Faraday's and Lenz's law application

AI Thread Summary
The discussion focuses on calculating the magnetic flux and electromotive force (emf) for a rotating rectangular coil in a magnetic field. When the coil is orthogonal to the axis AA', the magnetic flux is zero because the angle θ is 90°. The emf is calculated using the equation ε = -dφ_B/dt, resulting in ε = NBabω sin(ωt), which reaches its maximum when sin(θ) is 1, indicating maximum rate of change of flux. The key point is that the maximum emf occurs at the point of maximum rate of change of flux, not when the flux itself is at a maximum. The procedure and understanding of the concepts are confirmed to be correct.
cseil
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Homework Statement


A rectangular coil with a and b sides can rotate around the axis AA' with angular velocity ω. It is in a magnetic field as in figure.

esercizio spira circolare.jpg


Calculate the flux of B when the coil is orthogonal to the axis AA'. Calculate the \epsilon_{max} and express the position of the coil.

Homework Equations


\epsilon = -\frac{d}{dt} \phi_B

The Attempt at a Solution



The flux when the coil is orthogonal to the axis AA' is 0.

\phi_B = NBabcos\theta

\theta is 90°, so the flux is 0.

Now I calculate the emf.
\epsilon = -\frac{d}{dt} \phi_B = NBab\omega sin(\omega t)

The emf is max when sin(\theta) is 1, so when \theta is 90°.

Is it correct my procedure?
If yes, why emf is maximum when flux is 0?
 
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cseil said:
Calculate the flux of B when the coil is orthogonal to the axis AA'.
Since the coil rotates about that axis, it is always orthogonal to it. I assume you mean when the plane of the coil is perpendicular to the surface of the drawing.

cseil said:
Is it correct my procedure?
Yes.

cseil said:
If yes, why emf is maximum when flux is 0?
What matters is not where the emf is maximum but where its rate of change is maximum.

Edit: Oops... I meant flux, not emf, of course.
 
Last edited:
Doc Al said:
Since the coil rotates about that axis, it is always orthogonal to it. I assume you mean when the plane of the coil is perpendicular to the surface of the drawing.

Yes, sorry, I meant that.
Thank you!
 
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