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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, 蠅 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

A rectangular conducting loop with dimensions \(a\) and \(b\) and resistance \(R\), is placed in the \(x y\) -plane. A magnetic field of magnitude \(B\) passes through the loop. The magnetic field is in the positive \(z\) -direction and varies in time according to \(B=B_{0}\left(1+c_{1} t^{3}\right),\) where \(c_{1}\) is a constant with units of \(1 / \mathrm{s}^{3}\) What is the direction of the current induced in the loop, and what is its value at \(t=1 \mathrm{~s}\) (in terms of \(a, b, R, B_{0},\) and \(\left.c_{1}\right) ?\)

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You have a light bulb, a bar magnet, a spool of wire that you can cut into as many pieces as you want, and nothing else. How can you get the bulb to light up? a) You can't. The bulb needs electricity to light it, not magnetism. b) You cut a length of wire, connect the light bulb to the two ends of the wire, and pass the magnet through the loop that is formed. c) You cut two lengths of wire and connect the magnet and the bulb in series.

A square conducting loop with sides of length \(L\) is rotating at a constant angular speed, \(\omega\), in a uniform magnetic field of magnitude \(B\). At time \(t=0\), the loop is oriented so that the direction normal to the loop is aligned with the magnetic field. Find an expression for the potential difference induced in the loop as a function of time.

A circular coil of wire with 20 turns and a radius of \(40.0 \mathrm{~cm}\) is laying flat on a horizontal table as shown in the figure. There is a uniform magnetic field extending over the entire table with a magnitude of \(5.00 \mathrm{~T}\) and directed to the north and downward, making an angle of \(25.8^{\circ}\) with the horizontal. What is the magnitude of the magnetic flux through the coil?
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