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Chapter 21: Problem 85

A charge of \(4.0 \times 10^{-6} \mathrm{C}\) is placed on a small conducting sphere that is located at the end of a thin insulating rod whose length is \(0.20 \mathrm{m}\). The rod rotates with an angular speed of \(\omega=150 \mathrm{rad} / \mathrm{s}\) about an axis that passes perpendicularly through its other end. Find the magnetic moment of the rotating charge. (Hint: The charge travels around a circle in a time equal to the period of the motion.)

Short Answer

Expert verified
The magnetic moment is approximately \(1.20 \times 10^{-5} \mathrm{A\cdot m^2}\."

Step by step solution

01

Determine the Period of Rotation

To find the period (T) of the motion, we use the relation between angular speed \(\omega\) and period: \( T = \frac{2\pi}{\omega} \). Given \( \omega = 150 \, \mathrm{rad/s} \), the period T is calculated as follows:\[ T = \frac{2\pi}{150} \approx 0.0419 \, \mathrm{s} \]
02

Determine the Current Equivalent

The charge \( Q = 4.0 \times 10^{-6} \, \mathrm{C} \) completes one full revolution in time \( T \). The current \( I \) is expressed as charge divided by time, \( I = \frac{Q}{T} \). Substituting the values, we have:\[ I = \frac{4.0 \times 10^{-6}}{0.0419} \approx 9.55 \times 10^{-5} \, \mathrm{A} \]
03

Magnetic Moment Formula

The magnetic moment \( \mu \) for a rotating charge is given by the formula \( \mu = I \cdot A \), where \( A \) is the area of the circle described by the rotating charge. Since the radius \( r \) of the circle is equal to the rod's length (0.20 m), the area \( A \) is \( \pi \cdot r^2 \). So, \( A = \pi \cdot (0.20)^2 \approx 0.1257 \, \mathrm{m^2} \).
04

Calculate the Magnetic Moment

Now, substitute the current \( I \) and area \( A \) into the formula for magnetic moment:\[ \mu = 9.55 \times 10^{-5} \times 0.1257 \approx 1.20 \times 10^{-5} \, \mathrm{A\cdot m^2} \]
05

Conclusion

The calculated magnetic moment of the rotating charge is \( \mu \approx 1.20 \times 10^{-5} \, \mathrm{A\cdot m^2} \). This describes the strength and the orientation of the magnetic field due to the charge's motion.

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Key Concepts

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

rotating charges
When charges move in a circular path, they form a loop of current, similar to a tiny circuit. This motion of charges is crucial because it generates a magnetic field. This is the fundamental idea behind rotating charges.
When a charge rotates, it mimics the behavior of a loop of current, creating what is known as a magnetic moment. The magnetic moment gives us a measure of how strong the magnetic effect is. It depends on the speed at which the charge moves and the size of the loop it traces.
You can think of it like a tiny magnet created by the charge itself. Thus, rotating charges help us understand the principles behind electromagnetism and are essential for technologies that depend on magnetic fields.
  • The charge moves along the outer edge of a circle (the path it follows).
  • Its rotation speed and the size of the circle decide the magnetic moment it produces.
  • In this example, the charge is attached to a rod, moving predictably around a central point.
angular speed
Angular speed, represented by the symbol \( \omega \), is a way of describing how fast something moves around in a circle. It's measured in radians per second (rad/s), which tells us how much of the circle's total angle the object covers in a second.
Imagine it as the speedometer for rotation! For a full rotation, an object travels through an angle of \( 2\pi \) radians. If it takes one second to complete this, the angular speed is \( 2\pi \) rad/s.
In our example, the angular speed of \( 150 \) rad/s indicates that the charge completes the circle more than 20 times in just one second.
  • Angular speed is pivotal for computing the period of motion, relating rotation to time.
  • Higher angular speed means quicker rotations.
  • It is used to assess how swiftly the charge completes one revolution.
period of motion
The period of motion is the time it takes for the rotating charge to make one full circle around its path. It's related to angular speed through a simple formula: \( T = \frac{2\pi}{\omega} \). This relationship highlights how speed and time are inversely linked.
Our exercise shows us that with an angular speed of \( 150 \) rad/s, the period is approximately \( 0.0419 \) seconds. This short timeframe indicates a rapid motion!
  • The period is how long it takes to go completely around the circle once.
  • Finding the period helps determine how often the charge returns to its starting point.
  • It is an essential step in computing the current and hence the magnetic moment.
area of a circle
The area of a circle, symbolized as \( A \), is crucial in understanding how the rotating charge creates a magnetic moment. The formula to find the area of a circle is \( A = \pi r^2 \), where \( r \) is the radius, or the distance from the circle's center to its edge.
In our case, with the radius being the length of the rod, \( 0.20 \) meters, the area works out to approximately \( 0.1257 \) square meters. This area acts like the loop through which the charge moves, influencing the strength of the generated magnetic field.
  • The area helps compute the magnetic moment by providing the size of the field-creating loop.
  • The larger the area, the more substantial the potential magnetic effect.
  • This is why the radius is squared in the formula, making it a crucial factor in determining the circle's size.

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Most popular questions from this chapter

In the operating room, anesthesiologists use mass spectrometers to monitor the respiratory gases of patients undergoing surgery. One gas that is often monitored is the anesthetic isoflurane (molecular mass \(=3.06 \times 10^{-25} \mathrm{kg}\) ). In a spectrometer, a singly ionized molecule of isoflurane (charge \(=+e)\) moves at a speed of \(7.2 \times 10^{3} \mathrm{m} / \mathrm{s}\) on a circular path that has a radius of \(0.10 \mathrm{m} .\) What is the magnitude of the magnetic field that the spectrometer uses?

At New York City, the earth's magnetic field has a vertical component of \(5.2 \times 10^{-5} \mathrm{T}\) that points downward (perpendicular to the ground) and a horizontal component of \(1.8 \times 10^{-5} \mathrm{T}\) that points toward geographic north (parallel to the ground). What are the magnitude and direction of the magnetic force on a \(6.0-\mathrm{m}\) long, straight wire that carries a current of 28 A perpendicularly into the ground?

Suppose that an ion source in a mass spectrometer produces doubly ionized gold ions \(\left(\mathrm{Au}^{2+}\right),\) each with a mass of \(3.27 \times 10^{-25} \mathrm{kg} .\) The ions are accelerated from rest through a potential difference of \(1.00 \mathrm{kV}\). Then, a 0.500-T magnetic field causes the ions to follow a circular path. Determine the radius of the path.

An electron is moving through a magnetic field whose magnitude is \(8.70 \times 10^{-4} \mathrm{T}\). The electron experiences only a magnetic force and has an acceleration of magnitude \(3.50 \times 10^{14} \mathrm{m} / \mathrm{s}^{2} .\) At a certain instant, it has a speed of \(6.80 \times 10^{6} \mathrm{m} / \mathrm{s}\). Determine the angle \(\theta\) (less than \(90^{\circ}\) ) between the electron's velocity and the magnetic field.

A charge of \(-8.3 \mu \mathrm{C}\) is traveling at a speed of \(7.4 \times 10^{6} \mathrm{m} / \mathrm{s}\) in a region of space where there is a magnetic field. The angle between the velocity of the charge and the field is \(52^{\circ} .\) A force of magnitude \(5.4 \times 10^{-3} \mathrm{N}\) acts on the charge. What is the magnitude of the magnetic field?
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