/*! 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 58 A plastic circular loop has radi... [FREE SOLUTION] | 91Ó°ÊÓ

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A plastic circular loop has radius \(R,\) and a positive charge \(q\) is distributed uniformly around the circumference of the loop. The loop is then rotated around its central axis, perpendicular to the plane of the loop, with angular speed \(\omega .\) If the loop is in a region where there is a uniform magnetic field \(\vec{B}\) directed parallel to the plane of the loop, calculate the magnitude of the magnetic torque on the loop.

Short Answer

Expert verified
The magnetic torque on the loop is \(\tau = \frac{q\omega R^2 B}{2}\)

Step by step solution

01

Calculate the Induced Current

Let's start by finding the current. In this case we have a circular loop with charge \(q\) rotating with an angular speed \(\omega\). This rotating charge is equivalent to a current \(I\), which is given by the the charge passing through a point in the loop per unit time. Therefore, we have \(I = \frac{q}{T}\) where \(T\) is the period of rotation. But we know that \(T = \frac{2\pi}{\omega}\). Thus, we get \(I = \frac{q\omega}{2\pi}\)
02

Calculate the Magnetic Moment

The magnetic moment \(\mu\) of the loop can be defined as the current through the loop times the area of the loop. So, we have \(\mu = I \cdot Area\). The loop is circular with radius \(R\), which means the area is \(\pi R^2\). Substituting our earlier found current into it, we get \(\mu = \frac{q\omega}{2\pi} \cdot \pi R^2 = \frac{q\omega R^2}{2}\)
03

Calculate the Magnetic Torque

The magnetic torque \(\tau\) on a loop is given by \(\tau = \mu \cdot B \cdot sin\theta\) where \(\theta\) is the angle between the magnetic moment and the field. However, given that the magnetic field is parallel to the plane of the loop (perpendicular to the magnetic moment), \(\theta=90^\circ\) and \(sin\theta = 1\). This gives us \(\tau = \mu \cdot B = \frac{q\omega R^2}{2} \cdot B\)

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Key Concepts

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

Induced Current
Understanding induced current is essential when examining the behavior of electrically charged objects in a magnetic field. When a charged object, like our plastic circular loop with radius \(R\), moves through a magnetic field or has a magnetic field passing through it, a current may be induced. In our example, the loop itself is not moving through the magnetic field, but the charge distributed on it is, due to the loop's rotation.

As the loop spins with an angular speed \(\omega\), the charges are continuously passing through a point in space, much like water flowing through a chute. This movement of charge constitutes an electric current. To clarify, the induced current \(I\) is effectively the charge \(q\) passing a given point during one full rotation, which is the period \(T\). By understanding the relationship between the angular speed and the period of rotation, \(T = \frac{2\pi}{\omega}\), we calculate the current as \(I = \frac{q}{T} = \frac{q\omega}{2\pi}\).

This understanding of induced current is crucial when dealing with rotating systems in magnetic fields and is the first step in explaining other phenomena, like magnetic torque and its effects on the system.
Magnetic Moment
The magnetic moment is a vector quantity fundamental to magnetism and electromagnetism. It represents the strength and direction of a magnetic source, and in the case of our spinning loop, it arises due to the induced current.

Defining the magnetic moment \(\mu\) as the product of the current \(I\) and the area enclosed by the current, we use the formula \(\mu = I \cdot Area\). For our circular loop, this area is the area of a circle, \(\pi R^2\), considering \(R\) is the radius. Using the current found earlier, the magnetic moment for our rotating loop is \(\mu = \frac{q\omega}{2\pi} \cdot \pi R^2 = \frac{q\omega R^2}{2}\).

This magnetic moment is aligned perpendicular to the plane of the loop due to the right-hand rule, which indicates the direction of magnetic moment vectors in relation to the direction of current. The magnitude of the magnetic moment affects the magnitude of other magnetic phenomena, such as torque, which we will examine next.
Angular Speed
Angular speed, denoted by \(\omega\), is a measure of how quickly an object rotates or spins around a fixed axis. It is essentially the rotational equivalent of linear speed. In our example, angular speed quantifies how fast the plastic circular loop rotates.

