The magnetic field within a long, straight solenoid with a circular cross section and radius R is increasing at a rate of dB/dt. (g) What is the magnitude of the induced emf if the radius in part (e) is 2R?

Video transcript

Hi, everyone in this particular problem, we have a magnetic field which is increasing at a rate of 0.5 tesla per seconds to a circular solenoid of a radius of 10 centimeters. There's a square loop of line 30 centimeters center about the access of the solenoid. And we want to determine the E M F induced in the loop itself. So first, we want to identify all the different information given first, we have the increasing rate of the magnetic field which is D B over D T, which is equals to 0.5 tesla per seconds. We also have the radius of the solenoid. I'm just gonna write it down as our because we only have one are here, which is going to be 10 centimeters, essentially 10.1 m and then the square loop of line 30 centimeters. I'm just gonna write down as L Which is also 0.3 m. Okay. So we want to determine the EMF induced in the loop itself. And we know that the magnetic field of the solenoid is confined to the region inside of the solenoid itself. So because of that the magnetic field as a function of our will equals 204 and are bigger than our, which this are, is essentially going to be the radius of the solenoid itself. Increasing our beyond the radius of the solenoid are won't affect the flux because the B is essentially going to be zero outside of the solenoid. So I'm just going to write down B is zero outside of solenoid. That essentially is what we're saying here. Next, you want to apply the faraday's law to just find the induce E M F. So we wanna apply, induce E M F equals E multiplied by D L which will equals two minus D flux B over D T. Um Because we are looking into just the magnitude here, I'm just gonna put an absolute value throughout. So this minus sign is just essentially negligible and then at our, of E equals to 0.1 m, then we can calculate the indec E M F to B D flux over DT, which will then be recall that the magnetic flux will equals to be multiplied by A. So this is essentially D B A over D T. And in this particular example, the A is going to be constant in the solenoid and the B is the one that is changing the magnetic field is increasing. So we can pull the A out. So this will be a multiplied by D B over D T and we can actually plug everything in. So we know that the Indie CMF is going to be the area is pi R square multiplied by the B over D T. So this will be pi multiplied by 0.1 meters squared multiplied by the D B over D T of 0. Tesla per seconds just like. So, so to induce E M F is actually going to be 1.57 times 10 to the power of minus three fault. And that will essentially be the answer to this practice problem, which will correspond to option B that we have here and that will be all for this problem. So if you guys have any sort of confusion, please make sure to check out our other lessons, videos regarding this similar topics on this and that will be all. Thank you.