# Electromagnetic induction

1. Jun 2, 2009

### eoghan

1. The problem statement, all variables and given/known data
Hi! I have a problem with this exercise:
A squared loop of length L and mass M can rotate around one of its vertices and it's on a vertical plane. In the region below the suspension point there is a uniform magnetic field B perpendicular to the plane of the loop.
If the loop rotates at a constant angular velocity w, find the electromagnetically induced EMF.

Here's a picture:
http://www.allfreeportal.com/imghost/thumbs/756421Untitled1.png [Broken]

3. The attempt at a solution
$$\phi(B)=\frac{L^2B}{2}tan(\omega t)$$
$$EMF=-\dot{\phi}(B)$$

Now the question: why does the text tell me the mass of the loop?
Has it something to do with the fact that the angular velocity is constant?

Last edited by a moderator: May 4, 2017
2. Jun 2, 2009

### Hao

I don't think M is necessary. If there is a second part to the question, it may be relevant there.

However, you should take a second look at your expression for φ.

Based on the units you are using, φ is the magnetic flux.
$$\phi = \int B dA$$
For a uniform field and an area A inside that field,
$$\phi = B \int dA = B A_{inside\hspace B}$$

According to your φ, it goes to infinity every half period, which would imply that either L or B goes to infinity, which is not the case.

To find φ properly, it is necessary to express the area of the loop that is in the magnetic field as a function of angle, and hence a function of time. As the loop is a square, the expression may be a little complicated.

Next, Farady's Law of Induction tells us:
$$EMF = -\frac{d \phi}{d t} \neq -\phi$$

Note that when the entire loop is completely above or below the suspension point (a whole π radians of its motion per cycle), φ is constant, and hence EMF = 0.

Last edited: Jun 2, 2009
3. Jun 2, 2009

### eoghan

Yes, there is a second part: calculate the energy dissipated in one oscillation by the induced current.

4. Jun 2, 2009

### Hao

I can only imagine energy being dissipated if the loop has a finite resistance.

Last edited: Jun 2, 2009
5. Jun 2, 2009

### eoghan

oh sorry.. the loop has a resistance R.

6. Jun 2, 2009

### Hao

Once you have found the EMF as a function of time, you will be able to find power, and hence, by integration over one cycle, the energy dissipated.

7. Jun 5, 2009

Thank you!