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Chapter 29: Problem 17

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: \(V = \frac{2\pi}{3} B \omega R^3\)

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Most popular questions from this chapter

Having just learned that there is energy associated with magnetic fields, an inventor sets out to tap the energy associated with the Earth's magnetic field. What volume of space near Earth's surface contains \(1 \mathrm{~J}\) of energy, assuming the strength of the magnetic field to be \(5.0 \cdot 10^{-5} \mathrm{~T} ?\)

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An electromagnetic wave propagating in vacuum has electric and magnetic fields given by \(\vec{E}(\vec{x}, t)=\vec{E}_{0} \cos (\vec{k} \cdot \vec{x}-\omega t)\) and \(\vec{B}(\vec{x}, t)=\vec{B}_{0} \cos (\vec{k} \cdot \vec{x}-\omega t)\) where \(\vec{B}_{0}\) is given by \(\vec{B}_{0}=\vec{k} \times \vec{E}_{0} / \omega\) and the wave vector \(\vec{k}\) is perpendicular to both \(\vec{E}_{0}\) and \(\vec{B}_{0} .\) The magnitude of \(\vec{k}\) and the angular frequency \(\omega\) satisfy the dispersion relation, \(\omega /|\vec{k}|=\left(\mu_{0} \epsilon_{0}\right)^{-1 / 2},\) where \(\mu_{0}\) and \(\epsilon_{0}\) are the permeability and permittivity of free space, respectively. Such a wave transports energy in both its electric and magnetic fields. Calculate the ratio of the energy densities of the magnetic and electric fields, \(u_{B} / u_{E}\), in this wave. Simplify your final answer as much as possible.
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