinder is given by the formula \( V = \pi r^2 h \), where \( r \) is the radius and \( h \) is the height (or length) of the cylinder. Given the diameter is 1.00 cm, the radius is 0.50 cm, and the length is 5.00 cm. Convert these measurements to meters: \( r = 0.005 m \) and \( h = 0.05 m \). Now, compute the volume: \[ V = \pi (0.005)^2 (0.05) \].
02

Calculate the Volume

Substitute the values into the formula: \[ V = \pi (0.005)^2 (0.05) = \pi (0.000025) (0.05) = 3.93 \times 10^{-6} \; m^3 \].
03

Find the Magnetic Dipole Moment

The magnetic dipole moment \( m \) can be calculated using the formula \( m = MV \), where \( M \) is the magnetization and \( V \) is the volume. Given \( M = 5.30 \times 10^3 \; A/m \), substitute the values: \[ m = (5.30 \times 10^3) \times (3.93 \times 10^{-6}) \; A\cdot m^2 \].
04

Calculate the Magnetic Dipole Moment

Perform the multiplication: \[ m = 5.30 \times 10^3 \times 3.93 \times 10^{-6} = 2.08 \times 10^{-2} \; A\cdot m^2 \]. The magnitude of the magnetic dipole moment is \( 0.0208 \; A\cdot m^2 \).

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.

cylinder volume calculation
To calculate the volume of a cylinder, you need the formula: \[ V = \pi r^2 h \]. Here, \( r \) is the radius and \( h \) is the height (or length) of the cylinder.
In this exercise, we have a cylindrical magnet with a diameter of 1.00 cm and a length of 5.00 cm.
First, convert these measurements to meters:
  • Radius: \( r = \frac{1.00}{2} = 0.50 \) cm = 0.005 m
  • Length: \( h = 5.00 \) cm = 0.05 m
Now, substitute these values into the volume formula: \[ V = \pi (0.005)^2 (0.05) \ V = \pi (0.000025) (0.05) \ V = 3.93 \times 10^{-6} \, m^3 \].
magnetization
Magnetization is the measure of the magnetic moment per unit volume of a material.
It tells us how strongly a material is magnetized.
The formula for magnetization is \( M = \frac{m}{V} \), where \( m \) is the magnetic moment and \( V \) is the volume.
In this problem, we need to find the magnetic dipole moment using the given magnetization.
Given: \( M = 5.30 \times 10^3 \, A/m \) and \( V = 3.93 \times 10^{-6} \, m^3 \).
We use the formula: \( m = M \times V \) to find the magnetic dipole moment:
Substitute the values: \[ m = (5.30 \times 10^3) \times (3.93 \times 10^{-6}) \, A \cdot m^2 \] Calculate the result:
\[ m = 2.08 \times 10^{-2} \, A \cdot m^2 \].
unit conversion in physics
Unit conversion is very important in physics, as it helps us use standard units.
In this exercise, we converted centimeters to meters.
Here's a quick guide for common conversions:
  • 1 cm = 0.01 m
  • 1 mm = 0.001 m
  • 1 km = 1000 m
For our problem:
  • The diameter of 1.00 cm was converted to 0.01 m, and then the radius was found to be 0.005 m
  • The length of 5.00 cm was converted to 0.05 m
magnetic properties of materials
Magnetic properties of materials describe how a material responds to a magnetic field.
The key properties include magnetization, magnetic dipole moment, and magnetic permeability.
Magnetization indicates how easily a material becomes magnetized.
The magnetic dipole moment is the torque experienced by a magnet in a magnetic field.
In our exercise, we calculated the magnetic dipole moment using the magnetization (\( M = 5.30 \times 10^3 \, A/m \)) and the volume of the cylinder (\( 3.93 \times 10^{-6} \, m^3 \)).
These properties are vital for understanding and designing magnetic materials.

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

Two coils are at fixed locations. When coil 1 has no current and the current in coil 2 increases at the rate \(15.0 \mathrm{~A} / \mathrm{s}\), the emf in coil 1 is \(25.0 \mathrm{mV}\). (a) What is their mutual inductance? (b) When coil 2 has no current and coil 1 has a current of \(3.60 \mathrm{~A}\), what is the flux linkage in coil \(2 ?\)

A \(12 \mathrm{H}\) inductor carries a steady current of \(2.0 \mathrm{~A}\). How can a \(60 \mathrm{~V}\) self-induced emf be made to appear in the inductor?

In New Hampshire the average horizontal component of the Earth's magnetic field in 1912 was \(16 \mu \mathrm{T}\) and the average inclination or "dip" was \(73^{\circ} .\) What was the corresponding magnitude of the Earth's magnetic field?

are part of the spark coil of an automobile. When the current in one solenoid falls from \(6.0\) A to zero in \(2.5 \mathrm{~ms}\), an emf of \(30 \mathrm{kV}\) is induced in the other solenoid. What is the mutual inductance \(M\) of the solenoids?

The saturation magnetization \(M^{\max }\) of the ferromagnetic metal nickel is \(4.70 \times 10^{5} \mathrm{~A} / \mathrm{m} .\) Calculate the magnetic moment of a single nickel atom. (The density of nickel is \(8.90 \mathrm{~g} / \mathrm{cm}^{3}\) and its molar mass is \(58.71 \mathrm{~g} / \mathrm{mol}\).)
See all solutions

Recommended explanations on Physics Textbooks

Energy Physics

Read Explanation

Kinematics Physics

Read Explanation

Modern Physics

Read Explanation

Radiation

Read Explanation

Waves Physics

Read Explanation

Famous Physicists

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.