Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

A wedding ring (of diameter \(1.95 \mathrm{~cm}\) ) is tossed into the air and given a spin, resulting in an angular velocity of 13.3 rev/s. The rotation axis is a diameter of the ring. If the magnitude of the Earth's magnetic field at the ring's location is \(4.77 \cdot 10^{-5} \mathrm{~T}\), what is the maximum induced potential difference in the ring?

Short Answer

Expert verified
Answer: The maximum induced potential difference in the wedding ring is approximately \(6.0\times10^{-6}\) V.

Step by step solution

01

Find the area of the ring

The diameter of the ring is given as \(1.95 \mathrm{~cm}\), so we can find the radius by dividing this by 2: $$r = \frac{1.95}{2} = 0.975\ \mathrm{cm}$$ This should be converted to meters to match the unit of the magnetic field: $$r = 0.975\ \mathrm{cm} \times \frac{1\ \mathrm{m}}{100\ \mathrm{cm}} = 0.00975\ \mathrm{m}$$ Now we can find the area A of the ring: $$A = \pi r^2 = \pi(0.00975\ \mathrm{m})^2 \approx 2.98 × 10^{-4}\mathrm{~m}^2$$
02

Find the linear velocity of the ring

The angular velocity is given as 13.3 rev/s which can be converted to radians per second using the following relationship, with \(2\pi\) radians per revolution: $$\omega = 13.3\ \mathrm{rev/s} \times \frac{2\pi \mathrm{~rad}}{\mathrm{rev}} \approx 83.65\ \mathrm{rad/s}$$ Now, we can find the linear velocity using the formula \(v = r \omega\): $$v = (0.00975\ \mathrm{m})(83.65\ \mathrm{rad/s}) \approx 0.815\ \mathrm{m/s}$$
03

Determine the change in magnetic flux

The change in magnetic flux through the ring is given by the product of the area, magnetic field, and the component of linear velocity perpendicular to the magnetic field. Since the magnetic field is aligned along the plane of the ring in this case, the maximum change in magnetic flux occurs when the velocity vector is perpendicular to the plane of the ring. Thus, we have: $$\Delta \Phi = BA\sin \theta = BA\sin 90^\circ = BA$$ where \(\theta = 90^\circ\) (perpendicular to the magnetic field) for the maximum change in magnetic flux.
04

Calculate the maximum induced potential difference

We can now use Faraday's law of electromagnetic induction to find the maximum induced potential difference in the ring. Faraday's law states that the induced potential difference is equal to the negative rate of change of magnetic flux through the ring: $$\epsilon = -\frac{d \Phi}{d t}$$. Since we are calculating the maximum induced potential difference, we can ignore the negative sign. Now we have: $$\epsilon_\mathrm{max} = \frac{\Delta \Phi}{\Delta t}$$ Since the linear velocity, v is constant, we can write \(\Delta t\) as the time taken for the ring to move a distance equal to its diameter, which is: $$\Delta t = \frac{2r}{v} = \frac{2(0.00975\ \mathrm{m})}{0.815\ \mathrm{m/s}} \approx 0.0239\ \mathrm{s}$$ Now we find the maximum induced potential difference: $$\epsilon_\mathrm{max} = \frac{BA}{\Delta t} = \frac{(4.77\times10^{-5}\ \mathrm{T})(2.98\times10^{-4}\ \mathrm{m}^2)}{0.0239\ \mathrm{s}} \approx 6.0\times10^{-6}\ \mathrm{V}$$ So, the maximum induced potential difference in the wedding ring is approximately \(6.0\times10^{-6}\) V.

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 Vaia!

