/*! 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 17 A solid metal disk of radius \(R... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A solid metal disk of radius \(R\) is rotating around its center axis at a constant angular speed of \(\omega .\) The disk is in a uniform magnetic field of magnitude \(B\) that is oriented normal to the surface of the disk. Calculate the magnitude of the potential difference between the center of the disk and the outside edge.

Short Answer

Expert verified
Answer: The potential difference between the center and the outside edge of a rotating metal disk in a uniform magnetic field can be calculated using the formula \(V = \frac{2\pi}{3} B \omega R^3\), where V is the potential difference, B is the magnetic field strength, ω is the angular speed of the disk, and R is the radius of the disk.

Step by step solution

01

Understand the problem

The metal disk is rotating in a magnetic field, which will generate an induced EMF. We are asked to calculate the potential difference between the center and the outside edge of the disk.
02

Write down the expression of induced EMF

Faraday's law states that the induced EMF is equal to the rate of change of magnetic flux through a closed loop. Since the magnetic field is perpendicular to the disk's surface, we can take the circular rings as the closed loops. For a circular ring of radius r, the induced EMF is given by: \(\epsilon = -\frac{d\phi}{dt}\)
03

Find the magnetic flux

The magnetic flux through each ring is: \(\phi = B\cdot A = B\cdot 2\pi r dr\)
04

Calculate the rate of change of magnetic flux with respect to time

Differentiate the magnetic flux with respect to time, considering that the magnetic field is uniform and dr/dt is the linear speed of the ring, which can be expressed as \(v = r\omega\). We get: \(\frac{d\phi}{dt} = B \cdot 2\pi r \frac{dr}{dt} = B \cdot 2\pi r^2 \omega\)
05

Determine the induced EMF

Using Faraday's law, the induced EMF in the ring is: \(\epsilon = -\frac{d\phi}{dt} = -B \cdot 2\pi r^2 \omega\) Since the negative sign indicates the direction of the induced EMF, and we are interested in the magnitude, we can ignore it for further calculations.
06

Integrate the induced EMF to find the total potential difference

To calculate the total potential difference between the center of the disk (r=0) and the outside edge (r=R), we integrate the induced EMF from 0 to R: \(V = \int_0^R \epsilon dr = \int_0^R B \cdot 2\pi r^2 \omega dr = B \cdot 2\pi \omega \int_0^R r^2 dr\) \(V = B \cdot 2\pi \omega \left[\frac{1}{3} r^3\right]_0^R = B \cdot 2\pi \omega \cdot \frac{1}{3}R^3\)
07

Write down the final expression for potential difference

The magnitude of the potential difference between the center of the disk and the outside edge is: \(V = \frac{2\pi}{3} B \omega R^3\)

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 91Ó°ÊÓ!

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

Which of the following statements regarding self induction is correct? a) Self-induction occurs only when a direct current is flowing through a circuit. b) Self-induction occurs only when an alternating current is flowing through a circuit. c) Self-induction occurs when either a direct current or an alternating current is flowing through a circuit. d) Self-induction occurs when either a direct current or an alternating current is flowing through a circuit as long as the current is varying.

A popular demonstration of eddy currents involves dropping a magnet down a long metal tube and a long glass or plastic tube. As the magnet falls through a tube, there is changing flux as the magnet falls toward or away from each part of the tube. a) Which tube has the larger voltage induced in it? b) Which tube has the larger eddy currents induced in it?

A rectangular wire loop (dimensions of \(h=15.0 \mathrm{~cm}\) and \(w=8.00 \mathrm{~cm}\) ) with resistance \(R=5.00 \Omega\) is mounted on a door. The Earth's magnetic field, \(B_{\mathrm{E}}=2.6 \cdot 10^{-5} \mathrm{~T}\), is uniform and perpendicular to the surface of the closed door (the surface is in the \(x z\) -plane). At time \(t=0,\) the door is opened (right edge moves toward the \(y\) -axis) at a constant rate, with an opening angle of \(\theta(t)=\omega t,\) where \(\omega=3.5 \mathrm{rad} / \mathrm{s}\) Calculate the direction and the magnitude of the current induced in the loop, \(i(t=0.200 \mathrm{~s})\).

A long solenoid with length \(3.0 \mathrm{~m}\) and \(n=290\) turns \(/ \mathrm{m}\) carries a current of \(3.0 \mathrm{~A} .\) It stores \(2.8 \mathrm{~J}\) of energy. What is the cross-sectional area of the solenoid?

When a magnet in an MRI is abruptly shut down, the magnet is said to be quenched. Quenching can occur in as little as \(20.0 \mathrm{~s}\). Suppose a magnet with an initial field of \(1.20 \mathrm{~T}\) is quenched in \(20.0 \mathrm{~s},\) and the final field is approximately zero. Under these conditions, what is the average induced potential difference around a conducting loop of radius \(1.00 \mathrm{~cm}\) (about the size of a wedding ring) oriented perpendicular to the field?

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.