The magnetic field B at all points within the colored circle shown in Fig. E29.15 has an initial magnitude of 0.750 T. (The circle could represent approximately the space inside a long, thin solenoid.) The magnetic field is directed into the plane of the diagram and is decreasing at the rate of -0.0350 T/s. (e) If the ring is cut at some point and the ends are separated slightly, what will be the emf between the ends?

Video transcript

Hi, everyone. In this particular practice problem, we are asked to determine the induced EMF in between the ends where we have a square loop of side 25 cm where it is being inserted into a magnetic field directed into the page. Initially, the magnetic field has a magnitude of 0.95 tesla and the rate of decrease of the magnetic field is minus 0.65 tesla per seconds. There is an insulator in the loop separating the two ends of the loop from touching. And we want to calculate the induced CMF in between those two ends. So first, we want to identify that the different um information given in our problem statement. So first, we have the length of the side. I'm just gonna write it down as side which will equal to centimeters or essentially 0.25 m because we wanted an S I and then we have the magnitude of the magnetic field which is 0.95 tesla initially. And then we have the rate of which the magnetic field is decreasing, which is D B over D T because it is decreasing. We wanna include the minus minus 0.65 tesla per second, just like that. Okay. Now that we have all the information listed, we want to recall that to solve for this particular problem here, we wanna use Faraday's Law. So in Faraday's law, um we can recall that the formula is going to be for the slaw, the integral of E multiplied by D L equals minus the derivative of the magnetic flux D five D over DT. OK. So we want to input everything into this formula here. So I'm just gonna start with working on the left side of things and the right side. So first on the left, we have the integral of E multiplied by D L. And in this case, using the sign convention for Faraday's law, we get that the direction of induced electric field E is going to be clockwise and around the square loop. And this is because we want to use the right hand rule where the magnetic field is pointing into the page. And therefore we have our other fingers rotating clockwise. Therefore, induced electric field is E is going clockwise. So in this case, the integral of E D L because it is a square loop, we wanted to be equal to the integral of E multiplied or yeah, E multiplied by D of the sites essentially because we have sites here. And this will just be E multiplied by four times the sides just like. So, and then for the right side, we have the minus D flux over D T and recall that a flux can be calculated by multiplying B with A or the area. And this will then be over the T and this will then be minus D B R minus D be multiplied by the area of the square loop is side squared D T. And recall that the side or the side squared or the area is going to be constant, so we can pull it out. So minus the flux B over D T will equal to minus side squared multiply by D B over D T just like. So, so we can then combine the left side and the right side into our faraday's law. So the integral of E D L equals minus D Phi B over D T will equals to our left side. That will be E multiplied by four side, four times site. And the right side will then equal to minus side squared over deeply over D T. And we essentially can calculate all of this right now for finding the E here. Okay. So the way we want to do that is we can just first across one of the sides and then we want to remove the four to the other side. So essentially E is just going to be minus 1/4, multiplied by the side multiplied by D B over D T. And earlier, we have determined from the problem statement. The site is 0.25 m and the DB over DT is going to be 0.0.065, that's like per seconds. There are 65 Tesla per seconds. And therefore the E It's going to be 4.06 times 10 to the power of -3 full per meter just like so okay. So now that we found the E, we can actually calculate the induced E M F. So we know that the ends of the loop are separated by a distance equal to the perimeter of the square loop due to the presence of the insulator. So because of that, we know that induced E M F is going to equals to the E multiplied by essentially the perimeter of the loop four times the site which will equals to four point oh six stem stem to the part of minus 34 per meter Um multiplied by four, multiplied by 0.25 m. And this will give us an induced EMF or four or 4.06 times to the power of -3 fault. And that will essentially be the answer to this problem which will corresponds to option B right here. So that will be pretty much all for this particular problem. If you guys still have any sort of confusion, please make sure to check out our other lesson planned. Videos and that will be all for this particular example. Thank you. Yeah.