/*! 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 circular loop of wire with are... [FREE SOLUTION] | 91Ó°ÊÓ

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A circular loop of wire with area \(A\) lies in the xy-plane. As viewed along the z-axis looking in the \(-z\)-direction toward the origin, a current \(I\) is circulating clockwise around the loop. The torque produced by an external magnetic field \(\overrightarrow{\boldsymbol{B}}\) is given by \(\vec{\tau}=D(4 \hat{i}-3 \hat{J})\), where \(D\) is a positive constant, and for this orientation of the loop the magnetic potential energy \(U=-\overrightarrow{\boldsymbol{\mu}} \cdot \overrightarrow{\boldsymbol{B}}\) is negative. The magnitude of the magnetic field is \(B_{0}=13 D / I A\). (a) Determine the vector magnetic moment of the current loop. (b) Determine the components \(B_{x}, B_{y}\), and \(B_{z}\) of \(\overrightarrow{\boldsymbol{B}} . a^{x}\)

Short Answer

Expert verified
Magnetic moment: \\( \overrightarrow{\mu} = IA \hat{k} \\\); \\( B_x = 0 \\), \\( B_y = 0 \\), \\( B_z = \frac{13D}{IA} \\\).

Step by step solution

01

Understand the relationship between torque and magnetic moment

The torque experienced by a magnetic moment \( \overrightarrow{\mu} \) in an external magnetic field \( \overrightarrow{B} \) is given by \( \overrightarrow{\tau} = \overrightarrow{\mu} \times \overrightarrow{B} \). Therefore, if we have \( \vec{\tau} = D(4\hat{i} - 3\hat{j}) \), we can compare it to the expression from the cross product to draw relations between \( \overrightarrow{\mu} \) and \( \overrightarrow{B} \).
02

Calculate magnetic moment direction

Since the torque \( \vec{\tau} = \overrightarrow{\mu} \times \overrightarrow{B} = D(4\hat{i} - 3\hat{j}) \), this implies that \( \overrightarrow{\mu} \) must lie along the z-axis due to the fact that crossing vectors result in components existing in the xy-plane only. Therefore, \( \overrightarrow{\mu} = \mu \hat{k} \).
03

Use potential energy relationship

Given that the magnetic potential energy \( U = -\overrightarrow{\mu} \cdot \overrightarrow{B} \) is negative, it follows that the angle between \( \overrightarrow{\mu} \) and \( \overrightarrow{B} \) must be less than 90 degrees. Since \( \overrightarrow{\mu} \hat{k} \), this implies both vectors align along the z-axis, allowing us to assume a relationship between their magnitudes.
04

Calculate the magnitude of the magnetic field

The magnitude of the magnetic field is given by \( B_0 = \frac{13D}{IA} \). Given \( \overrightarrow{B} = (B_x, B_y, B_z) \), the component along the z-axis (noted from the conditions where \( \overrightarrow{\mu} \hat{k} \)) determines that \( B_z = B_0 = \frac{13D}{IA} \).
05

Determine magnetic moment

The magnitude of the magnetic moment from the relationship \( \overrightarrow{\mu} = I \vec{A} \hat{k} \) is \( \mu = IA \), assuming the area vector of the loop is along the z-axis.
06

Resolve the rest of \\(\overrightarrow{B}\) components

Since the potential energy is negative and \( \overrightarrow{\mu} = \mu \hat{k} \), \( B_x \) and \( B_y \) must equal zero to ensure only the aligned z-axis component contributes to the magnitude of the magnetic field. Thus, \( B_x = 0 \), \( B_y = 0 \).

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Key Concepts

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

Magnetic Field
The concept of a magnetic field is central to understanding how magnets and currents interact. In this exercise, a magnetic field is present, denoted by the vector \( \overrightarrow{\boldsymbol{B}} \), within which a circular loop carrying a current \( I \) is placed. The field exerts forces on moving charges or currents, which can lead to various phenomena such as torque.

A magnetic field can be visualized as lines extending from the north pole of a magnet to its south pole. These lines are denser where the field is stronger. The strength of the magnetic field is usually represented by the magnitude \( B \), and its direction by the orientation of the field lines.

In this exercise, we learn that the magnitude of the magnetic field is calculated using \( B_0 = \frac{13D}{IA} \). This tells us that the field strength is inversely proportional to the product of the current \( I \) and the area \( A \) of the loop.
Torque
Torque describes the twisting or rotational effect produced on an object. It is defined here by the relationship \( \overrightarrow{\tau} = \overrightarrow{\mu} \times \overrightarrow{B} \). This means the torque \( \overrightarrow{\tau} \) on the loop stems from the interaction of the magnetic moment \( \overrightarrow{\mu} \) with the magnetic field \( \overrightarrow{B} \).

