A long, straight solenoid with a cross-sectional area of 8.00 cm^2 is wound with 90 turns of wire per centimeter, and the windings carry a current of 0.350 A. A second winding of 12 turns encircles the solenoid at its center. The current in the solenoid is turned off such that the magnetic field of the solenoid becomes zero in 0.0400 s. What is the average induced emf in the second winding?

Video transcript

Hi, everyone in this particular practice problem, we are asked to calculate the mean E M F induced in the second winding where we have a current of 0.45 M those through a winding in the form of a linear solenoid, having 80 turns of wire per centimeter and a cross sectional area of seven centimeters squared. It is going to be bounded by a secondary winding of only 15 turns, turning off the current in the solenoid will reduce the magnetic field of the sunlight to zero in 0.7 seconds. And we are asked to calculate the mean E M F induced in the second winding. So we want to recall that the magnetic field of a large straight solenoid is going to be calculated by um or large ST. So B is going to be calculated by me not times N times I just like. So inside the solenoid and going to be equals to zero outside after Solomon itself. So the magnetic field is going to equal to this inside and zero outside. And in this case, we have a five B that will go into equals two B times A and we know that in this case, the A is going to be seven centimeters squared or seven times 10 to the power of minus four m squared just like. So, and that will be the cross sectional area for the large long straight solenoid. And then we know that an E M F is going to be induced in the second winding, even though the magnetic field of the solenoid is going to be zero at the location of the second winding. Because the change in magnetic field will actually induce an electric field outside the solenoid. And that induced electric field will produce the E M F I am going to summarize that. So the change in the magnetic field will induce an electric field outside of solenoid that will actually induce an E M F. So the way we want to solve this is to by applying faraday's law for calculating the average or mean E M F induced in the second winding. And that will just essentially be E F or in average induce E M F equals to N multiplied by the change in the magnetic flux over the change in time just like. So and recall that the flux be can be calculated by this formula right here. So essentially this will just be end, I'm gonna write this down first just like this. And then I and then after this, I am going to actually include an input this formula here. And this then will be, and it's applied by the first one is going to be uh a multiplied by B F minus a multiplied by B I. And we know that the A or the cross sectional area is going to actually be the same after the magnetic field of the solid turns to zero. So we can plot that out and this will just then be an A times B F minus B I over delta T just like so okay. So then we know that the be here can actually be calculated for a large straight solenoid using substitute, using this formula right here equals you not. And I, so I'm going to include that and exchange that substitute that into this equation right here. So we have E F induce CMF average into CMF and a multiplied by mu naught and I F minus mu naught and I initial over delta T and we know that the mu not and and is essentially just constant you not terms and recall that this end here is going to be the number of turns that is going to actually be different from this one right here. So this is going to be the number of turns of the second winding and this one as the number of turns equals on the first winding or essentially the really long linear solenoid. I'm just going to write that the number of turns linear solenoid just like. So and this will actually be I over delta T just like so okay. So this is exactly the formula that we need to actually solve this problem. So we can input all the information that we know. So the E F or the, the average induced E M F induced E M F is going to be equals two, four Pi Times 10 to the power of -7. Uh This is going to be the new, not here, first T M over M And then we have the first end which is going to be 15, Given we have 15 turns here and then we have the cross sectional area which is going to be sure Seven times 10 to the part of -4 m squared. And then we have the last end which is going to be 80 turns of wire per centimeter Just like so, so that will actually equal to eight Because we have centim to the power of -1 instead of defining it by 100 to get the meter, we wanna multiply it by 100. So this is going to be 8000 m to the part of minus one. And then we have the I we're just going to be 0.45 amp And we have the Delta T which is going to be 0.07 seconds. And that will be all for this one and this relation Chris Pond to a 6.7, 8 times 10 to the power of minus four fault, that will be the answer to demean or the average induced E M F in the second winding, which will actually correspond to option A here. And that will be all for this particular problem. If you guys have any sort of confusion on this example, please make sure to check out our other lesson videos on similar topics and that will be all for this problem. Thank you.