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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.

Short Answer

Expert verified
The magnetic moment is \( 1.2 \times 10^{-5} \text{ A} \cdot \text{m}^2 \).

Step by step solution

01

Understanding the Concept

The magnetic moment \( \mu \) of a rotating charge is analogous to the magnetic moment of a current loop. When charge \( q \) moves in a circular path of radius \( r \), it creates an effective current \( I \), and the magnetic moment is given by \( \mu = I \times A \), where \( A \) is the area of the circular path.
02

Finding the Angular Relationship

Since the charge is moving in a circle, identify the radius of the circle as the length of the rod, \( r = 0.20 \text{ m} \). The charge travels around the circle with angular speed \( \omega = 150 \text{ rad/s} \). The speed \( v \) of the charge is related to the angular speed by \( v = r \cdot \omega \).
03

Calculating the Speed of the Charge

Calculate the speed \( v \) of the charge using the relationship \( v = r \cdot \omega \): \[ v = 0.20 \times 150 = 30 \text{ m/s} \]
04

Determining the Period of Motion

The period \( T \) is the time taken for the charge to complete one full circle. It is related to speed \( v \) and circumference \( 2\pi r \) by \( T = \frac{2\pi r}{v} \).
05

Finding the Period

Calculate the period \( T \) using the formula:\[ T = \frac{2\pi \times 0.20}{30} = \frac{0.4\pi}{30} \]\[ T = \frac{\pi}{75} \text{ seconds} \]
06

Calculating the Equivalent Current

The effective current \( I \) is the charge \( q \) divided by the period \( T \): \[ I = \frac{q}{T} = \frac{4.0 \times 10^{-6}}{\pi/75} = \frac{4.0 \times 10^{-6} \times 75}{\pi} \] \[ I = \frac{300 \times 10^{-6}}{\pi} \text{ Amperes} \]
07

Area of the Circle

The area \( A \) of the circle is given by \( A = \pi r^2 \). Substitute \( r = 0.20 \text{ m} \): \[ A = \pi \times (0.20)^2 = \pi \times 0.04 = 0.04\pi \text{ m}^2 \]
08

Calculating the Magnetic Moment

Substitute \( I \) and \( A \) into the formula for magnetic moment \( \mu \):\[ \mu = I \times A = \frac{300 \times 10^{-6}}{\pi} \times 0.04\pi \] \[ \mu = 300 \times 10^{-6} \times 0.04 = 12 \times 10^{-6} \] \[ \mu = 1.2 \times 10^{-5} \text{ A} \cdot \text{m}^2 \]
09

Result Interpretation

Therefore, the magnetic moment of the rotating charge is \( 1.2 \times 10^{-5} \text{ A} \cdot \text{m}^2 \).

