/*! 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 40 The magnetic dipole moment of Ea... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 28: Problem 40

The magnetic dipole moment of Earth has magnitude \(8.00 \times 10^{22} \mathrm{~J} / \mathrm{T}\). Assume that this is produced by charges flowing in Earth's molten outer core. If the radius of their circular path is \(3700 \mathrm{~km}\), calculate the current they produce.

Short Answer

Expert verified
The current is approximately \(1.86 \times 10^{9} \mathrm{~A}\).

Step by step solution

01

Understanding the Problem

We need to find the electric current produced by Earth's molten outer core, which acts as a circular loop that has a magnetic dipole moment. The given magnetic dipole moment is \(8.00 \times 10^{22} \mathrm{~J} / \mathrm{T}\) and the radius of the circular path is \(3700 \mathrm{~km}\).
02

Relating Magnetic Dipole Moment and Current

The magnetic dipole moment \(\mu\) of a circular current loop is related to the current \(I\) and the area \(A\) of the loop by the formula \(\mu = I \cdot A\), where \(A = \pi r^2\).
03

Calculate the Area of the Loop

First convert the radius into meters: \(3700 \mathrm{~km} = 3.7 \times 10^6 \mathrm{~m}\). Then calculate the area \(A\): \[A = \pi (3.7 \times 10^6)^2 = 4.30 \times 10^{13} \mathrm{~m}^2.\]
04

Solve for Current

Rearrange the formula to solve for \(I\): \(I = \frac{\mu}{A}\). Substitute the given values: \[I = \frac{8.00 \times 10^{22} \mathrm{~J} / \mathrm{T}}{4.30 \times 10^{13} \mathrm{~m}^2} = 1.86 \times 10^{9} \mathrm{~A}.\]
05

Conclusion

Thus, the current produced by the charges flowing in Earth's molten outer core is approximately \(1.86 \times 10^{9} \mathrm{~A}\).

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.

Electric Current
Electric current is a flow of electric charge. In most cases, this charge is carried by moving electrons through a conductor like a wire. However, in the case of Earth's molten outer core, it is the flow of charged particles within molten iron that produces electric current.
  • Electric current is measured in amperes (A).
  • It is typically described by the direction in which positive charges would flow.
  • In the context of Earth's core, these currents are massive, creating a large magnetic field.
Electric current plays a crucial role in generating magnetic fields. It is the key to understanding phenomena like Earth's magnetism, which shields the planet from harmful solar and cosmic radiation.
Circular Current Loop
A circular current loop is formed when electric current flows along a circular path. It can be thought of as a looped wire that carries electricity.
  • When current flows in such a loop, it creates a magnetic field.
  • The magnetic moment is a crucial property of a circular current loop.
  • This concept underpins devices like electromagnets and is also similar to the current in Earth's core.
In the case of Earth, the molten iron in the core flows in circular patterns, akin to a gigantic circular current loop. This large-scale looping current generates a significant magnetic dipole moment.
Area Calculation
Calculating the area of a circular current loop is vital to finding the magnetic dipole moment and electric current.
  • The area of a circle is given by the formula: \(A = \pi r^2\).
  • This area is crucial for calculations involving the magnetic properties of the loop.
For example, to find the area of Earth's molten outer core, first convert the radius to meters, as in the solution: \(3700 \text{ km} = 3.7 \times 10^6 \text{ m}\). Once converted, apply the area formula to get \(4.30 \times 10^{13} \text{ m}^2\).
This calculated area helps in determining the strength and characteristics of the magnetic field produced by the electric currents.
Earth's Magnetic Field
Earth's magnetic field is a fascinating and complex subject. It is generated by the electric currents flowing through its molten outer core.
  • This field is similar in structure to the magnetic field created by a bar magnet.
  • The magnetic dipole moment of Earth, which relates to its magnetic field strength, originates from these core currents.
  • Earth's magnetic field plays a crucial role in navigation and protecting the planet.
The magnetic dipole moment "\(8.00 \times 10^{22} \text{ J/T}\)" provided in the exercise helps understand the intensity of this field. By examining how these currents flow within a gigantic circular loop, scientists gain insights into the workings of Earth's protective magnetic bubble.

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 that has an instantaneous velocity of $$ \vec{v}=\left(-5.0 \times 10^{6} \mathrm{~m} / \mathrm{s}\right) \hat{\mathrm{i}}+\left(3.0 \times 10^{6} \mathrm{~m} / \mathrm{s}\right) \hat{\mathrm{j}} $$ is moving through the uniform magnetic field \(\vec{B}=(0.030 \mathrm{~T}) \hat{\mathrm{i}}-\) \((0.15 \mathrm{~T}) \hat{j}\). (a) Find the force on the electron due to the magnetic field. (b) Repeat your calculation for a proton having the same velocity.

A conducting rectangular solid of dimensions \(d_{x}=5.00 \mathrm{~m}, d_{y}=3.00 \mathrm{~m}\), and \(d_{z}=2.00 \mathrm{~m}\) moves with a constant velocity \(\vec{v}=(20.0 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{i}}\) through a uniform magnetic field \(\vec{B}=(40.0 \mathrm{mT}) \hat{\mathrm{j}}\) (Fig. 28-22). What are the resulting (a) electric field within the solid, in unit- vector notation, and (b) potential difference across the solid? (c) Which face becomes negatively charged?

A wire \(2.30 \mathrm{~m}\) long carries a current of \(13.0 \mathrm{~A}\) and makes an angle of \(35.0^{\circ}\) with a uniform magnetic field of magnitude \(B=1.50 \mathrm{~T}\). Calculate the magnetic force on the wire.

A circular loop of wire having a radius of \(8.0 \mathrm{~cm}\) carries a current of \(0.20 \mathrm{~A}\). A vector of unit length and parallel to the dipole moment \(\vec{\mu}\) of the loop is given by \(0.60 \hat{\mathrm{i}}-0.80 \mathrm{j}\). (This unit vector gives the orientation of the magnetic dipole moment vector.) If the loop is located in a uniform magnetic field given by \(\vec{B}=(0.50 \mathrm{~T}) \hat{\mathrm{i}}+(0.20 \mathrm{~T}) \hat{\mathrm{k}}\), find (a) the torque on the loop (in unit-vector notation) and (b) the orientation energy of the loop.

(a) Find the frequency of revolution of an electron with an energy of \(189 \mathrm{eV}\) in a uniform magnetic field of magnitude \(70.0 \mu \mathrm{T}\). (b) Calculate the radius of the path of this electron if its velocity is perpendicular to the magnetic field.
See all solutions

Recommended explanations on Physics Textbooks

Modern Physics

Read Explanation

Electrostatics

Read Explanation

Astrophysics

Read Explanation

Nuclear Physics

Read Explanation

Conservation of Energy and Momentum

Read Explanation

Energy Physics

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.