At the instant when the current in an inductor is increasing at a rate of 0.0640 A/s, the magnitude of the self-induced emf is 0.0160 V. (b) If the inductor is a solenoid with 400 turns, what is the average magnetic flux through each turn when the current is 0.720 A?

Video transcript

Welcome back, everybody. We are running a series of tests on an induct. Er and we are told a couple of different things here. We are told that it has a self induced E M F with a magnitude of 0.185 volts. We're told that it has a change of current with respect to time of 0.58 amps per second. In total. There are 550 turns and we are tasked with finding what the mean magnetic flux in each turn is when the induct er draws a current of 5500.77 amps. Luckily, we have a formula for this, this is just equal to our induct ints times our current that is drawn divided by the number of turns here. Now, what is this inductive value? Well, we know that induced E M F is just the induct inserts times our change of current with respect to time. So what I'm gonna do is I'm gonna divide both sides by that value and you will see that these terms cancel out, leaving us with that are inducted is equal to R induced E M F divided by the change of current with respect to time. So let's calculate our induct instead. So that way we can go ahead and then Find our mean magnetic flux here we have that are inducted using this formula is equal to .185 divided by .58. This gives us an inductive of .319 Hz Great. So now we are ready to go ahead and find the mean magnetic flux. This is equal to 0.319 times 0. divided by 5 50 which when we plug into our calculator gives us a final answer of 4.47 times 10 to the negative fourth Weber's, which of course corresponds to our final answer. Choice of c Thank you all so much for watching. Hope this video helped. We will see you all in the next one.