/*! 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 58 The magnetic dipole moment of Ea... [FREE SOLUTION] | 91Ó°ÊÓ

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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 \(3500 \mathrm{~km}\), calculate the current they produce.

Short Answer

Expert verified
The current produced is approximately 732 kA.

Step by step solution

01

Understand the relationship

The magnetic dipole moment \(M\) is related to the current \(I\) and the area \(A\) of the circular path by the formula \(M = I \times A\). We need to find the current \(I\).
02

Calculate the area of the circular path

The area \(A\) of a circle is given by \(A = \pi r^2\), where \(r\) is the radius. Here, \(r = 3500 \times 10^3\) meters (converted from kilometers to meters). So, \(A = \pi \times (3500 \times 10^3)^2\).
03

Apply the relationship formula

Rearrange the formula for magnetic dipole moment to find current: \(I = \frac{M}{A}\).
04

Compute the current

Plug in the values: the magnetic dipole moment \(M = 8.00 \times 10^{22} \mathrm{~J/T}\) and the area \(A = \pi \times (3500 \times 10^3)^2\). Calculate \(I = \frac{8.00 \times 10^{22}}{\pi \times (3500 \times 10^3)^2}\). Evaluate this expression to find the current.

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Key Concepts

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

Magnetism
Magnetism is a physical phenomenon produced by the motion of electric charges, which results in attractive and repulsive forces between objects. It is one of the fundamental forces of nature and plays a critical role in many physical processes. In the context of the Earth's magnetic field, magnetism is generated by electric currents in the Earth's molten outer core.
One common way to quantify magnetism is through the magnetic dipole moment. This is a vector quantity that represents the strength and orientation of a magnet's magnetic field. For Earth, this is attributed to movements of molten iron and other metals that flow like a slowly circulating ocean below the crust. These movements act like a dynamo, generating Earth's magnetic field. The given magnetic dipole moment of the Earth is a considerable value, revealing how powerful and substantial these processes are.
Current Calculation
Calculating the current involved in generating a magnetic dipole moment requires using the relationship between the magnetic dipole moment, current, and the area of the path where the charge flow takes place. The formula used is:
  • \[ M = I \times A \]
where \(M\) is the magnetic dipole moment, \(I\) is the current, and \(A\) is the area of the circular path.
First, we find the area of the circular path using the formula for the area of a circle:
  • \[ A = \pi r^2 \]
Given that the radius \(r\) is \(3500\) kilometers, convert it to meters as \(3500 \times 10^3\) meters, then plug this into the area formula. After finding the area, rearrange the magnetic dipole moment formula to solve for current \(I\):
  • \[ I = \frac{M}{A} \]
Insert the given values, calculate \(I\), and evaluate to find the current generated by the charges flowing in Earth's outer core.
Physics Problem Solving
Solving physics problems, like calculating current from Earth's magnetic dipole moment, involves a series of clear steps. First, understand the problem by identifying what is given and what needs to be found. In this exercise, the magnetic dipole moment and the radius of the path are known, and the current is to be calculated.
Translate the words into physics concepts and formulas. Use the formula connecting magnetic dipole moment to area and current. Ensure all units are consistent (such as converting kilometers to meters) to avoid mistakes. This understanding ensures that calculations using physics laws are accurate and meaningful.
Breaking down complex problems into smaller, manageable steps is useful. This involves computing area first, simplifying equations iteratively, and finally plugging in values to obtain a solution. Checking units and doing a sanity check on the solution ensures the final answer makes sense in the real-world context.

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Most popular questions from this chapter

A wire lying along a \(y\) axis from \(y=0\) to \(y=0.250 \mathrm{~m}\) carries a current of \(2.00 \mathrm{~mA}\) in the negative direction of the axis. The wire fully lies in a nonuniform magnetic field that is given by \(\vec{B}=(0.300 \mathrm{~T} / \mathrm{m}) y \hat{\mathrm{i}}+(0.400 \mathrm{~T} / \mathrm{m}) y \mathrm{j} .\) In unit-vector notation, what is the magnetic force on the wire?

An electron with kinetic energy \(2.5 \mathrm{keV}\) moving along the positive direction of an \(x\) axis enters a region in which a uniform electric field of magnitude \(10 \mathrm{kV} / \mathrm{m}\) is in the negative direction of the \(y\) axis. A uniform magnetic field \(\vec{B}\) is to be set up to keep the electron moving along the \(x\) axis, and the direction of \(\vec{B}\) is to be chosen to minimize the required magnitude of \(\vec{B}\). In unit-vector notation, what \(\vec{B}\) should be set up?

In a Hall-effect experiment, a current of \(3.0 \mathrm{~A}\) sent lengthwise through a conductor \(1.0 \mathrm{~cm}\) wide, \(4.0 \mathrm{~cm}\) long, and \(10 \mu \mathrm{m}\) thick produces a transverse (across the width) Hall potential difference of \(10 \mu \mathrm{V}\) when a magnetic field of \(1.5 \mathrm{~T}\) is passed perpendicularly through the thickness of the conductor. From these data, find (a) the drift velocity of the charge carriers and (b) the number density of charge carriers. (c) Show on a diagram the polarity of the Hall potential difference with assumed current and magnetic field directions, assuming also that the charge carriers are electrons.

A positron with kinetic energy \(2.00 \mathrm{keV}\) is projected into a uniform magnetic field \(\vec{B}\) of magnitude \(0.100 \mathrm{~T}\), with its velocity vector making an angle of \(89.0^{\circ}\) with \(\vec{B}\). Find (a) the period, (b) the pitch \(p\), and (c) the radius \(r\) of its helical path.

A current loop, carrying a current of \(5.0 \mathrm{~A}\), is in the shape of a right triangle with sides 30,40 , and \(50 \mathrm{~cm}\). The loop is in a uniform magnetic field of magnitude \(80 \mathrm{mT}\) whose direction is parallel to the current in the \(50 \mathrm{~cm}\) side of the loop. Find the magnitude of (a) the magnetic dipole moment of the loop and (b) the torque on the loop.

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