A single loop of wire with an area of 0.0900 m^2 is in a uniform magnetic field that has an initial value of 3.80 T, is perpendicular to the plane of the loop, and is decreasing at a constant rate of 0.190 T/s. (b) If the loop has a resistance of 0.600 Ω, find the current induced in the loop.

Video transcript

Hi, everyone. In this particular problem, we are asked to determine the current induced in the loop if we have a copper wire, which has a resistance of 0.5 M, which also produces a coil that only has one complete turn with area of 0.4 m squared. So the turn is placed in a uniform magnetic field oriented parallel to the axis of the loop and has a magnitude of 6.21 tesla. And the magnetic field actually increases at a rate of 0.315 tesla per seconds. So from this information, we know that the only thing that's changing Is the magnetic field strength itself which increases at a rate of 0.315 per second. So let's start with creating a list of everything that is given. So first, we have the resistance which is 0.5 hope. And then we also have the area which is 0. m square. And then we also have the actual magnetic field itself with a magnitude of 6.21 Tesla. And the rate I'm gonna write it down as DP over DT the rate at which it increases is at 0.315 tesla per seconds. Okay. Just like. So now that we know all this information, we want to recall that the changing magnetic field or the D B over D T is actually is what's causing uh a change in the magnetic flux through the loop because of there's a change in the magnetic flux recall that there will be an induced E M F. So because there's a change in the magnetic flux, this actually induces an E M F in the loop which causes a current to flow in it. So I'm just gonna kind of summarize that. So there is a change in B which introduces a change in the magnetic flux which induced uh E M F which kind of causing the wire to have a current. Okay. This is all because we can, we know that. So the induced E M F and the current is related with this formula right here, which is essentially just the normal vehicles are. And then next, we know that the Indian E M F can be calculated using the change in magnetic flux over time. So delta Phi B over delta time just like that. And we also know I'm just gonna lastly, we also know that the flux itself is calculated by multiplying be multiplied by area multiplied by cosine Phi or Koh sine of an angle or cosine theta Where in this case, the five equals to 0° because it is oriented parallel to the axis of the loop just like. So okay, I'm just gonna rewrite it here to make it clearer for you guys. Okay. So we know that there's a change in the magnetic field which causes a change in the flux which cost us an indu CMF and there's a current running into it. So we know that in this case D A is constant and B is changing. So with this formula right here, we want to input all of this information into this formula to actually essentially get differentiating B with respect to T. Okay. So I'm just gonna write rewrite this formula first, the flux over the T which will equals two D of B multiplied by a multiplied by cosine of zero, cosine of zero is gonna be one Right over the T and this is gonna be close to one. So essentially the induced E M F is going to be recalled that the A is going to be constant so we can pull it out. So this is going to be a multiplied by D B over D T just like so okay. So now we can actually start plugging in the values to get the Indie CMF. So plugging in all the values we get to induce E M F to be a 0.4 m meters squared, 0.04 m two. And then the DB over DT is listed here, which is going to be 0.315 tesla per seconds. So the inducing MF is actually going to be 0.0126 fold. So from the India CMF, we can plug this into our initial formula here to find what the I S because we are given what the resistance is of the system. So I is equals to E over R so E S 0.12 or epsilon is 0. fold while the resistance is given to be 0.5. So this gives us a value of I 0.2 amp. And that will be the answer to this particular problem. So That will corresponds to option C that we have here, which is 0.0252 app. And that will be pretty much all for this particular problem. If you guys have any sort of confusion on this, make sure to check out our other lesson plan or lesson videos and that will be all for this particular video. Thank you.