Question

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)

Answer 1

Answer:

$i = 7.83 \mu A$

Explanation:

Induced EMF in the coil is given by the equation

$EMF = M\frac{di}{dt}$

so we have

$M = 31 \mu H$

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

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

so induced EMF of coil is given as

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

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

now induced current in the coil will be given as

$i = \frac{EMF}{R}$

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

$i = 7.83 \mu A$