A long, straight solenoid has 800 turns. When the current in the solenoid is 2.90 A, the average flux through each turn of the solenoid is 3.25 * 10^-3 Wb. What must be the magnitude of the rate of change of the current in order for the self-induced emf to equal 6.20 mV?

Video transcript

Welcome back, everybody. We are making observations about a solenoid and we are told a couple of different things. We're told that in total, there are 650 turns were told that it has a mean flux of 6.8 times 10 to the negative fourth Weber's W B. Uh And we are tasked with finding here what is going to be the change of current with respect to time. And we are told a couple different things about this system as well. We're told that it has a self induced E M F of 0.8 volts. And we want to find this change of current with respect to time with a current of 3.2 amps. So what is the formula for this right here? Well, the formula for this is going to be our E M F divided by our conductance in our inductive is going to be given by the number of turns times our flux divided by our current. Let's go ahead and calculate our inductions first so that we can find our change of current with respect to time. Alright, so are inducted is equal to 650 times 6.8 times 10 to the negative fourth, all divided by 3.2, giving us an inductive of 0.138 Hertz. Perfect. So now we are ready to go ahead and find this change of current with respect to time. This is just going to be 0.8 divided by 0.138, giving us a final answer of 0.58 amps per second corresponding to our final answer choice of D Thank you all so much for watching. Hope this video helped we will see you all in the next one.