91Ó°ÊÓ

$R=0.25\text{m}$

$f=4000\text{Hz}$

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

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:

$F=q\left(\stackrel{→}{v}×\stackrel{→}{B}\right)$

$v=r\mathrm{Ó\neg }$

$\mathrm{Ó\neg }=2\mathrm{Ï€}f$

$w=∫F·dr$

$P=\mathrm{Î\mu }i$

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

$\stackrel{→}{F}=q\left(\stackrel{→}{v}×\stackrel{→}{B}\right)$

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 $v=r\mathrm{Ó\neg }$.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 $v=r\mathrm{Ó\neg }$,and$\mathrm{Ó\neg }=2\mathrm{Ï€}f$

$$\begin{gathered} \mathrm{Ó\neg }=2\mathrm{Ï€}×4000 \text{Hz} \\ =8000\mathrm{Ï€} /\text{s} \end{gathered}$$

So, the velocity of the electron is$v=r\mathrm{Ó\neg }$

$$\begin{gathered} v=\frac{0.250}{2}×8000\mathrm{π} \\ v=1000\mathrm{π} \text{m/s} \end{gathered}$$

Calculate the magnitude of the force as

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

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

$\frac{w}{q}=∫\frac{F}{q}·dr$

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

By substituting the value of R, we can get

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

$\frac{w}{q}=47.1 \text{V}$

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

$\mathrm{Î\mu }=\frac{w}{q}$

From equation (i),

$\mathrm{Î\mu }=47.1\text{V}$

Power is emf times the current; therefore,

$P=\mathrm{Î\mu }i$

By substituting the value, we can calculate

$P=47.1  \text{V}×50 \text{A}$

$P=2355~2.36×{10}^{3}W$