Problem 21.11 A flat, square coil with 17.0 turns has sides of length 0.140 m . The coil rotates in a magnetic field of 2.00×10−2 T . |
Part A What is the angular velocity of the coil if the maximum emf produced is 20.0 mV ?
SubmitHintsMy AnswersGive UpReview Part Correct Part B What is the average emf at the angular velocity found in part A?
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length of the coil = l = 0.14 m
area of cross section the coil = A = l^2 = 0.14^2
magnetic field B = 2*10^-2 T
magnetic flux phi = B*A*coswt
induced emf = rate of change in flux
emf = (d/dt)*B*A*coswt
emf = B*A*sinwt*w
maximum emf = B*A*w
given maximum emf = 20*10^-3 v
w = (20*10^-3)/(0.14^2*2*10^-2)
w = 51.02 rad/s
++++++
partB
average value of sinwt = 1/2
average emf = maxx emf/2
average = 20/2 mv
average = 10 mV
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Problem 21.11 A flat, square coil with 17.0 turns has sides of length 0.140 m ....
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