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230 Wrneand ridius of - circular coil that has N IJ 0.0755 directcd perpendicular to the plane of the coil.liesmagnelic lield that has magnitude ofWhat is thc magni...

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230 Wrneand ridius of - circular coil that has N IJ 0.0755 directcd perpendicular to the plane of the coil.liesmagnelic lield that has magnitude ofWhat is thc magnitude of the mugnclie Ilux $ through the coil?neneen steadily 0.165 orcr time intcrval ol 0.435 Thc magnelic field through the coil What = magnitude |E| of the emf induced in the coil during the time intcrval?

230 Wrneand ridius of - circular coil that has N IJ 0.0755 directcd perpendicular to the plane of the coil. lies magnelic lield that has magnitude of What is thc magnitude of the mugnclie Ilux $ through the coil? neneen steadily 0.165 orcr time intcrval ol 0.435 Thc magnelic field through the coil What = magnitude |E| of the emf induced in the coil during the time intcrval?



Answers

A circular coil enclosing an area $A=0.0100 \mathrm{m}^{2}$ is made of 200 turns of copper wire as shown in Figure $\mathrm{P} 31.45 .$ Initially, a uniform magnetic field of magnitude $B=1.10 \mathrm{T}$ points upward in a direction perpendicular to the plane of the coil. The direction of the field then reverses in a time interval $\Delta t$ . Determine how much charge enters one end of the resistor during this time interval if $R=5.00 \Omega .$

Hi integration problem. In a square shaped coil, the daughter turns are 50. The magnetic field is making an angle of 30 degrees. With the area vector initial magnetic clearly magnitude off magnetic field link. Initially with the coil is 200 micro Tessler. And then over the time it is increased to 600 micro Tesla. And that time duration is given a 0.400 seconds. And E. M. F induced is also given to us which is having a value of 80.0 millie volt. So using paradise laws of electromagnetic induction EMF induced has given us and times of defy by DT. The time rate of change of magnetic flux linked through the coil and here this is magnet. You don't leave. We are ignoring the sign. So for um I've induced this is 18 million world or 85,000 world is equal to the number of terms. And into here we can write it like by two minus 51 My final magnetic flux minus initial magnetic flux divided by time. Then for the flux disease, B to into A into cost three to minus B one A. Was theater divided by time. So finally here it is and a cost to divided by time into B two minus B one. So finally if we plug in all other known values, this is 80 by 1000 is equal to. For the number of terms this is 50 area we have to find this is not given to us area of the coil into B two which is 600 micro Tesla minus 200 micro Tesla. And the conversion factor for Michael, the sustained is one minus six into cost 30 degree divided by time, which is given as 0.400 So finally an expression for the area will be 18, multiplied by 0.400 divided by 1000 into 50 in two, 402 10 days par minus six into cost 30 which is ruled three by two. And finally this area if side of this square shaped coil is eight, then it's area with a square. And that a square here finally comes out to be 1.85 metarie square. So side of this square shaped coil will be a is equal to square root of 1.5. And side of the coil comes out to be 1.36 m. No perimeter of coil means we have to find the total length of the wire used to meet this coil, and that will be equal to 50 turns, multiplied by parameter of the coil means for a So here this length of the wire will be 50 in 24 in two side, which is 1.36 m. So finally, this length of the wire comes out to be 272 meter, which is the answer for this given problem. Thank you.

