How Does Mass Affect EMF in a Rotating Loop with Constant Angular Velocity?

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SUMMARY

The discussion centers on calculating the electromotive force (EMF) induced in a rotating square loop of length L and mass M within a uniform magnetic field B. The magnetic flux φ is defined as φ = B * A, where A is the area of the loop in the magnetic field, and the EMF is derived from Faraday's Law of Induction as EMF = -dφ/dt. The mass of the loop is deemed unnecessary for the EMF calculation but may be relevant for subsequent energy dissipation calculations, particularly when considering the loop's resistance R.

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  • Understanding of Faraday's Law of Induction
  • Knowledge of magnetic flux and its calculation
  • Familiarity with angular velocity and its implications in rotational dynamics
  • Basic principles of electrical resistance in circuits
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eoghan
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Homework Statement


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

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?:confused:
Has it something to do with the fact that the angular velocity is constant?
 
Last edited by a moderator:
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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:
Hao said:
I don't think M is necessary. If there is a second part to the question, it may be relevant there.

Yes, there is a second part: calculate the energy dissipated in one oscillation by the induced current.
 
I can only imagine energy being dissipated if the loop has a finite resistance.

I hope the question gives more information on what the loop is made of.
 
Last edited:
oh sorry.. the loop has a resistance R.
 
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.
 
Thank you!
 

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