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October 30, 2014

October 30, 2014

Posted by **Mary** on Thursday, October 11, 2007 at 11:09pm.

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) (B-B0/t-t0)

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 5:14amYour 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.phy-astr.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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