/*! 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 88 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 \(x y\) -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 \(\vec { B }\) is given by \(\vec { \tau } = D ( 4 \hat { \imath } - 3 \hat { \jmath } ) ,\) where \(D\) is a positive constant, and for this orientation of the loop the magnetic potential energy \(U = - \vec { \mu } \cdot B\) 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 \(\vec { B } .\)

Short Answer

Expert verified
(a) \(\vec{\mu} = -IA\hat{k}\); (b) \( B_x = \frac{3D}{IA}, B_y = \frac{4D}{IA}, B_z = \frac{12D}{IA} \).

Step by step solution

01

Express the Magnetic Moment

The magnetic moment \( \vec{\mu} \) of a current-carrying loop is given by \( \vec{\mu} = I \cdot \vec{A} \). Since the loop lies in the \(xy\)-plane and has area \(A\), the magnetic moment will be aligned along the \(z\)-axis in the direction of the surface normal. However, since the current is circulating clockwise when viewed from the \(z\)-axis looking in the \(-z\) direction, \(\vec{\mu} = - I \cdot A \hat{k} \).
02

Relate Torque to Magnetic Moment and Magnetic Field

The torque \( \vec{\tau} \) on a current loop in a magnetic field is given by \( \vec{\tau} = \vec{\mu} \times \vec{B} \). We have \( \vec{\tau} = D(4\hat{\imath} - 3\hat{\jmath}) \). Substitute \( \vec{\mu} = -IA\hat{k} \) into this equation: \( -IA\hat{k} \times \vec{B} = D(4\hat{\imath} - 3\hat{\jmath}) \).
03

Determine Magnetic Field Components using Cross Product

Perform the cross product \( -IA\hat{k} \times (B_x\hat{i} + B_y\hat{j} + B_z\hat{k}) = D(4\hat{i} - 3\hat{j}) \). Simplifying gives: \( IA(B_y)\hat{i} - IA(B_x)\hat{j} = D(4\hat{i} - 3\hat{j}) \). Thus, \( IA B_y = 4D \) and \( -IA B_x = -3D \). Solve these to get \( B_y = \frac{4D}{IA} \) and \( B_x = \frac{3D}{IA} \).
04

Find Magnetic Field Along z-axis

Given the magnitude of the magnetic field \( B_0 = 13D/IA \), relate it to the components: \( B_0 = \sqrt{B_x^2 + B_y^2 + B_z^2} \). Substitute \( B_x = \frac{3D}{IA}, \ \; B_y = \frac{4D}{IA} \) into the equation to solve for \( B_z \).
05

Solve for Component B_z

Substitute the expressions for \( B_x \) and \( B_y \) into the magnitude equation: \( \left( \frac{3D}{IA} \right)^2 + \left( \frac{4D}{IA} \right)^2 + B_z^2 = \left( \frac{13D}{IA} \right)^2 \). Simplify to find \( B_z^2 = \left( \frac{13D}{IA} \right)^2 - \left( \frac{3D}{IA} \right)^2 - \left( \frac{4D}{IA} \right)^2 \).\Solve for \( B_z \).
06

Calculate Magnetic Field Components

Perform the calculations: \( \left( \frac{3}{IA} \right)^2 = \frac{9}{I^2A^2}, \; \left( \frac{4}{IA} \right)^2 = \frac{16}{I^2A^2} \), so sum these and subtract from \( \left( \frac{13}{IA} \right)^2 = \frac{169}{I^2A^2} \) to find \( B_z^2 \). Simplification shows \( B_z^2 = \frac{144}{I^2A^2} \), thus \( B_z = \frac{12D}{IA} \).
07

Summarize Results for Magnetic Moment

Using previous findings of \( \vec{\mu} = -IA\hat{k} \), we can deduce its direction is negative \(z\)-axis and magnitude is \(IA\), given the relationship with the torque and external magnetic field.
08

Confirm Answers for Magnetic Field Components

The magnetic field components that satisfy the problem conditions are \( B_x = \frac{3D}{IA} \), \( B_y = \frac{4D}{IA} \), and \( B_z = \frac{12D}{IA} \). These values correctly solve the torque equation and match the field magnitude specified.

