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A 2.40cm

A 2.40cm

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Answer #1

Let 'i' be the current at any instant flowing through the long wire
Flux, 'Phi' linked with the loop is given by
phi = [(2*10^-7)*i]*Integral of 2.40dx/x] for x = 1.10 to (1.10+2.40)
or
Phi = [(2*10^-7)ln(3.50/1.10)]*i = (2.314*10^-7)i
Induced emf = d(phi)/dt = (2.314*10^-7)*di/dt = 2.314*120*10^-7 V = 2.77*10^-5 V
or Induced current = (2.77*10^-5)/(1.10*10^-2) = 2.5181*10^-3 A or 2.5181 mA

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