the answers are right but please show all steps and work The area of an elastic
ID: 1644558 • Letter: T
Question
the answers are right but please show all steps and work
The area of an elastic circular loop decreases at a constant rate, dA/dt = -3.50 times 10^-2 m^2/s. The loop is in a magnetic field B = 0.28 T whose direction is perpendicular to the plane of the loop. At t = 0, the loop has area A = 0.285m^2. Determine the induced emf at t = 0, and at t = 2.00 s. Answer: 9.8 mV Suppose the radius of the elastic loop in part (a) increases at a constant rate, dr/dt = 4.30 cm/s. Determine the emf induced in the loop at t = 0 and at t = 1.00 s. Answer: 23 mV 26 mVExplanation / Answer
A) induced emf = d(flux) /dt
= B dA/dt
= 0.28*3.5e-2
= 0.0098 V
= 9.8mV
Induced emf will be same at both times.
B) induced emf = d(flux) /dt
= B dA/dt
= B* pi*2r* dr/dt
Initial r = sqrt(A/pi) = sqrt (0.285/pi) = 0.301 m
Induced emf =0.28*pi*2*0.301*0.043 = 0.023 V = 23mV
After 1s, induced emf = 0.28*pi*2*(0.301+0.043)*0.043
= 0.026 V = 26mV
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