The average emf induced in the secondary coil is 0.12 V when the current in the primary coil changes from 3.8 to 1.1 A in 0.14 s. What is the mutual inductance of the coils?
V2 = M12*dI1/dt
You want the mutual inductance M12.
The rate of change of current in loop 2 is dI1/dt = 2.7/0.14 = 19.3 amps/s
You do the numbers.
0.12 V = M12*19.3 amps/s
M12 will be in Henries
It is somewhat unusual to talk of "primary" and "secondary" coils when dealing with mutual inductance. That language is usually applied to transformers, but would be OK there, too.
Note that M12 = M21, although that is not obvious.
To find the mutual inductance of the coils, we can use Faraday's law of electromagnetic induction, which states that the induced electromotive force (EMF) in a circuit is equal to the rate of change of magnetic flux through the circuit.
The formula for the induced EMF is given by:
EMF = -M * (delta I / delta t),
where EMF is the induced electromotive force, M is the mutual inductance of the coils, delta I is the change in current in the primary coil, and delta t is the change in time.
In this case, the induced EMF (EMF) is 0.12 V, the change in current (delta I) is 3.8 A - 1.1 A = 2.7 A, and the change in time (delta t) is 0.14 s. We can plug these values into the formula and solve for the mutual inductance (M):
0.12 V = -M * (2.7 A / 0.14 s).
To solve for M, we can rearrange the equation:
M = -(0.12 V) / (2.7 A / 0.14 s).
Calculating the right side of the equation gives us:
M = -(0.12 V) / (19.2857 A/s).
Simplifying further:
M = -0.006211 s * A / V.
Therefore, the mutual inductance of the coils is approximately -0.006211 H (henries). The negative sign indicates that the induced EMF has opposite polarity to the change in current.