# Homework Help: Rotating Conducting Cylinder in B - Induced Voltage

1. Apr 19, 2010

### xcvxcvvc

Rotating Conducting Cylinder in B -- Induced Voltage

[PLAIN]http://img690.imageshack.us/img690/2600/47007002.jpg [Broken]

I understand how to use motional emf to solve this problem.
$$\int_C U \times B\, dl$$
$$U = \omega R$$
and
$$\int_C dl = H$$
$$\omega RHB$$
where $$\omega$$ is the spinning in rad/s, R is the radius of the cylinder, H is the cylinder's height, and B is the radial magnetic field.

Could someone help me to use Faraday's law to derive the same answer? In my mind, the magnetic flux is constant since if you take snapshots of the spinning cylinder, you always have the same magnetic field flowing through the same surface area (which I believe to be the cylinder's surface area minus to two circular tops).

Last edited by a moderator: May 4, 2017
2. Apr 19, 2010

### kuruman

Re: Rotating Conducting Cylinder in B -- Induced Voltage

Think of it this way: The conducting cylinder has all those electrons in it that are free to move. They have velocity v = ω x r and they are in a magnetic field, which means they experience a Lorentz force F = qvxB, which means they will move in response to that force. This is another way of saying that we have motional emf.

Having said that, I hasten to add that this problem does not sit well with me because, strictly speaking, you cannot have a purely radial B field. Such a field violates the "No magnetic monopoles" Maxwell equation.

Last edited: Apr 19, 2010
3. Apr 19, 2010

### xcvxcvvc

Re: Rotating Conducting Cylinder in B -- Induced Voltage

I understand that interpretation in solving the problem as I did provide a solution using motional emf. I seek to understand more completely Faraday's law, which does not use Lorentz force as far as I can tell.

Do you know how correctly to apply the rate of change in flux to arrive to the same answer?

4. Apr 20, 2010

### kuruman

Re: Rotating Conducting Cylinder in B -- Induced Voltage

Consider a sliver of an area element on the surface of the cylinder that runs down the cylinder's length. It has "width" Rdθ and "length" H. The area is
dA = HRdθ
This area element takes time dt to rotate by amount dθ, so the rate of change of flux through it is
dΦ/dt = B dA/dt = B HRdθ/dt = BHRω.