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Key Concepts

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

Torque in Magnetic Fields
When a loop carrying a current is placed in a magnetic field, it experiences a torque. This torque, denoted as \(\vec{\tau}\), is directly related to the magnetic moment of the loop. The concept of torque in magnetic fields can be understood through its formula:
  • \(\vec{\tau} = \vec{\mu} \times \vec{B} \)
Here, \(\vec{\mu}\) is the magnetic moment, and \(\vec{B}\) is the magnetic field. The cross product implies that the torque direction is perpendicular to both the magnetic moment and the magnetic field.
The torque will try to align the magnetic moment with the magnetic field. This is analogous to the way a compass needle aligns with Earth's magnetic field, seeking a minimum energy state. In our scenario, the given torque \(D(4\hat{\imath} - 3\hat{\jmath})\) showcases how a current loop undergoes a rotational effect due to the magnetic field's components.
Magnetic Field Components
The magnetic field \(\vec{B}\) is characterized by its components in space: \(B_x\), \(B_y\), and \(B_z\). These components define the field's intensity in the \(x\), \(y\), and \(z\) directions respectively. Each component has a unique influence on the magnetic moment in terms of torque and potential energy.
The steps to find these components involve using the cross product from the torque equation:
  • Given \(\vec{\tau} = D(4\hat{\imath} - 3\hat{\jmath})\)
  • By relating it to \(-IA\hat{k} \times (B_x \hat{i} + B_y \hat{j} + B_z \hat{k})\), each component is determined.
Through solving the equations, we find:
  • \(B_x = \frac{3D}{IA}\)
  • \(B_y = \frac{4D}{IA}\)
  • \(B_z = \frac{12D}{IA}\)
These components are essential to describe how the magnetic field influences both the loop's torque and alignment.
Magnetic Potential Energy
Magnetic potential energy relates to the orientation of a magnetic moment within a magnetic field. The potential energy \(U\) of a magnetic moment \(\vec{\mu}\) in a magnetic field \(\vec{B}\) is expressed as:
  • \(U = - \vec{\mu} \cdot \vec{B} \)
Here, the negative sign indicates a desire for the system to reach lower energy states by aligning the moment with the field.
In our specific scenario, the negative potential energy implies that the loop is already near this stable orientation, despite resisting forces like torque that might seek to misalign it. The components \(B_x\), \(B_y\), and \(B_z\) contribute to this energy, dictating how strong the alignment tendency is in each spatial direction.
Understanding magnetic potential energy is vital for comprehending why objects like magnets and loops rotate within fields—they are seeking the lowest possible energy configuration.

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Most popular questions from this chapter

A particle with charge 7.80\(\mu \mathrm{C}\) is moving with velocity \(\vec{\boldsymbol{v}}=-\left(3.80 \times 10^{3} \mathrm{m} / \mathrm{s}\right) \hat{\boldsymbol{J}}\) . The magnetic force on the particle is measured to be \(\vec{\boldsymbol{F}}=+\left(7.60 \times 10^{-3} \mathrm{N}\right) \hat{\boldsymbol{\imath}}-\left(5.20 \times 10^{-3} \mathrm{N}\right) \hat{\boldsymbol{k}}\) (a) Calculate all the components of the magnetic field you can from this information. (b) Are there components of the magnetic field that are not determined by the measurement of the force? Explain. (c) Calculate the scalar product \(\vec{B} \cdot \vec{F} .\) What is the angle between \(\vec{B}\) and \(\vec{\boldsymbol{F}} ?\)

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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 \(\vec { B }\) fills the region between the rails (Fig. \(P 27.74 )\) (a) Find the magnitude and direction of the net force on the con- ducting 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 \(B = 0.80 \mathrm { T } , \quad I = 2.0 \times 10 ^ { 3 } \mathrm { A } , \quad m = 25 \mathrm { kg }\) and \(L = 50 \mathrm { cm } .\) For simplicity assume 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.

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