The angular speed has a direct impact on the induced current; the faster the loop spins, the greater the induced current and consequently the larger the magnetic moment. The formula we employ to find it is based on the relationship between the period of rotation \(T\) and angular speed, \(\omega\). Remember, \(T = \frac{2\pi}{\omega}\), where \(T\) is the time it takes to complete one full rotation. The angular speed provides a crucial piece of information in calculating both the induced current and, as we have seen, the magnetic moment - elements that directly influence the magnetic torque on the loop in the presence of a magnetic field.

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Most popular questions from this chapter

An alpha particle (a He nucleus, containing two protons and two neutrons and having a mass of \(6.64 \times 10^{-27} \mathrm{~kg}\) ) traveling horizontally at \(35.6 \mathrm{~km} / \mathrm{s}\) enters a uniform, vertical, \(1.80 \mathrm{~T}\) magnetic field. (a) What is the diameter of the path followed by this alpha particle? (b) What effect does the magnetic field have on the speed of the particle? (c) What are the magnitude and direction of the acceleration of the alpha particle while it is in the magnetic field? (d) Explain why the speed of the particle does not change even though an unbalanced external force acts on it.

A mass spectrograph is used to measure the masses of ions, or to separate ions of different masses (see Section 27.5 ). In one design for such an instrument, ions with mass \(m\) and charge \(q\) are accelerated through a potential difference \(V\). They then enter a uniform magnetic field that is perpendicular to their velocity, and they are deflected in a semicircular path of radius \(R .\) A detector measures where the ions complete the semicircle and from this it is easy to calculate \(R\). (a) Derive the equation for calculating the mass of the ion from measurements of \(B, V, R,\) and \(q\). (b) What potential difference \(V\) is needed so that singly ionized \({ }^{12} \mathrm{C}\) atoms will have \(R=50.0 \mathrm{~cm}\) in a 0.150 T magnetic field? (c) Suppose the beam consists of a mixture of \({ }^{12} \mathrm{C}\) and \({ }^{14} \mathrm{C}\) ions. If \(v\) and \(B\) have the same values as in part \((\mathrm{b}),\) calculate the separation of these two isotopes at the detector. Do you think that this beam separation is sufficient for the two ions to be distinguished? (Make the assumption described in Problem 27.53 for the masses of the ions.)

Magnetic Moment of the Hydrogen Atom. In the Bohr model of the hydrogen atom (see Section 39.3 ), in the lowest energy state the electron orbits the proton at a speed of \(2.2 \times 10^{6} \mathrm{~m} / \mathrm{s}\) in a circular orbit of radius \(5.3 \times 10^{-11} \mathrm{~m}\). (a) What is the orbital period of the electron? (b) If the orbiting electron is considered to be a current loop, what is the current \(I ?\) (c) What is the magnetic moment of the atom due to the motion of the electron?

A flat circular coil carrying a current of 8.80 A has a magnetic dipole moment of \(0.194 \mathrm{~A} \cdot \mathrm{m}^{2}\) to the left. Its area vector \(A\) is \(4.0 \mathrm{~cm}^{2}\) to the left. (a) How many turns does the coil have? (b) An observer is on the coil's axis to the left of the coil and is looking toward the coil. Does the observer see a clockwise or counterclockwise current? (c) If a huge \(45.0 \mathrm{~T}\) external magnetic field directed out of the paper is applied to the coil, what torque (magnitude and direction) results?

A circular area with a radius of \(6.50 \mathrm{~cm}\) lies in the \(x y\) -plane. What is the magnitude of the magnetic flux through this circle due to a uniform magnetic field \(B=0.230 \mathrm{~T}\) (a) in the \(+z\) -direction; (b) at an angle of \(53.1^{\circ}\) from the \(+z\) -direction; (c) in the \(+y\) -direction?

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