) A 73-mH solenoid inductor is wound on a form that is 0.80 m long and 0.10 m in diameter. A coil having a resistance of is tightly wound around the solenoid at its center. The mutual inductance of the coil and solenoid is At a given instant, the current in the solenoid is and is decreasing at the rate of At the given instant, what is the induced current in the coil

Complete question is;. A 73mH solenoid inductor is wound on a form that is 0.80m long and 0.10m in diameter a coil having a resistance of 7.7 ohms is tightly wound around the solenoid at its center the mutual inductance of the coil and solenoid is 19μH at a given instant the current in the solenoid is 820mA and is decreasing at the rate of 2.5A/s at the given instant what is the induced current in the coil

6.169 μA

Explanation:

Formula for induced EMF is given by the equation;

EMF = M(di/dt). We are given;

di/dt = 2.5 A/s

M = 19μH = 19 × 10^(-6) H

Thus;

EMF = 19 × 10^(-6) × 2.5.

EMF = 47.5 × 10^(-6) V

Formula for current is;

i = EMF/R. R is resistance given as 7.7 ohms.

Thus; i = 47.5 × 10^(-6)/7.7

i = 6.169 μA

c, 1a hope this

answer: b.

explanation: