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gtnhenbr
3 months ago
12

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 solenoi

d 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 × 10-7 T ∙ m/A)
Physics
1 answer:
Sav [3.1K]3 months ago
4 0

Answer:

i = 7.83 \mu A

Explanation:

The induced EMF in the coil can be expressed with the following formula

EMF = M\frac{di}{dt}

leading to

M = 31 \mu H

Additionally, the rate of change in the current flowing through the solenoid is expressed by

\frac{di}{dt} = 2.5 A/s

therefore, the induced EMF for the coil can be written as

EMF = (31 \times 10^{-6})(2.5)

EMF = 77.5 \times 10^{-6} A/s

and the induced current within the coil can be derived from

i = \frac{EMF}{R}

i = \frac{77.5 \times 10^{-6}}{9.9}

i = 7.83 \mu A

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