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 × 10-7 T ∙ m/A)

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aristocles aristocles · Answer 1 · 2019-08-16T03:48:30+02:00

Answer:

[tex]i = 7.83 \mu A[/tex]

Explanation:

Induced EMF in the coil is given by the equation

[tex]EMF = M\frac{di}{dt}[/tex]

so we have

[tex]M = 31 \mu H[/tex]

also we know that rate of change in current in solenoid is given as

[tex]\frac{di}{dt} = 2.5 A/s[/tex]

so induced EMF of coil is given as

[tex]EMF = (31 \times 10^{-6})(2.5)[/tex]

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

now induced current in the coil will be given as

[tex]i = \frac{EMF}{R}[/tex]

[tex]i = \frac{77.5 \times 10^{-6}}{9.9}[/tex]

[tex]i = 7.83 \mu A[/tex]