A 200-loop coil of cross sectional area 8.5 cm- lies in the plane of the page. A
ID: 1510644 • Letter: A
Question
Explanation / Answer
n = 200
A = 8.5*10^-4 m^2
B = 0.06 T
Bf = 0.02 T
t = 12*10^-3 s
a) Flux change = dB*A = (0.06-0.02)*8.5*10^-4 = 0.34*10^-4
b) V = nd(phi)/dt = 68*10^-4 V
c) I = V/R = 17*10^-4 A
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