In our problem, it’s given by \( \vec{\tau} = D(4\hat{i} - 3\hat{j}) \). The torque direction is significant because it hints at the relative orientations of \( \overrightarrow{\mu} \) and \( \overrightarrow{B} \).

The resulting torque is solely in the xy-plane, implying that the magnetic moment must lie in a direction (along the z-axis) perpendicular to this plane. Torque is larger when the magnetic moment is perpendicular to the field, as suggested by the cross product geometry.
Magnetic Potential Energy
Magnetic potential energy \( U \) is the energy stored due to an object's position in a magnetic field. It is described by \( U = -\overrightarrow{\mu} \cdot \overrightarrow{B} \). This formula highlights that the potential energy depends on the relative alignment of \( \overrightarrow{\mu} \) and \( \overrightarrow{B} \).

When \( U \) is negative, like in our exercise, it signifies that \( \overrightarrow{\mu} \) and \( \overrightarrow{B} \) face in the same general direction, with an angle less than 90 degrees between them.

Due to this configuration, the torque and potential energy conditions indicate the alignment and zero-out other components \( B_x \) and \( B_y \), emphasizing a simplification to focus on \( B_z \), hence \( \overrightarrow{B} = (0, 0, B_0) \). This alignment minimizes energy, stabilizing the physical system.

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Most popular questions from this chapter

A beam of protons traveling at \(1.20 \mathrm{~km} / \mathrm{s}\) enters a uniform magnetic field, traveling perpendicular to the field. The beam exits the magnetic field, leaving the field in a direction perpendicular to its original direction (Fig. E7.15). The beam travels a distance of \(1.18 \mathrm{~cm}\) while in the field. What is the magnitude of the magnetic field?

A group of particles is traveling in a magnetic field of unknown magnitude and direction. You observe that a proton moving at \(1.50 \mathrm{~km} / \mathrm{s}\) in the \(+x\)-direction experiences a force of \(2.25 \times 10^{-16} \mathrm{~N}\) in the \(+y\)-direction, and an electron moving at \(4.75 \mathrm{~km} / \mathrm{s}\) in the \(-z\)-direction experiences a force of \(8.50 \times\) \(10^{-16} \mathrm{~N}\) in the \(+y\)-direction. (a) What are the magnitude and direction of the magnetic field? (b) What are the magnitude and direction of the magnetic force on an electron moving in the \(-y\)-direction at \(3.20 \mathrm{~km} / \mathrm{s} ?\)

Torque on a Current Loop in a Nonuniform Magnetic Field. In Section \(7.7\) the expression for the torque on a current loop was derived assuming that the magnetic field \(\vec{B}\) was uniform. But what if \(\vec{B}\) is not uniform? Figure P7.49 shows a square loop of wire that lies in the \(x y\)-plane. The loop has comers at \((0,0)\), \((0, L),(L, 0)\), and \((L, L)\) and carries a constant current \(I\) in the clockwise direction. The magnetic field has no z-component but has both \(x\) - and \(y\)-components: \(\overrightarrow{\boldsymbol{B}}=\left(B_{0} y / L\right) \hat{i}+\left(B_{0} x / L\right) \hat{j}\), where \(B_{0}\) is a positive constant. (a) Sketch the magnetic field lines in the \(x y-\) plane. (b) Find the magnitude and direction of the magnetic force exerted on each of the sides of the loop by integrating Eq. (7.20). (c) If the loop is free to rotate about the \(x\)-axis, find the magnitude and direction of the magnetic torque on the loop. (d) Repeat part (c) for the case in which the loop is free to rotate about the \(y\)-axis. (e) Is Eq. (7.26), \(\overrightarrow{\boldsymbol{\tau}}=\overrightarrow{\boldsymbol{\mu}} \times \overrightarrow{\boldsymbol{B}}\), an appropriate description of the torque on this loop? Why or why not?

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. P7.41). (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 \(B=0.80 \mathrm{~T}, I=2.0 \times 10^{3} \mathrm{~A}, m=25 \mathrm{~kg}\), and \(L=50 \mathrm{~cm}\). For simplicity asssume 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.

You wish to hit a target from several meters away with a charged coin having a mass of \(4.25 \mathrm{~g}\) and a charge of \(+2500 \mu \mathrm{C}\). The coin is given an initial velocity of \(12.8 \mathrm{~m} / \mathrm{s}\), and a downward, uniform electric field with field strength \(27.5 \mathrm{~N} / \mathrm{C}\) exists throughout the region. If you aim directly at the target and fire the coin horizontally, what magnitude and direction of uniform magnetic field are needed in the region for the coin to hit the target? (take \(\left.g=9.8 \mathrm{~m} / \mathrm{s}^{2}\right)\)

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