Question: 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 $\mathit{R}\mathbf{}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{250}\mathbf{ }\mathbf{\text{m}}$and rotation frequency $\mathit{f}\mathbf{}\mathbf{=}\mathbf{}\mathbf{4000}\mathbf{}\mathbf{\text{Hz}}$, and the device is in a uniform magnetic field of magnitude $\mathit{B}\mathbf{}\mathbf{=}\mathbf{}\mathbf{60}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{\text{mT}}$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 $\mathbf{50}\mathbf{.}\mathbf{0}\mathbf{ }\mathbf{\text{A}}$, what is the power at which electrical energy is being produced?

$R=0.25\text{m}$

$f=4000\text{Hz}$

$B=60\text{mT} \text{or} 60\times {10}^{-3}\text{T}$

The influence of a magnetic field produced by one charge on another is what is known as the magnetic force between two moving charges.

First we use the concept of magnetic force for the direction of the electron. After that, we use the relation between angular velocity and linear velocity to find the force, and by using the force, we can find the work done. The is the work done per unit charge. Using this concept, we find the Finally, we find power, which depends on emf.

Formula:

$F=q\left(\overrightarrow{v}\times \overrightarrow{B}\right)$

$v=r\omega$

$\omega =2\pi f$

$w=\int F·dr$

$P=\epsilon i$

As the electron moves due to the magnetic force, the force is

$\overrightarrow{F}=q\left(\overrightarrow{v}\times \overrightarrow{B}\right)$

As the field is pointing towards left, electrons, which are negatively charged, are forced to move clockwise. So using the Right Hand rule, we can conclude that the direction of the magnetic force is upwards in the figure.

Force depends on the velocity of a moving electron, where velocity can be written as $v=r\omega$.So the magnitude of the force depends on the radial distance. The radial distance is more at the rim, so the force is more at the rim.

As we know $v=r\omega$,and$\omega =2\pi f$

$$\begin{gathered} \omega =2\pi \times 4000 \text{Hz} \\ =8000\pi /\text{s} \end{gathered}$$

So, the velocity of the electron is$v=r\omega$

$$\begin{gathered} v=\frac{0.250}{2}\times 8000\pi \\ v=1000\pi \text{m/s} \end{gathered}$$

Calculate the magnitude of the force as

$$\begin{gathered} F=qvB \\ =\left(1.6\times {10}^{-19} \text{C}\right)\times \left(1000\pi \text{m/s} \right)\left(\times 60\times {10}^{-3} \text{T}\right) \\ =3.02\times {10}^{-17}\text{N} \end{gathered}$$

Now we have to find the work done per unit charge by the formula as

$\frac{w}{q}=\int \frac{F}{q}·dr$

$$\begin{gathered} \frac{w}{q}={\int }_R^0\frac{3.02\times {10}^{-17}\text{N}}{-1.6\times {10}^{-19} \text{C}}·dr \\ =\frac{3.02\times {10}^{-17}\text{N}}{-1.6\times {10}^{-19} \text{C}}{\int }_R^{0}dr \\ =\frac{3.02\times {10}^{-17}\text{N}}{-1.6\times {10}^{-19} \text{C}}{\left[r\right]}_R^0 \\ =\frac{3.02\times {10}^{-17}\text{N}}{-1.6\times {10}^{-19} \text{C}}\left[-R\right] \end{gathered}$$

By substituting the value of R, we can get

$\frac{w}{q}=188.75 \text{N/C}\times 0.250 \text{m}$

$\frac{w}{q}=47.1 \text{V}$

Now find the emf as the work done per unit charge.

$\epsilon =\frac{w}{q}$

From equation (i),

$\epsilon =47.1\text{V}$

Power is emf times the current; therefore,

$P=\epsilon i$

By substituting the value, we can calculate

$P=47.1 \text{V}\times 50 \text{A}$

$P=2355~2.36\times {10}^{3}W$