What is the EMF generated in the loop?

iggybaseball
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" A circular loop of wire 10 cm in diameter is placed with its normal making an angle of 30 degrees with the direction of a uniform 0.50T magnetic field. The loop is "wobbled" so that its normal rotates in a cone about the field direction at the constant rate of 100rev/min; the angle between the normal and the field direction (= 30 degrees) remains unchanged during the process. What emf appears in the loop?"

Alright I know that I am supposed to use the equation:
E(potential difference) = BDv
With B as the magnetic field, D as the area with a magnetic field that is not equal to 0 and v is the velocity. The back of the book says that the emf is zero. Is this because the loop is not being displaced ( it is wobbling in the same spot?) so it's velocity is zero?
 
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iggybaseball said:
Alright I know that I am supposed to use the equation:
E(potential difference) = BDv
That equation applies to a special case: That equation will calculate the motional EMF along a wire segment of length D that moves with speed v perpendicular to a magnetic field B.

Rather than struggle to apply this equation, consider the more general Faraday's law. (Look it up if you need to.) What's the flux through the loop? Does the flux change?
 
To solve this, I first used the units to work out that a= m* a/m, i.e. t=z/λ. This would allow you to determine the time duration within an interval section by section and then add this to the previous ones to obtain the age of the respective layer. However, this would require a constant thickness per year for each interval. However, since this is most likely not the case, my next consideration was that the age must be the integral of a 1/λ(z) function, which I cannot model.
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