Drop of voltage between terminals while conductive rod is rotating

[Antoha1]

Homework Statement

A conducting rod rotates in a plane perpendicular to the lines of induction $B$ of a 1 T magnetic field.
The axis of rotation passes through one end of the rod. The other end of the rod slides in an arc with a radius $r$ of 10 cm. Between
C and the point O of the rod's axis of rotation, a source is turned on
(see figure). Find the voltage drop between the
terminals of the source.
It is known that the resistance of the rod and the arc is
significantly less than the internal resistance of the source, the angular velocity $\omega$ of rotation of the rod is
300 Hz.
Note.
The figure shows the direction of magnetic
induction B and the rotation of the rod.

Relevant Equations

$\varepsilon=Blv$
$\omega=\frac{v}{r}$
IDK what else for now...

I do not really know how to solve this problem. For now I just have mathematical expression for $\varepsilon$ EMF. I think this is sources EMF. Correct me if I'm wrong.
$\varepsilon=Blv$. $l$ here is lenght of the rod so $l=r$. $v$ here is linear speed of the rod.
$\omega=\frac{v}{r}\implies v=\omega r$
$\varepsilon=B\omega r^{2}$.
What should be going on further?

 

[kuruman]

The equation that you quoted, $\varepsilon=Blv$, is applicable to the case where all the points on the rod have the same velocity $v$. Then you can say that the voltage drop from one end to the other is given by that formula.

Here, the linear velocity is $v=0$ at the axis of rotation, $v=\frac{1}{2}\omega L$ at the mid point and $v=\omega L$ at the end. You have to consider the voltage drop $d\varepsilon$ across a small length $dr$ and add all such voltage drops from one end of the rod to the other. In other words, you have to do an integral.

(Edited for typos.)

 

[Steve4Physics]

The axis of rotation passes through one end of the rod. The other end of the rod slides in an arc with a radius $r$ of 10 cm. Between C and the point O of the rod's axis of rotation, a source is turned on (see figure).

I think the description is not quite right. Guess there is a conducting circular ring (shown red below):

the angular velocity $\omega$ of rotation of the rod is 300 Hz.

Angular velocity (speed), $\omega$, is usually measured in units of rad/s. Frequency, $f$, is usually measured in Hz. You need to check if the angular velocity is 300 rad/s or something else,. Note: $\omega = 2\pi f$.

Relevant Equations: $\varepsilon=Blv$

When does this apply? Why doesn't it apply to a rotating conductor?

$\omega=\frac{v}{r}$

In the above formula, what point is moving with speed v?

I do not really know how to solve this problem.

A couple of questions to consider

Q1. If the supply (the cell) is removed, what is the size and polarity of the induced emf between the ends of the rotating rod?
Hint. You could consider the average speed off different points on the rod. Or (better) use Faraday's law and consider the rate of cutting of flux ($\frac {d\phi}{dt}$).

Q2 If the rod stationary (but the supply is present) what is the size and direction of the current?

EDIT. I was too slow. @kuruman replied while I preparing my reply.

 

[Antoha1]

The equation that you quoted, $\varepsilon=Blv$, is applicable to the case where all the points on the rod have the same velocity $v$. Then you can say that the voltage drop from one end to the other is given by that formula.

Here, the linear velocity is $v=0$ at the axis of rotation, $v=\frac{1}{2}\omega L$ at the mid point and $v=\omega L$ at the end. You have to consider the voltage drop $d\varepsilon$ across a small length $dr$ and add all such voltage drops from one end of the rod to the other. In other words, you have to do an integral.

(Edited for typos.)

Hi, I don't have the basics of integral and differential calculus yet, but I dug a little deeper and could this be an expression for the voltage drop between the terminals? $\varepsilon=\int_{0}^{L}B\omega r\ dr$

 

[kuruman]

Hi, I don't have the basics of integral and differential calculus yet, but I dug a little deeper and could this be an expression for the voltage drop between the terminals? $\varepsilon=\int_{0}^{L}B\omega r\ dr$

That is indeed the expression. If you don't know what to do with it, you have to wait until you learn how to integrate. You may wish to inform the person who assigned this problem to you that you lack the basics of calculus. It is important that you understand how this expression was put together in terms of what is going on physically.

 

[Steve4Physics]

Hi, I don't have the basics of integral and differential calculus

It may be worth adding that you can find the induced emf without calculus using (a simplified version of) Faraday’s law.

Consider the emf produced while the rod is 'cutting flux' at constant angular speed $\omega$, during a rotation.

$\mathcal E = - \frac {\Delta \Phi}{\Delta t}$ (Faraday’s law)
where
$\Delta \Phi = \Delta BA = B \Delta A = B \pi l^2$ and $\Delta t = \frac {2\pi}{\omega}$.

You would need $\omega$ to be in units of rad/s to do the numerical calculation.

Minor edit.