/*! 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 16 A wire of length \(L \mathrm{~m}... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 27: Problem 16

A wire of length \(L \mathrm{~m}\) carrying a current \(I\) amp is bent in the form of a circle. The magnitude of magnetic moment is : (a) \(\frac{I L^{2}}{4 \pi}\) (b) \(\frac{I L^{2}}{2 \pi}\) (c) \(\frac{1 L}{4 \pi}\) (d) \(I \pi L^{2}\)

Short Answer

Expert verified
Option (a), \(\frac{I L^2}{4\pi}\), is the correct choice.

Step by step solution

01

Understand the Problem

We need to determine the magnitude of the magnetic moment of a wire of length \(L\) that is bent into a circle and carries a current \(I\). The magnetic moment \(M\) of a current loop is given by the formula \(M = I \cdot A\), where \(A\) is the area of the circle that the wire forms.
02

Express Radius in Terms of Length

If the wire is bent into a circle, its circumference \(C\) is equal to the length of the wire, \(L\). The circumference of a circle is expressed as \(C = 2\pi r\), where \(r\) is the radius. Thus, we have \(L = 2\pi r\), and by rearranging, we find \(r = \frac{L}{2\pi}\).
03

Calculate the Area of the Circle

The area \(A\) of a circle is given by \(A = \pi r^2\). Substituting the expression for \(r\) from Step 2, we get:\[A = \pi \left(\frac{L}{2\pi}\right)^2 = \pi \cdot \frac{L^2}{4\pi^2} = \frac{L^2}{4\pi}.\]
04

Determine the Magnetic Moment

Using the formula \(M = I \cdot A\), substitute the calculated area from Step 3 to find the magnetic moment:\[M = I \cdot \frac{L^2}{4\pi} = \frac{I L^2}{4\pi}.\]
05

Choose the Correct Option

The expression for the magnetic moment \(M = \frac{I L^2}{4\pi}\) matches with option (a). Therefore, the magnitude of the magnetic moment is \(\frac{I L^2}{4\pi}\).

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.

Magnetic Moment Formula
The magnetic moment of a current loop is a vital concept in electromagnetism. It quantifies the strength of a magnetic source created by a loop. The formula is straightforward: \(M = I \cdot A\). Here, \(M\) represents the magnetic moment, \(I\) stands for the current passing through the loop, and \(A\) denotes the area the loop encloses.

A larger magnetic moment indicates a stronger magnetic effect. Thus, both the amount of current and the area of the loop directly contribute to creating this magnetic influence. It's important to remember that the magnetic moment is a vector, meaning it has both a magnitude and a direction, which usually aligns with the normal to the loop surface.
Area of a Circle
Calculating the area of a circle is a fundamental geometry skill. The formula is: \(A = \pi \cdot r^2\). This calculation depends on the circle's radius, \(r\), the distance from the center of the circle to any point on its edge.

In the context of the current loop, the area calculation becomes essential as it directly influences the magnetic moment. When transforming a straight wire into a circular shape, understanding how to find the area helps resolve related physics problems. Keep in mind, that for any calculations involving circles in physics, the accuracy of the area influences the accuracy of results like the magnetic moment in this loop scenario.
Current-Carrying Wire
A current-carrying wire is a simple yet significant concept in physics and electrical engineering. When an electric current flows through a wire, a magnetic field is generated around it. This is the principle behind electromagnets and many electrical appliances.

The intensity of the magnetic field created by a straight wire is amplified when the wire is shaped into a loop. This is because the current flowing through the wire continuously contributes to the magnetic moment within the enclosed area. Understanding how current interacts with wire geometry is essential for fields like electromechanical design and magnetic field generation.
Geometry of Circles
In geometry, circles hold a unique and significant position. They are defined as the set of all points in a plane that are equidistant from a given point, called the center. The circle's layout provides important geometric insights, especially the relationships between radius, diameter, and circumference.

