/*! 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 85 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 extemal magnetic field \(\vec{B}\) is given by \(\vec{\tau}=D(4 \hat{z}-3 \hat{y}),\) where \(D\) is a positive constant, and for this orientation of the loop the magnetic potential energy \(U=-\vec{\mu} \cdot \vec{B}\) is negative. The magnitude of the magnetic field is \(B_{0}=13 D / L A\) (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) \( \vec{\mu} = \mu \hat{z}, \text{magnitude } \mu = \frac{3D}{B_x} \). (b) \(B_x = \text{\text{solve force equ potential}} \( B_y = 0, B_z = B_0\).}

Step by step solution

01

Understanding Torque and Magnetic Moment

In this exercise, we need to relate the torque \( \vec{\tau} \) experienced by a loop of wire in a magnetic field to its magnetic moment \( \vec{\mu} \) and the magnetic field \( \vec{B} \). The relationship is \( \vec{\tau} = \vec{\mu} \times \vec{B} \). We're given that \( \vec{\tau} = D(4 \hat{z} - 3 \hat{y}) \). The magnetic moment \( \vec{\mu} \) is associated with the current and the geometry of the loop.
02

Determining the Magnetic Moment

Since the torque is given by the expression \( \vec{\tau} = \vec{\mu} \times \vec{B} \), we know \( \vec{\tau} \) is perpendicular to both \( \vec{\mu} \) and \( \vec{B} \). Given the loop's alignment in the \( xy \)-plane and current's clockwise direction from \(-z\), the magnetic moment \( \vec{\mu} \) is directed along \(\hat{z}\). Hence, \( \vec{\mu} = \mu \hat{z} \).
03

Solve for Magnetic Moment Magnitude

Using the equation for torque, \( D(4\hat{z} - 3\hat{y}) = \vec{\mu} \times \vec{B} \), we assume \( \vec{B} = B_x \hat{x} + B_y \hat{y} + B_z \hat{z} \). The cross-product expression simplifies to \(-\mu B_y \hat{x} + \mu B_x \hat{y} + 0\, \hat{z}\) = \(D(4\hat{z} - 3\hat{y})\). Therefore, \( \mu B_x=3D \) and \( \mu B_y=0 \).
04

Solving for Magnetic Field Components

The magnitude of \( \vec{B} \) is \( B_0 = \frac{13D}{LA} \). Checking for consistency with \( \mu B_x=3D \) and \( \mu B_y=0 \), we also use energy \( U = -\vec{\mu} \cdot \vec{B} < 0 \). Thus, \( \vec{\mu} = 3D/ B_x \hat{z} \) is negative if \( B_z > 0 \) and negative with \( \mu \).
05

Express Magnetic Field Components

The vector \( \mu \) across \(\hat{x}, \hat{y}, \hat{z}\) terms imply cartesian bounds on \( \vec{B} \) so that \( B_x = \frac{3D}{\mu}, B_y = 0, B_z = B_0\). The \( z\)-component maintains energy condition under field affect.

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Key Concepts

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

Torque and Magnetic Field
When dealing with a current loop in a magnetic field, one important concept is the torque produced on the loop. Torque is a twisting force that tends to cause rotation. The exact way torque works here involves both the loop's magnetic moment and the magnetic field itself. The equation that describes this relationship is \( \vec{\tau} = \vec{\mu} \times \vec{B} \). Here, \( \vec{\tau} \) represents torque, \( \vec{\mu} \) is the magnetic moment, and \( \vec{B} \) is the magnetic field. The cross product (\( \times \)) signifies that the torque is perpendicular to both the magnetic moment and the magnetic field.
  • Torque depends on the orientation of the magnetic moment relative to the magnetic field. If they are aligned, torque is minimized.
  • In our exercise, the torque is given as \( D(4 \hat{z} - 3 \hat{y}) \), implying specific aligns of the components.
The magnitude of torque will give you insight into how the loop tries to rotate within the field. It's a crucial part of understanding magnetic interactions.
Current Loop
Current loops are an essential part of electromagnetism and physics. A current loop is simply a loop made of wire through which electric current flows. The current loop in this exercise lies flat in the \( xy \)-plane, forming a vital part of understanding the magnetic moment.Any loop carrying a current will have a magnetic moment \( \vec{\mu} \). The direction of this magnetic moment is given by the right-hand rule. For a current loop viewed from the positive \( z \)-axis as moving clockwise, the magnetic moment extends along the negative \( z \)-axis.
  • The magnetic moment is proportional to the current \( I \) and the area \( A \) of the loop: \( \vec{\mu} = I \times A \times \hat{n} \), where \( \hat{n} \) is the unit vector perpendicular to the loop.
  • In this case, because the loop lies in the \( xy \)-plane, \( \vec{\mu} \) aligns with the \( z \)-axis.
The current loop is central to deriving outcomes like torque or the interactions of the loop in fields. These loops underpin a lot of electromagnetic concepts you'll encounter.
Cross Product in Magnetostatics
The cross product is a key mathematical tool used extensively in physics to handle vector quantities, especially in magnetostatics. It helps determine directions and magnitudes useful in understanding how magnetic moments and fields interact.In the expression \( \vec{\tau} = \vec{\mu} \times \vec{B} \), the cross product \( \times \) calculates the torque by determining the perpendicular vector resulting from \( \vec{\mu} \) and \( \vec{B} \).
  • The direction of the resultant vector (like torque) is perpendicular to the plane formed by the two original vectors.
  • The magnitude of this vector is the product of the magnitudes of the two vectors and the sine of the angle between them.
This math insight helps solve physics problems by clearly showing the resultant directions, especially when analyzing forces and fields in electromagnetism. The cross product not only provides direction but helps deduce component alignments as seen in the exercise.

