Answer:
94.05 mV
Step-by-step explanation:
N = Number of turns = 149
r = Radius of coil = 2.45 cm = 0.0245 m
t = Time = 0.141 s
Initial magnetic field = 50.9 mT
Final magnetic field = 98.1 mT
A = Area = πr² = π×0.0245²

∴ Magnitude of the average EMF, that is induced in the coil during this time interval is 94.05 mV
solution:
solid spindle ab:
c= 1/2ds = ½(1.5) = 0.75in.
j= π/2c1= π/2(0.75)2 = 0.49701 in4
jmax = Tc/j
tab=jtal/c = (0.49701)(12)/0.75 = 7.952
sleeve cd = c2 = 1/2d2 = ½(3.0) = 1.5in
j = π/2(c24 – c14) = π/2(1.54-1.254) = 4.1172 in4
tab = jtl/c2 = (4.1172)(7)/1.5 = 19.213 kip.in