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Chapter 32: Problem 40

You place a magnetic compass on a horizontal surface, allow the needle to settle, and then give the compass a gentle wiggle to cause the needle to oscillate about its equilibrium position. The oscillation frequency is \(0.312 \mathrm{~Hz}\). Earth's magnetic field at the location of the compass has a horizontal component of \(18.0 \mu \mathrm{T}\). The needle has a magnetic moment of \(0.760 \mathrm{~mJ} / \mathrm{T}\). What is the needle's rotational inertia about its (vertical) axis of rotation?

Short Answer

Expert verified
The needle's rotational inertia is approximately \(3.55 \times 10^{-9} \mathrm{kg} \cdot \mathrm{m}^2\).

Step by step solution

01

Understand the Problem

We are asked to find the rotational inertia of the compass needle about its vertical axis of rotation. We know the frequency of oscillation, the Earth's magnetic field at the location, and the needle's magnetic moment.
02

Formula for Oscillation Frequency

The formula that relates the frequency of oscillation to the magnetic moment, magnetic field, and rotational inertia is \( f = \frac{1}{2\pi} \sqrt{\frac{mB}{I}} \), where \( f \) is the frequency, \( m \) is the magnetic moment of the needle, \( B \) is the horizontal component of Earth's magnetic field, and \( I \) is the rotational inertia.
03

Rearrange the Formula

To find the rotational inertia \( I \), rearrange the formula: \( I = \frac{mB}{(2\pi f)^2} \). This equation will allow us to calculate \( I \) using the given values.
04

Substitute Known Values

Substitute the given values into the rearranged formula: \( m = 0.760\, \mathrm{mJ}/\mathrm{T} \), \( B = 18.0 \times 10^{-6} \mathrm{T} \), and \( f = 0.312 \mathrm{Hz} \). So, the expression becomes: \( I = \frac{0.760 \times 18.0 \times 10^{-6}}{(2\pi \times 0.312)^2} \).
05

Calculate Rotational Inertia

First, calculate the denominator: \((2\pi \times 0.312)^2\). Then, compute the entire expression to find \( I \). After calculation: \( I = \frac{0.760 \times 18.0 \times 10^{-6}}{(1.963)^2} \).
06

Final Calculation and Result

Calculate the actual number: \( 1.963^2 = 3.853 \). Then \( I = \frac{0.760 \times 18.0 \times 10^{-6}}{3.853} = \frac{0.01368 \times 10^{-6}}{3.853} \approx 3.55 \times 10^{-9} \mathrm{kg} \cdot \mathrm{m}^2 \).

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Key Concepts

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

Magnetic Moment
In this exercise involving a compass needle, the magnetic moment is a pivotal concept. The magnetic moment essentially measures the strength and direction of a magnet's ability to produce or be affected by magnetic forces. Here, the magnetic moment of the needle is given as \(0.760\, \mathrm{mJ}/\mathrm{T}\).

This value tells us how the needle will align itself with a magnetic field, like Earth's.
  • Definition: The magnetic moment \(m\) is a vector quantity, representing the magnetic strength and orientation of a magnet or other object that produces a magnetic field.
  • Impact: The larger the magnetic moment, the stronger the needle's response to a magnetic field, such as Earth's.
When a needle with a magnetic moment is exposed to a magnetic field, it experiences a torque, causing it to align with the field. This torque influences the needle's oscillation as described by the exercise.
Understanding the relationship between the magnetic moment, torque, and resulting motion is crucial for analyzing oscillations in magnetic fields.
Oscillation Frequency
Oscillation frequency is key to understanding the movement of the compass needle in this exercise. It represents how many oscillations happen per second as the needle swings back and forth around its equilibrium position, influenced by Earth's magnetic field. Here, the frequency is \(0.312\, \text{Hz}\).

Mathematically, the frequency \(f\) can be determined by the equation \( f = \frac{1}{2\pi} \sqrt{\frac{mB}{I}} \).
  • Frequency \(f\): Measured in Hertz (Hz), it indicates the number of complete oscillations per second.
  • Relates to Rotational Inertia \(I\): The frequency helps us calculate the needle's rotational inertia, as seen in the solution steps.
This frequency is tied to both the needle's magnetic moment and Earth's horizontal magnetic field component. As the needle oscillates due to torque, its behavior is dictated by these quantities.
When approaching problems involving oscillation, it is crucial to understand the role each variable plays, as they collectively determine the frequency and nature of motion.
Earth's Magnetic Field
Earth's magnetic field plays a significant role in this exercise, as it influences how the compass needle behaves. The horizontal component of Earth's magnetic field at the compass location is given as \(18.0\, \mu\mathrm{T}\), which affects the torque exerted on the needle.

Earth's magnetic field is a vast magnetic field generated deep within the planet's core.
  • Horizontal Component \(B\): Refers to the component of Earth's magnetic field that lies parallel to the Earth's surface, affecting how the needle aligns.
  • Relationship with Magnetic Moment: The magnetic moment interacts with this field, causing the needle to rotate and align along the field lines.
The interaction between the needle's magnetic moment and Earth's magnetic field results in a torque that leads to oscillation. This interaction is crucial for determining the needle's movement and its rotational inertia. It is essential to recognize how Earth's magnetic properties directly influence such systems, helping to predict and understand the behavior of magnetic needles and compasses.
When analyzing compass behavior, recognizing how Earth's magnetic properties affect the system is crucial.

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Most popular questions from this chapter

A capacitor with parallel circular plates of radius R 1.20 cm is discharging via a current of 12.0 A. Consider a loop of radius R/3 that is centered on the central axis between the plates. (a) How much displacement current is encircled by the loop? The maximum induced magnetic field has a magnitude of 12.0 mT. At what radius (b) inside and (c) outside the capacitor gap is the magnitude of the induced magnetic field 6.00 mT?

Measurements in mines and boreholes indicate that Earth's interior temperature increases with depth at the average rate of \(30 \mathrm{C}^{\circ} / \mathrm{km}\). Assuming a surface temperature of \(10^{\circ} \mathrm{C}\), at what depth does iron cease to be ferromagnetic? (The Curie temperature of iron varies very little with pressure.)

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A silver wire has resistivity \(\rho=1.62 \times 10^{-3} \mathrm{R} \cdot \mathrm{m}\) and a crosssectional area of \(5.00 \mathrm{~mm}^{2}\). The current in the wire is uniform and changing at the rate of 2000 A/s when the current is \(54.0 \mathrm{~A}\). (a) What is the magnitude of the (uniform) electric field in the wire when the current in the wire is \(100 \mathrm{~A}\) ? (b) What is the displacement current in the wire at that time? (c) What is the ratio of the magnitude of the magnetic field due to the displacement current to that due to the current at a distance \(r\) from the wire?

The magnitude of the magnetic dipole moment of Earth is 8.0 1022 J/T. (a) If the origin of this magnetism were a magnetized iron sphere at the center of Earth, what would be its radius? (b) What fraction of the volume of Earth would such a sphere occupy? Assume complete alignment of the dipoles. The density of Earth鈥檚 inner core is 14 g/cm3 . The magnetic dipole moment of an iron atom is 2.1 1023 J/T. (Note: Earth鈥檚 inner core is in fact thought to be in both liquid and solid forms and partly iron, but a permanent magnet as the source of Earth鈥檚 magnetism has been ruled out by several considerations. For one, the temperature is certainly above the Curie point.)
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