Key Concepts

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

Faraday's Law of Electromagnetic Induction
Faraday's law of electromagnetic induction is a fundamental principle that explains how electric currents are induced in conductors as a result of changing magnetic fields. It states that the induced electromotive force (EMF) in any closed circuit is equal to the negative of the time rate of change of the magnetic flux through the circuit. In simpler terms, whenever the magnetic environment of a circuit changes, an EMF is generated.
When we consider a spinning ring tossed into the air, Faraday's law is what determines the induced potential difference (voltage) in the ring as it passes through the Earth's magnetic field. The key point to understand here is the '-dPhi/dt' part of the law, representing the change in magnetic flux over time. The minus sign indicates the direction of the induced EMF (as per Lenz's law), which creates a current that opposes the change in flux — a topic that has deep implications in understanding electromagnetic fields.
For practical uses, when we calculate induced EMF like in the wedding ring's example, and we're interested in the magnitude rather than the direction, we can disregard the negative sign, focusing on the absolute value of the EMF produced.
Angular Velocity
Angular velocity is a measure of the rate at which an object rotates around a specified axis. It is typically expressed in radians per second (rad/s) or revolutions per second (rev/s). For example, when the wedding ring spins with an angular velocity of 13.3 rev/s, we're seeing how many times it completes a full 360-degree spin each second. To connect this to linear velocity or to calculate changes in the ring's position, we need to convert revolutions per second to radians per second since radians are the standard unit for angular measurements in physics.
The formula to convert revolutions per second to rad/s is
\begin{center}\( \text{{angular velocity in rad/s}} = \text{{angular velocity in rev/s}} \times 2\text{{π rad/rev}} \).\br>\br>\br>When dealing with induced EMF problems, understanding angular velocity is crucial because it helps us to calculate the linear speed at which the ring cuts through the magnetic flux lines, this being a key factor in Faraday's law.
Magnetic Flux
Magnetic flux represents the quantity of the magnetic field that passes through a given area. It is mathematically defined as the product of the magnetic field strength, the area through which the magnetic field lines pass, and the cosine of the angle between the magnetic field lines and the normal (perpendicular line) to the area's surface. The SI unit for magnetic flux is the weber (Wb).
In the context of our spinning wedding ring, the maximum change in magnetic flux occurs when the area of the ring is aligned perpendicularly to the magnetic field, like when the ring is flat parallel to the ground and moves through the Earth's magnetic field vertically. Under these conditions, we calculate the change in magnetic flux simply as the product of the area and the magnetic field strength because the cosine of 90 degrees (cos(90°)) is equal to 1.
The calculation of the change in magnetic flux is key to determining the induced potential difference through Faraday's law, as a greater change in flux over a short period will result in a higher EMF.
Ring Area Calculation
The area of a ring, which in physics is often treated as a simple circle, is found using the formula \(A = \text{π} \times r^2\), where \(r\) is the radius of the circle. In order to calculate accurately, all units must be consistent, so if the magnetic field is given in teslas, a unit of meters squared (m²) for area will be compatible.
After converting the diameter of the wedding ring to radius and then to meters, we utilized this formula to determine the area through which the magnetic flux will pass. Correctly calculating the area is essential to finding out the induced potential difference because if this initial value is incorrect, all subsequent calculations would also be incorrect, leading to an inaccurate determination of the potential difference. Thus, precision in fundamental aspects like area calculation is vital for reaching the right answer in physics problems.

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 long solenoid with cross-sectional area \(A_{1}\) surrounds another long solenoid with cross-sectional area \(A_{2}

A wedding ring is tossed into the air and given a spin, resulting in an angular velocity of 13.5 rev/s. The rotation axis is a diameter of the ring. The magnitude of the Earth's magnetic field is \(4.97 \cdot 10^{-5} \mathrm{~T}\) at the ring's location. If the maximum induced voltage in the ring is \(1.446 \cdot 10^{-6} \mathrm{~V},\) what is the diameter of the ring?

A metal loop has an area of \(0.100 \mathrm{~m}^{2}\) and is placed flat on the ground. There is a uniform magnetic field pointing due west, as shown in the figure. This magnetic field initially has a magnitude of \(0.123 \mathrm{~T}\) which decreases steadily to \(0.075 \mathrm{~T}\) during a period of \(0.579 \mathrm{~s}\). Find the potential difference induced in the loop during this time.

A solenoid with 200 turns and a cross-sectional area of \(60 \mathrm{~cm}^{2}\) has a magnetic field of \(0.60 \mathrm{~T}\) along its axis. If the field is confined within the solenoid and changes at a rate of \(0.20 \mathrm{~T} / \mathrm{s}\), the magnitude of the induced potential difference in the solenoid will be a) \(0.0020 \mathrm{~V}\). b) \(0.02 \mathrm{~V}\). c) \(0.001 \mathrm{~V}\). d) \(0.24 \mathrm{~V}\).

Faraday's Law of Induction states that a) a potential difference is induced in a loop when there is a change in the magnetic flux through the loop. b) the current induced in a loop by a changing magnetic field produces a magnetic field that opposes this change in magnetic field. c) a changing magnetic field induces an electric field. d) the inductance of a device is a measure of its opposition to changes in current flowing through it. e) magnetic flux is the product of the average magnetic field and the area perpendicular to it that it penetrates.

See all solutions

Recommended explanations on Physics Textbooks

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.

Sign-up for free