Figure 28-56 shows a homopolar generator, which has a solid...

Figure $28-56$ shows a homopolar generator, which has a solid conducting disk as rotor and which is rotated by a motor (not shown). Conducting brushes connect this emf device to a circuit through which the device drives current. The device can produce a greater emf than wire loop rotors because they can spin at a much higher angular speed without rupturing. The disk has radius $R=$ $0.250 \mathrm{~m}$ and rotation frequency $f=4000 \mathrm{~Hz}$, and the device is in a uniform magnetic field of magnitude $B=60.0 \mathrm{mT}$ that is perpendicular to the disk. As the disk is rotated, conduction electrons along the conducting path (dashed line) are forced to move through the magnetic field. (a) For the indicated rotation, is the magnetic force on those electrons up or down in the figure? (b) Is the magnitude of that force greater at the rim or near the center of the disk? (c) What is the work per unit charge done by that force in moving charge along the radial line, between the rim and the center? (d) What, then, is the emf of the device? (e) If the current is $50.0 \mathrm{~A}$, what is the power at which electrical energy is being produced?

Figure $28-56$ shows a homopolar generator, which has a solid conducting disk as rotor and which is rotated by a motor (not shown). Conducting brushes connect this emf device to a circuit through which the device drives current. The device can produce a greater emf than wire loop rotors because they can spin at a much higher angular speed without rupturing. The disk has radius $R=$ $0.250 \mathrm{~m}$ and rotation frequency $f=4000 \mathrm{~Hz}$, and the device is in a uniform magnetic field of magnitude $B=60.0 \mathrm{mT}$ that is perpendicular to the disk. As the disk is rotated, conduction electrons along the conducting path (dashed line) are forced to move through the magnetic field. (a) For the indicated rotation, is the magnetic force on those electrons up or down in the figure? (b) Is the magnitude of that force greater at the rim or near the center of the disk?
(c) What is the work per unit charge done by that force in moving charge along the radial line, between the rim and the center?
(d) What, then, is the emf of the device? (e) If the current is $50.0 \mathrm{~A}$, what is the power at which electrical energy is being produced?

Key Concepts

Lorentz Force

This concept describes the force experienced by a moving charged particle in a magnetic field. The force is given by the cross product of the particle’s velocity and the magnetic field vector, which results in a direction perpendicular to both. It is crucial for understanding how the motion of conduction electrons in a moving conductor produces a force that is directed either up or down, depending on the orientation of the velocity and magnetic field.

Motional Electromotive Force (emf)

When a conductor moves through a magnetic field, an emf is induced along its length because the magnetic force does work on the charges. This process, which involves the conversion of mechanical energy into electrical energy, is fundamental to the operation of devices like homopolar generators. Calculating the emf involves integrating the induced electric field (related to the Lorentz force) along the path of the moving charge.

Homopolar Generator

A homopolar generator is a type of direct current generator that consists of a rotating conductive disk or rotor exposed to a perpendicular magnetic field. It produces a constant potential difference (emf) between its center and edge due to the continuous separation of charges by the Lorentz force. Its design allows high angular speeds, enabling the generation of significant emf values without mechanical failure.

Rotational Kinematics

This concept links angular motion to linear motion and is used to relate the angular speed of the rotating disk to the tangential velocity of charges at various radii. In the context of the generator, the tangential speed determines the magnitude of the Lorentz force acting on the electrons, which is proportional to the distance from the center of rotation.

Work per Unit Charge and Energy Conversion

The work done per unit charge moving through an electric potential difference is central to the concept of emf. It represents the energy per charge that is converted from mechanical work into electrical energy as the electrons travel along a path in the magnetic field. This is derived by integrating the force over the displacement, which is key to determining the generated voltage.

Electrical Power

Electrical power in this context is the rate at which electrical energy is produced from the mechanical energy input into the system. It is calculated by multiplying the emf by the current in the circuit. This concept is essential for understanding the efficiency and performance of energy conversion devices like the homopolar generator.

Transcript

00:01 This is the solution to problem 87 in chapter 28 in the 10th volume of fundamentals of physics.

