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
Answer:
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
To solve this problem we will apply the concepts related to magnetic flux and induced voltage. This last expression understood as the variation of the magnetic flux over time and, in turn, the magnetic flux expressed as the variation of the magnetic field in a certain area.
Magnetic flux through the circular coil is given as
The induced voltage is the change of the magnetic flux across the time, then
At the same time the magnetic flux through the square coil would be given as,
And the induced voltage EMF will be
Equating both expression we have
Therefore the emf induced in the square coil is 0.23355V
Explanation:
answer is that i think???
Answer:
<u><em>A. wavelength</em></u>
Explanation:
The others are about sound and how high it is. That has nothing to do with time.
Answer:
The answer would be Saturn.
Explanation:
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