Question

The current through a coil as a function of time is represented by the equation

I(t) = Ae^−bt sin(ωt), where A = 5.25 A, b = 1.75 ✕ 10⁻² s⁻¹, and ω = 375 rad/s. At t = 0.880 s, this changing current induces an emf in a second coil that is close by. If the mutual inductance between the two coils is 4.35 mH, determine the induced emf. (Assume we are using a consistent sign convention for both coils. Include the sign of the value in your answer.)

I only have a few more tries to get this one correct. Please help and show how you got it so I can find what I'm doing incorrectly. Thank You!

Answer 1

Relation between induced emf and mutual inductance is given by,

induced emf(e) = -M*dI/dt

given, M = mutual inductance = 4.35 mH = 4.35*10^-3 H

I(t) =  Ae^(−bt)*sin(ωt)

using given values,

I(t) = 5.25*e^[-(1.75*10^-2)*t]*sin(375*t)

by derivating it,

dI/dt = -(1.75*10^-2)*5.25*e^[-(1.75*10^-2)*t]*sin(375*t) + 5.25*e^[-(1.75*10^-2)*t]*375*cos(375*t)

at t = 0.880 s,

dI/dt = -(1.75*10^-2)*5.25*e^[-(1.75*10^-2)*0.880]*sin(375*0.880) + 5.25*e^[-(1.75*10^-2)*0.880]*375*cos(375*0.880)

dI/dt = -1921.59 Amp/s

So, induced emf(e) = -(4.35*10^-3)*(-1924.59)

e = 8.37 V

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