The magnetic flux through each turn of a 180-turn coil is given by psi_B = 9.50
ID: 1570686 • Letter: T
Question
The magnetic flux through each turn of a 180-turn coil is given by psi_B = 9.50 times 10^-3 sin (omega t). where omega is the angular speed of the coil and phi_B is in webers. At one instant, the coil is observed to be rotating at a rate of 8.60 times 10^2 rev/min (Assume that t is in second) (a) What is the induced emf in the coil as a function of time for this angular speed? (Use the following as necessary: t Do not use other variables substitute numeric value Assume that element is in volts. Do not include units in your answer.) (b) What is the maximum induced emf in the coil for this angular speed?Explanation / Answer
N=180
flux=9.5*10^-3*sin(w*t)
w=8.6*10^2 rev/min
a)
emf=N*d(flux)/dt
=180*d/dt(9.5*10^-3*sin(w*t))
=180*(9.5*10^-3*w*cos(w*t))
=180*(9.5*10^-3*8.6*10^2*(2pi/60)*cos(w*t))
=154*cos(w*t) V
b)
maximum emf = 154 V
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