When the current in a toroidal solenoid is changing at a rate of 0.0260 A/s, the magnitude of the induced emf is 12.6 mV. When the current equals 1.40 A, the average flux through each turn of the solenoid is 0.00285 Wb. How many turns does the solenoid have?

Video transcript

Welcome back, everybody. We are making observations about a Tor Oid and we are told a couple of different things here. The induced E M F has a magnitude of 13.5 million volts. And we are told that the current changes at a rate of 0.3 to two amps per second, The mean flux per turn in the solenoid is .00372 Webbers. And this is at a current of 2.6 amps. And we are tasked with finding how many turns are in the solenoid here. Alright. Well, we know that the induct inside of a solenoid is equal to the number of turns times the mean flux. I'm actually gonna move this over just a little bit is the number of turns times the mean flux divided by the current. But we also know that the induct ints is equal to our induced E M F divided by the rate at which the current is changing. So what I'm gonna do, I'm gonna multiply both sides here by the current divided by the mean flux aren't divided by the mean flux. And you'll see that these terms cancel out. Leaving us with that the number of turns is equal to the induced E M F times the current divided by the rate at which the current changes times the mean flux. Great. So now that we have that we have all of our values, let's just go ahead and plug it in here. We have that our number of turns is equal to 13.5 million volts. But we need this in volts. So I'm gonna multiply this by 10 to the negative three times our current of 2.6 amps divided by our rate of change for current, which is 0.322 times our mean flux per turn of 0.372, which when you plug all of this in your calculator, you get 293 turns corresponding to our answer choice of D. Thank you all so much for watching. Hope this for your help. We will see you all in the next one.