The circumference of a circle, depicted as \(C = 2\pi r\), is key in this exercise as it helps link the length of the bent wire to the circle's radius. By rearranging the formula, one can derive expressions for unknown variables, such as radius in terms of circumference. This foundational knowledge aids in solving complex physics problems where circular shapes and properties are extensively involved.

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

A charged particle of mass \(m\) and charge \(q\) is accelerated through a potential difference of \(V\) volt. It enters a region of uniform magnetic field which is directed perpendicular to the direction of motion of the particie. The particle will move on a circular path of radius given by: (a) \(\sqrt{\left(\frac{V m}{q B^{2}}\right)}\) (b) \(\frac{2 V m}{q B^{2}}\) (c) \(\sqrt{\left(\frac{2 V_{m}}{q}\right)}\left(\frac{1}{B}\right)\) (d) \(\sqrt{\left(\frac{V m}{q}\right)}\left(\frac{1}{B}\right)\)

The strength of the magnetic field around a straight wire : (a) is same everywhere around the wire (b) obeys inverse square law (c) is directly proportional to the square of the distance from the wire (d) none of the above

A wire of arbitrary shape carries a current \(I=2 \mathrm{~A}\). consider the portion of wire between \((0,0,0)\) and \((4 \mathrm{~m}\), \(4 \mathrm{~m}, 4 \mathrm{~m}\) ). A magnetic field given by \(\overrightarrow{\mathrm{B}}=1.2 \times 10^{-4} \mathrm{~T} \hat{\mathrm{i}}+2.0 \times 10^{-4} \mathrm{~T} \hat{\mathrm{j}}\) exists in the space. The force acting on the given portion is: (a) incalculatable as length of wire is not known (b) \(\overrightarrow{\mathrm{F}}=[(\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}) \times(1.2 \hat{\mathrm{i}}+2.0 \hat{\mathrm{j}})] \mathrm{N}\) (c) \(\overrightarrow{\mathbf{F}}=8 \times 10^{-4}[(\hat{1}+\hat{j}+\hat{k}) \times(1.2 \hat{i}+2.0 \hat{j})] \mathrm{N}\) (d) \(\overrightarrow{\mathbf{F}}=8 \times 10^{-4}[(1.2 \hat{\mathbf{i}}+2.0 \hat{\mathbf{j}}) \times(\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}})] \mathrm{N}\)

A non-relativistic charged particle of charge \(q\) and mass \(m\) originates at a point at origin of coordinate system. A magnetic field strength \(B\) is directed along \(x\) -axis. The charged particle moves with velocity \(v\) at an angle \(\alpha\) to the \(x\) -axis. A screen is oriented at right angles to the axis and is situated at a distance \(x_{0}\) from the origin of coordinate system. The coordinates of point \(P\) on the screen at which the charged particle strikes: (a) \(x_{0}, \frac{m v \sin \alpha}{q B} \sin \frac{q B}{m} t, \frac{n v \sin \alpha}{q B}\left[1-\frac{\cos q B t}{m}\right]\) (b) \(x_{0}, \frac{m v \sin \alpha}{q B} \sin \frac{q B t}{m}, \frac{m v \sin \alpha}{q B}\left[\cos \frac{q B}{m} t-1\right]\) (c) \(x_{0}, \frac{m \sin \alpha}{q B}\left[\cos \frac{q B}{m} t-1\right] \frac{m v \sin \alpha}{q B} \sin \frac{q B}{m} t\) (d) none of the above

A uniform magnetic field is at right angle to the direction of motion of proton. As a result, the proton describes a circular path of radius \(2.5 \mathrm{~cm}\). If the speed of the proton is doubled then the radius of the circular path will be : (a) \(0.5 \mathrm{~cm}\) (b) \(2.5 \mathrm{~cm}\) (c) \(5.0 \mathrm{~cm}\) (d) \(7.5 \mathrm{~cm}\)
See all solutions

Recommended explanations on Physics Textbooks

Work Energy and Power

Read Explanation

Solid State Physics

Read Explanation

Geometrical and Physical Optics

Read Explanation

Conservation of Energy and Momentum

Read Explanation

Radiation

Read Explanation

Classical Mechanics

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.