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Most popular questions from this chapter

The plane of a \(5.0 \mathrm{cm} \times 8.0 \mathrm{cm}\) rectangular loop of wire is parallel to a \(0.19-\mathrm{T}\) magnetic field. The loop carries a current of 6.2 \(\mathrm{A}\) . (a) What torque acts on the loop? (b) What is the magnetic moment of the loop? (c) What is the maximum torque that can be obtained with the same total length of wire carrying the same current in this magnetic field?

An insulated wire with mass \(m=5.40 \times 10^{-5} \mathrm{kg}\) is bent into the shape of an inverted U such that the horizontal part has a length \(l=15.0 \mathrm{cm} .\) The bent ends of the wire are partially immersed in two pools of mercury, with 2.5 \(\mathrm{cm}\) of each end below the mercury's surface. The entire structure is in a region containing a uniform \(0.00650-\mathrm{T}\) magnetic field directed into the page (Fig. 27.71\()\) . An electrical connection from the mercury pools is made through the ends of the wires. The mercury pools are connected to a \(1.50-\mathrm{V}\) battery and a switch \(\mathrm{S}\) . When switch \(\mathrm{S}\) is closed, the wire jumps 35.0 \(\mathrm{cm}\) into the air, measured from its initial position. (a) Determine the speed \(v\) of the wire as it leaves the mercury. (b) Assuming that the current \(I\) through the wire was constant from the time the switch was closed until the wire left the mercury, determine \(I\) (c) Ignoring the resistance of the mercury and the circuit wires, determine the resistance of the moving wire.

A straight, vertical wire carries a current of 1.20 \(\mathrm{A}\) downward in a region between the poles of a large superconducting electromagnet, where the magnetic field has magnitude \(B=\) 0.588 \(\mathrm{T}\) and is horizontal. What are the magnitude and direction of the magnetic force on a \(1.00-\mathrm{cm}\) section of the wire that is in this uniform magnetic field, if the magnetic field direction is (a) east; (b) south; (c) \(30.0^{\circ}\) south of west?

The magnetic flux through one face of a cube is \(+0.120 \mathrm{Wb} .\) (a) What must the total magnetic flux through the other five faces of the cube be? (b) Why didn't you need to know the dimensions of the cube in order to answer part (a) 2 (c) Suppose the magnetic flux is due to a permanent magnet like that shown in Fig. \(27.11 .\) In a sketch, show where the cube in part (a) might be located relative to the magnet.

In a shunt-wound dc motor with the field coils and rotor connected in parallel (Fig. \(27.56 ),\) the resistance \(R_{t}\) of the field coils is \(106 \Omega,\) and the resistance \(R_{r}\) of the rotor is 5.9\(\Omega\) . When a potential difference of 120 \(\mathrm{V}\) is applied to the brushes and the motor is running at full speed delivering mechanical power, the current supplied to it is 4.82 \(\mathrm{A}\) (a) What is the current in the field coils? (b) What is the current in the rotor? (c) What is the induced emf developed by the motor? (d) How much mechanical power is developed by this motor?

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