Answer:
M = 0.45 mH
Explanation:
For the system described in the statement, you have that the induced emf of both coils are given by:
[tex]\epsilon_1=-M\frac{\Delta I_2}{\Delta t}\\\\\epsilon_2=-M\frac{\Delta I_1}{\Delta t}[/tex] (1)
ε1, ε2: induced emfs
M: mutual inductance
I1, I2: current on the wires
That is, the induced emf of one coil is generated by the change in the current of the other coil.
You can consider that the current of 8.0A belongs to the second coil, and the induced emf of 3.00V is generated in the first coil. The current of the second coil is switched off in 1.20ms.
With the previous information you have:
ε1 = 3.00V
ΔI2 = 0.00A - 8.00A = -8.00A
Δt = 1.20ms = 1.2*10^-3 s
You use the equation (1) to solve for M:
[tex]M=-\frac{(\Delta t)( \epsilon_1)}{\Delta I_2}=-\frac{(1.20*10^{-3}s)(3.00V)}{-8.00A}\\\\M=4.5*10^{-4}H=0.45mH[/tex]
The mutual inductance of the coils is 0.45mH
