An inductor has inductance of 0.280 HH and carries a current that is decreasing at a uniform rate of 15.0 mA/smA/s .Part AFind the self-induced emf in this inductor.Express your answer in millivolts.
Question
An inductor has inductance of 0.280 HH and carries a current that is decreasing at a uniform rate of 15.0 mA/smA/s .Part AFind the self-induced emf in this inductor.Express your answer in millivolts.
Solution
The self-induced emf in an inductor can be calculated using Faraday's law of electromagnetic induction, which states that the induced emf in a circuit is equal to the rate of change of magnetic flux through the circuit. In the case of an inductor, this can be simplified to:
emf = -L * (dI/dt)
where:
- L is the inductance of the inductor, and
- dI/dt is the rate of change of current.
Given that L = 0.280 H and dI/dt = -15.0 mA/s = -0.015 A/s (note that we need to convert from mA to A), we can substitute these values into the equation to find the emf:
emf = -0.280 H * -0.015 A/s = 0.0042 V = 4.2 mV
So, the self-induced emf in this inductor is 4.2 millivolts.
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