c = 0.01 m
side of the square loop a = b = 0.0190 m
resistance R = 1.00 * 10-2 ohm
current rate dI/dt = 130 A/s
loop current I loop = E /R
= (1/R)(d phi / dt) -------------- (1)
phi = (mu_0 I b / 2 pi)[ln(c + a / c)]
dphi / dt = (mu_0 b / 2 pi)(dI/dt)[ln(c + a / c)] ------------ (2)
substitute eq (2) in eq (1)
I loop = (mu_0 b / 2 pi R)(dI/dt)[ln(c + a / c)]
= [(4 pi * 10-7)(0.019) /(2)(pi)(1.00 * 10-2)](130)[ln(0.019 + 0.01 / 0.01)]
= 52.59 * 10-6 A = 52.59 muA