A 96-mH solenoid inductor is wound on a form 0.80 m in length and 0.10 m in diameter. A coil is tightly wound around the solenoid at its center. The coil's resistance is 9.9 ohms. The mutual inductance of the coil and solenoid is 31 μH. At a given instant, the current in the solenoid is 540 mA, and is decreasing at the rate of 2.5 A/s. At the given instant, what is the magnitude of the induced current in the coil? (μ₀ = 4π × 10^-7T ∙ m/A)

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Quizplus · Accepted Answer

The induced emf in the coil is given by:$$emf = -M \frac{di}{dt}$$where:* M is the mutual inductance (31 μH)* di/dt is the rate of change of current in the solenoid (-2.5 A/s)Substituting these values into the equation, we get:$$emf = -(31 \mu H)(-2.5 A/s) = 77.5 \mu V$$The induced current in the coil is then given by:$$i = \frac{emf}{R} = \frac{77.5 \mu V}{9.9 \Omega} = 7.8 \mu A$$Therefore, the magnitude of the induced current in the coil is 7.8 μA.

A 96-MH Solenoid Inductor Is Wound on a Form 0