00:09 So in this problem, i have a device, i'm going to show you to on the next page, called a homopolar generator, and it rotates through a constant magnetic field whose magnitude of 60 milletessla.

00:23 And as it rotates, it creates a current that flows in the direction shown in the figure, and i'll show you that in a minute, by means of an emf.

00:32 The outer radius of the device is 0 .25 meters and it has a rotational frequency of 4 ,000 hertz.

00:40 So i got a series of questions here.

00:42 Number one is the magnetic force acting on the electrons in the disk up or down in the figure.

00:49 Is the magnitude of the magnetic force greater at the rim or near the center of the rotating disc? what is the work per unit charge done by that force in moving electrons between the rim and the center? what is the emf of the device? and the last question is, if the current is 50 amps, what is the power level at which electric energy is produced by this? now, the first question is, is the magnetic force acting on the electrons up or down in the figure? and maybe i should take a minute here, first of all, to show you this figure, and explain how this device works, okay? because that's kind of critical than you're doing all this problem and all the rest of the problems.

01:31 So i have this green conducting disk shown here, and this rotates.

01:38 It's actually being driven by a motor, so it turns in the direction shown there with that sort of red curved arrow.

01:45 So it's rotating around, and it's immersed in a magnetic field, which points uniformly to the left.

01:53 You can see that.

01:53 That's the blue lines there.

01:55 Show the magnetic field.

01:58 And so the electrons, which are kind of embedded in this disk, they're rotating around.

02:04 So they have a velocity, and that velocity, in fact, is perpendicular to the direction of the magnetic field.

02:11 So they experience a magnetic force.

02:14 And we're going to show that the magnetic force acting on those electrons in the figure is up in the picture.

02:26 So, but we'll get to that in a minute.

02:29 So these electrons within the disk then have this force acting on them, which is up.

02:36 And because the electrons then have this force acting upwards on them, they move up from the center of the disk out towards the rim.

02:44 Now, these blue things that are shown here on the disk and then on the sort of rotor here, these are brushes.

02:52 And they're used to collect charge and a deposit charge.

02:55 And while the electrons are negatively charged, so since they go up the disk, that results in an actual flow of current, positive conventional current downward on that disk.

03:09 So that causes this current to flow through downward the disk and then through the rotor and then out to the right -hand side of the device.

03:18 So basically you're spinning this disk in this uniform magnetic field, and those electrons, embedded within the disk are experiencing magnetic force, which causes the electrons to move up the disk, and that generates a current flowing through this whole device.

03:34 And so let's look at the picture here of the magnetic field.

03:39 So if i just pick a sort of arbitrary electron that's moving along this vertical line shown by this red line along the center of that disk, the direction instantaneously of that electron out of the page because the disk is spinning around in a circle.

04:00 And so if i just take the direction of the cross product, v cross b, the direction of that cross product is down.

04:08 And i can see that just by using the right -hand rule.

04:11 V cross b is down.

04:13 But the electrons are negatively charged, which means the force on them is actually upward.

04:19 So the direction of the magnetic force on those electrons is upward, as i promise.

04:25 And as i said, that causes then this current to flow through the entire device.

04:31 That's how the whole thing works, okay? just so you know.

04:36 The next question is, is the magnitude of the magnetic force greater at the rim or near the center of the rotating disk? now, the magnitude of the velocity of a particle within the disk is given by v is equal to omega -r, where omega -mega is 2 -5, that's 2 -5 times the frequency.

04:57 That's called the angular frequency of the disk, and that's constant.

05:00 It's rotating at 4 ,000 hertz.

05:03 So since the velocity then depends linearly on the radius, the distance away from the center of the disc.

05:10 So at the very outer rim, that's when the magnitude of v is largest.

05:15 And therefore, the magnitude of the magnetic force is going to be greatest for electrons at the rim.

05:21 Okay, so that answers that question, at the rim.

05:26 Question c.

05:27 Now, what is the work per unit charge done by that magnetic force in moving electrons between the rim and the center? so the work done on an electron and moving it from the center of the disk to the rim is just given by the standard integral you have when you have a force acting on an object through a path, right? the work is equal to the integral over the path.

05:54 Of the scalar product of the force and the differential displacement ds, so f...

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