A problem with a magnetic field and a revolving stick

[fara0815]

Helle there!
Elektrodynamics is giving me a hard time. I have been trying to figure the following question out by using textkooks like Haliday, Schaum's Outline and even my professor's script ;)
But unfortunately, I do not know how to solve it.

Problem:

A stick of length 0,1 m is revolving around one of its ends in a constant magnetic field of B=(0;0;1) mT. The stick's angular velocity is $$\omega=(0;0;1) s^1$$.
What is the voltage between the two ends?
(5 microV).

Since I have absolutely no clue, I would appreciate if someone could give me a hint where to start and lead me in a few steps to the result.

Many thanks in advance to this great community!

 

[OlderDan]

You have probably already done the problem of finding the emf of a stick sliding on a rectangular loop of wire. Does that give you any ideas?

 

[fara0815]

Let's see if I am getting close:

According to Faraday's Law, the magnetic flux is $$\phi_b= BA$$ when B is normal to the surface A.
My problem here is that I first have to get the surface which the magnetic lines go through.
l=length of the stick
The surface of one radiant is $$A_{rad}=\frac{\pi l^2}{2\pi}=\frac{l^2}{2}$$
with that I get the surface related to the time:
$$A(t)=\frac{l^2\omega t}{2}$$

So the magnetic flux is
$$\phi_B=\frac{ Bl^2\omega t}{2}$$

And the EMF is
$$U_{ind}=\frac{\delta}{\delta t}\frac{ Bl^2\omega t}{2}=\frac{ Bl^2\omega}{2}= 0.05 microVolts$$

What am I missing? Is this even close?

 

[OlderDan]

Let's see if I am getting close:

According to Faraday's Law, the magnetic flux is $$\phi_b= BA$$ when B is normal to the surface A.
My problem here is that I first have to get the surface which the magnetic lines go through.
l=length of the stick
The surface of one radiant is $$A_{rad}=\frac{\pi l^2}{2\pi}=\frac{l^2}{2}$$
with that I get the surface related to the time:
$$A(t)=\frac{l^2\omega t}{2}$$

So the magnetic flux is
$$\phi_B=\frac{ Bl^2\omegat}{2}$$

And the EMF is
$$U_{ind}=\frac{\delta}{\delta t}\frac{ Bl^2\omegat}{2}=\frac{ Bl^2\omega}{2}= 0.05 microVolts$$

What am I missing? Is this even close?

You are just missing the $\omega t$ in a couple of places (flux and derivative of flux) but you got it back in the end.

 

[fara0815]

latex was leaving them out. Now they are there. Sorry about that ;)
and I got confused by the units. But thank you very much for your help!
I am soo glad I finally figured it out! Right in time for the class tomorrow morning ;)