Posted by Mary on Thursday, October 11, 2007 at 11:09pm.
I got the correct answer for part (a) but I am unsure of how to tackle part (b). Please help!!!
Interactive LearningWare 22.2 at wiley/college/cutnell reviews the fundamental approach in problems such as this. A constant magnetic field passes through a single rectangular loop whose dimensions are 0.35 m 0.55 m. The magnetic field has a magnitude of 2.1 T and is inclined at an angle of 75° with respect to the normal to the plane of the loop.
(a) If the magnetic field decreases to zero in a time of 0.46 s, what is the magnitude of the average emf induced in the loop?
emf = N (cos 75) (BB0/tt0)
emf = 1 (0.258819)(2.1T/0.46s)
emf = 1 (0.258819)(4.565217391)
emf = 1.18156V
(b) If the magnetic field remains constant at its initial value of 2.1 T, what is the magnitude of the rate A / t at which the area should change so that the average emf has the same magnitude as in part (a)?
__________ m2/s

Physic please help!  bobpursley, Friday, October 12, 2007 at 5:14am
Your use of significant digits baffles me. There is only two significant digits given in the problem.
This is a non trivial error in physics, especially in labs involving measurement. One cannot create precision with calculators.
http://hyperphysics.phyastr.gsu.edu/hbase/electric/farlaw.html#c1 is a good summary of Faradays law. You are dealing with changing generating an emf by a changing magnetic flux, magnetic flux is B*area
So Faradays law will be
emf=N d flux /dt
emf=N (A*dB/dt + B*dArea/dt)
in part a, dArea/dt was zero. In the second part, dB/dT is zero. So for the same emf in the second part, dArea/dt=dB/dt ,
B*dArea/dT=4.6 m^2/second
B=2.1T, and since area is tilted, the effective cross sectional area is
L*W*cosTheta or A cosTheta.
B*cosTheta* dA/dt=emf
I am not so comfortable with the "correct"response in a) Where is area in the calculations? One cannot ignore the area, as you are dealing with flux. Flux is B*area.
YOu may find some benefit in reviewing Faraday's law.
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