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Key Concepts

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

Rotating Charge
A rotating charge refers to a situation where a charged particle is moving along a circular path. In physics, when such motion occurs, it can produce a magnetic field similar to a loop of current. This is because, as the charge moves in a circle, it effectively acts like a loop of electric current. Imagine tying a small charge to the edge of a spinning wheel. If the wheel spins steadily, the charge rotates around a fixed center point. This movement is essential for calculating properties like the magnetic moment.
The exercise involves a charge of 4.0 × 10^-6 C on a conducting sphere, which is attached to a rod, rotating around an axis at the speed of 150 rad/s. Here, the path traced by the charge is a circle with a radius equal to the length of the rod, 0.20 m. These elements combine to form a current-like behavior in terms of physics, leading us to the concept of magnetic moment calculation.
Angular Speed
Angular speed (\( \omega \)) characterizes how fast an object rotates or revolves around a central point. It is typically measured in radians per second (\( \ ext{rad/s} \)). For a body rotating in a circle, like the charge in the problem, angular speed defines how quickly it moves along its circular path.
This concept plays a crucial role in understanding the behavior of the rotating charge. Given \( \omega = 150 \ ext{ rad/s} \), this tells us how many radians the charge covers each second while performing its circular movement.
  • If you think of the circle as a clock, a complete rotation would be \( 2\pi \) radians (since a circle has \( 360^{\circ} \), and in radians this is \( 2\pi \)).
  • Given this angular speed, you can calculate the time it takes for one full revolution by determining the period (T).
This relationship between angular speed, period, and radius is vital for computing other related phenomena, such as linear speed and magnetic moments.
Current Loop
In physics, a current loop is a concept where a steady flow of electric charge circulates in a closed path or loop. This creates a magnetic field through a principle known as Ampere's Law. Here, the rotating charge effectively simulates such a current loop.
To visualize, imagine the charged sphere at the tip of our 0.20 m rod. As it spins, the motion is akin to the charges moving through a circular wire. The speed and frequency of these rotations develop an equivalent current, even though there is no actual wire.
  • The equivalent current (\( I \)) is calculated using the formula \( I = \frac{q}{T} \), where \( q \) is the charge and \( T \) is the period.
  • The charge completes one rotation per period, creating a complete current loop effect.
This creates the foundation for further calculations of the magnetic moment, as this imaginary 'loop' mimics the effect of real current loops in magnetic field generation.
Magnetic Moment Calculation
The magnetic moment is a measure of the strength of a magnetic source. For our rotating charge, the magnetic moment (\( \mu \)) can be computed by considering the circle as both an area and a current source. The key formula here is \( \mu = I \times A \), where \( I \) is the effective current and \( A \) is the area of the circle.
The area of the circle is calculated as \( A = \pi r^{2} \). Using the formula, substitute \( r = 0.20 \ ext{ m} \):
  • Find area \( A = \pi \times 0.20^2 = 0.04 \pi \ ext{ m}^{2} \).
  • The effective current \( I \) comes from charge divided by the computed period \( T \).
  • \( \mu \) is found by substituting \( I \) and \( A \) into the magnetic moment formula.
Thus, by combining these calculations, we determine the magnetic moment of the rotating charge in the exercise as \( \mu = 1.2 \times 10^{-5} \ ext{ A} \cdot \ ext{m}^{2} \). This represents how effectively the rotating system simulates a magnetic field.

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Most popular questions from this chapter

A wire of length \(0.655 \mathrm{~m}\) carries a current of \(21.0 \mathrm{~A}\). In the presence of a \(0.470-\mathrm{T}\) magnetic field, the wire experiences a force of \(5.46 \mathrm{~N}\). What is the angle (less than \(90^{\circ}\) ) between the wire and the magnetic field?

The \(x, y,\) and \(z\) components of a magnetic field are \(B_{x}=0.10 \mathrm{~T}, B_{y}=0.15 \mathrm{~T},\) and \(B_{z}=0.17 \mathrm{~T}\). A \(25-\mathrm{cm}\) wire is oriented along the \(z\) axis and carries a current of \(4.3 \mathrm{~A}\) What is the magnitude of the magnetic force that acts on this wire?

Each of these problems consists of Concept Questions followed by a related quantitative Problem. The Concept Questions involve little or no mathematics. They focus on the concepts with which the problems deal. Recognizing the concepts is the essential initial step in any problem-solving technique. Concept Questions Particle 1 and particle 2 carry the same charge \(q,\) but particle has a smaller mass than particle \(2 .\) These two particles accelerate from rest through the same electric potential difference \(V\) and enter the same magnetic field, which has a magnitude \(B\). The particles travel perpendicular to the field on circular paths. Upon entering the field region, which particle, if either, has the greater (a) kinetic energy and (b) speed? Give your reasoning. Problem The masses of the particles are \(m_{1}=2.3 \times 10^{-8} \mathrm{~kg}\) and \(m_{2}=5.9 \times 10^{-8} \mathrm{~kg} .\) The radius of the circular path for particle 1 is \(r_{1}=12 \mathrm{~cm} .\) What is the radius of the circular path for particle \(2 ?\)

A charged particle with a charge-to-mass ratio of \(|q| / m=5.7 \times 10^{8} \mathrm{C} / \mathrm{kg}\) travels on a circular path that is perpendicular to a magnetic field whose magnitude is \(0.72 \mathrm{~T}\). How much time does it take for the particle to complete one revolution?

The 1200 -turn coil in a dc motor has an area per turn of \(1.1 \times 10^{-2} \mathrm{~m}^{2}\). The design for the motor specifies that the magnitude of the maximum torque is \(5.8 \mathrm{~N} \cdot \mathrm{m}\) when the coil is placed in a 0.20 -T magnetic field. What is the current in the coil?

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