/*! 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 152 An insulating rod of length \(\e... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 9: Problem 152

An insulating rod of length \(\ell\) carries a charge \(q\) distributed uniformly on it. The rod is pivoted at its mid point and is rotated at a frequency \(f\) about a fixed axis perpendicular to the rod and passing through the pivot. The magnetic mornent of the rod system is a. \(\frac{1}{12} \pi q f \ell^{2}\) b \(\pi_{q} f \ell^{2}\) c. \(\frac{1}{6} \pi q f \ell^{2}\) d. \(\frac{1}{3} \pi q f \ell^{2}\)

Short Answer

Expert verified
Option a: \( \frac{1}{12} \pi q f \ell^{2} \) is correct.

Step by step solution

01

Understand Magnetic Moment Formula

The magnetic moment \( \mu \) of a rotating charged body is expressed as \( \mu = \frac{1}{2} I \omega \), where \( I \) is the moment of inertia and \( \omega \) is the angular velocity. In this scenario, since the pivot point is at the middle, the rod rotates as if it were composed of two particles, each \( \frac{\ell}{2} \) from the pivot.
02

Calculate Moment of Inertia

The moment of inertia \( I \) for a rod of length \( \ell \) and mass \( m \), pivoted in the middle, is \( \frac{1}{12} m \ell^2 \). In this case, replace \( m \) with the linear mass distribution \( \lambda \), which is the charge per unit length \( \lambda = \frac{q}{\ell} \), thus making \( m = q \).
03

Determine Angular Velocity

The angular velocity \( \omega \) can be related to the frequency by \( \omega = 2 \pi f \), since the rod rotates through \( 2 \pi \) radians per complete cycle and completes \( f \) cycles per second.
04

Calculate Magnetic Moment

Substituting \( \omega \) and \( I \) into the formula for magnetic moment, we have \[ \mu = \frac{1}{2} \left( \frac{1}{12} q \ell^2 \right) (2 \pi f) = \frac{1}{12} \pi q f \ell^{2}. \]
05

Choose the Correct Option

The calculated magnetic moment \( \frac{1}{12} \pi q f \ell^{2} \) matches option a.

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.

Insulating Rod
An insulating rod is a type of rod that does not conduct electricity due to its non-conductive material. This means it can neither allow current to flow through it nor allows free movement of electrons across its surface. This property is crucial in this exercise because the charge remains uniformly distributed along the rod. For simplicity, we consider the rod as an ideal one-dimensional object when calculating physical properties.
  • Insulators have high electrical resistance.
  • They are often used to support or separate electrical conductors.
  • In this context, the rod ensures a consistent distribution of charge.
Understanding this is essential as it allows us to focus on the movement and distribution of charges without worrying about loss to conductivity.
Moment of Inertia
Moment of inertia is a property of any rotating body. It quantifies how difficult it is to change the object's rotational speed. For a rod pivoted at its middle, the moment of inertia considers the mass distribution about the pivot point. For a rod of length \( \ell \) and a total charge \( q \), the mass can analogously be considered as the charge for our electrical analogy. It is derived from the formula:\[ I = \frac{1}{12} m \ell^2 \]where for our purpose, \( m \) is replaced by \( q \), assuming the linear charge density as a factor. This value tells you how much the rotational motion resists changes and influences the evolution of the system's magnetic moment.
  • Higher moment of inertia implies greater resistance to angular acceleration.
  • Calculated differently for various geometries and mass/charge distributions.
Angular Velocity
Angular velocity describes how quickly an object rotates. It is measured in radians per second. In the context of our rotating rod, angular velocity \( \omega \) can be connected directly to the frequency of rotation \( f \), knowing that it completes one full cycle every \( 2 \pi \) radians:\[ \omega = 2 \pi f \].This formula highlights the relationship between linear frequency and circular motion for objects like the rod.This concept is crucial because it allows us to express the rotating motion linearly. It connects the rate of rotation to the physical properties of our system, like the charge and length of the rod.
  • Angular velocity clarifies the rate of rotation for any spinning object.
  • Tied deeply to the calculation of rotational kinetic energy and magnetic effects.
Charge Distribution
Charge distribution refers to how charge is spread over the insulating rod. In this exercise, the charge is evenly distributed along the rod's entire length. This uniformity simplifies calculations and leads to precise solutions for physical properties:- To quantify this, we use linear charge density \( \lambda \), which is charge per unit length: \( \lambda = \frac{q}{\ell} \).With a uniform distribution, each segment of the rod has the same charge density, making it straightforward to calculate other properties like the magnetic moment.
  • Uniform distribution means simplifications in mathematical modeling.
  • One can determine potential energy and electric fields based on given distribution.
  • It maintains a constant magnetic activity when rotating.

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 proton and an \(\alpha\) -particle enter a uniform magnetic field perpendicular with the same speed. If the proton takes \(25 \mu \mathrm{s}\) to make 5 revolutions, then the periodic time for the \(\alpha\) -particle would be a. \(50 \mu \mathrm{s}\) b. \(25 \mu \mathrm{s}\) c. \(10 \mathrm{us}\) d. \(5 \mu s\)

A steady current is flowing in a circular coil of radius \(R\), made up of a thin conducting wire. The magnetic field at the center of the loop is \(B_{L}\). Now, a circular loop of radius \(R / n\) is made from the same wire without chenging its length, by unfolding and refolding the loop, and the same current is passed through it. If new magnetic field at the center of the coil is \(B_{\mathcal{F}}\), then the ratio \(B_{L} / B_{C}\) is a. \(1: n^{2}\) b. \(n^{1 / 2}\) c. \(n: 1\) d. none of these

Two parallel wires carrying equal currents in opposite directions are placed at \(x=\pm a\) parallel to \(y\) -axis with \(z=0\). Magnetic field at origin \(O\) is \(B_{1}\) and at \(P(2 a, 0,0)\) is \(B_{2}\). Then, the ratio \(B_{1} / B_{2}\) is a. \(-3\) b. \(-\frac{1}{2}\) c. \(-\frac{1}{3}\) d. 2

A circular curren carrying coil has a radius \(R\). The distance from the center of the coil on the axis where the magnetic induction will be \((1 / 8)^{\mathrm{h}}\) of its value at the center of the coil, is a. \(R / \sqrt{3}\) b \(R \sqrt{3}\) c \(2 R \sqrt{3}\) d \((2 \sqrt{3}) R\)

Two particles \(X\) and \(Y\) having equal charges, after being accelerated through the same potential difference, enter a region of uniform magnetic field and describe circular paths of radii \(R_{1}\) and \(R_{2}\), respectively. The ratio of masses of \(X\) and \(Y\) is a. \(\left(R_{1} / R_{2}\right)^{1 / 2}\) b. \(\left(R_{2} / R_{1}\right)\) c. \(\left(R_{1} / R_{2}\right)^{2}\) d. \(\left(R_{1} / R_{2}\right)\)
See all solutions

Recommended explanations on Physics Textbooks

Atoms and Radioactivity

Read Explanation

Scientific Method Physics

Read Explanation

Further Mechanics and Thermal Physics

Read Explanation

Electromagnetism

Read Explanation

Fields in Physics

Read Explanation

Medical Physics

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.