In this example. We have a conductive loop of wire here in a uniform magnetic field, and we're going to change the strength of the field over time at a rate of 0.73 test less per second. I mean, what we want to know is what is the strength of the induced magnetic field at the center of this loop of wire? Even though we know all this information about the problem, So we have the internal resistance of the wires, the radius of the loop, the number of turns in the loop, and then just a change in the field strength over time. So this change in the fuel strength is going to correspond to a change in flux. Specifically, flux will increase, and because of that, will have a induced EMF and correspondingly and induced current. And from that induced current, we know we'll be able to calculate the magnetic field. If you remember for a loop of wire, the strength of the magnetic field at the center of the wire is going to be starting at the center of the loop. It's going to be n times you, not the permeability of free space times the current divided by two times the radius. So if we're able to get our induced current here, we'll be able to actually be. So to get the induced current, we're going to start with Faraday's law induction, which states that you change influx over the change in time equal to the induced e m f. And once we have the induced e m f, we can use homes law to calculate the induced current. So for our induced e M f here, the change in flux it's just going to be the change in our expression for the flocks, which is going to Maybe that is a times the co sign of the angle between the magnetic field and the surface normal, which in this case, they are parallel. So the angle is zero and the coastline of zero is just one, so we can ignore that. So we're just looking at the change in B times a day. And since the area is constant during this whole time, we're really just looking at the area times changing the magnetic field. Now, of course, for our loop here, we actually have 105 coils, and we only calculated the change influx for a single quail. So really, we're gonna have to tag on an extra in here for each coil trying to squeeze it in. Okay, so now we have that the change influxes, end times a times Delta bi. So we should have that. The induced your math is minus end. Times a time is still to be over gelati. Okay, so we have the number of loops. We almost have the area, right? We don't have it exactly, but we know that this is a circular loop and that for a circle, the area is just going to be high r squared. So again we can substitute out for a here with high r squared and 11 end times pi r squared times cell to be over Delta T. Okay, so now we have our and we have dealt to be over Delta T. So we will be able to calculate what this value is, and we really only care about with the magnitude of the induced in Memphis so we can just go ahead and ignore the sign. Here. You see the magnitude of the induced your mouth people to and pie r squared times still to be over lt. And if we plug in all of those values, we will find that the induced Ian Method 0.413 volts. Okay, so now that we have our induced your math we can use owns law V equals IR. We're in our case. Our voltage is just are induced e and f e We know the resistance that we know you do, CMS that we could just solve four i total divide both sides by our we'll have that induced enough over r is equal to the induced current. So if we plug in our value for the resistance and 0.4134 are induced enough, we'll get that the induced current is equal to 0.86 04 amps, just kind of an intermediate result. And now we're just about ready for the grand finale If we, uh, right down our expression again for the strength of the magnetic field at the center of a loop of wire, with current going through it end times mu not I over and twice are where are constantly you not the permeability of three spaces just or pi times 10 to the minus seven. Tesla's tens meters over amps. Just in case you forgot and we plug in all of our values here, nothing needs to be converted or anything. They're already in the right units. We will get that. The strength of our magnetic field is one point for to times 10 to the minus three amperes. So there you have it. We calculated the strength of the magnetic field at the center of a loop of wire, with the current going through it where that current was induced by changing the magnetic field in the region of our little choir.

In the given problem, we have been given us where Lou having 50 terms and no magnetic field. There is a magnetic full also so that the normal to the plane off course is making an angle of 30 degrees. Winter magnetic field means if I draw a normal here, so this will be done. Magnetic field, which is making an angle off 30 degree with a plane off boy and the magnetic here magnitude of this magnetosphere. Initially it ISS so 100 micro Tesla or I can see 210 days to par minus six Tesla. And finally it becomes 600 Michael Tesler or 600 into 10 Dressed bar minus six. Tesla, you know, given time, Interval off 0.400 seconds. And the enough in used also has been going to us, which is 18 million volt or I can see 80 in pretenders to par minus three Bull and we have to find the side off this square, Lou. Okay, so if I apply Faraday's laws off electromagnetic induction here, I'll get the magnitude of this e m f in news he's given us and de five by DT or I can see end times off fight to minus 51 divided by time and five is given as being do it or I can be a casita because the formula for flood is the dot product off magnet. Clear with the area which becomes be a cause, Peter. So here it will be big toe accosted er minus B one it goes Tito divided by time. So if you put all these values in place off enough, it is 50 into tended to par minus three. Number off earns 50. It costs it will come out so you don't become here. Cost 30 divided by time which waas 0.4 and B to minus leven. Remaining inside will be 600 minus 200 for micro then dished up one minus six common. So if I rearrange it, I'll get area is equal to it d into tenders for minus tree into a 0.4 zeros. It would be invited by 50 into for cost 30 it is rude tree by two and 600 minus wounded will give me 400 into penetration par minus six meter square. And if I put all these values here in place off area. I'm going to get one blind it for each. And as area is the squared off site. So it's why? Because 1.8 for it. So the site will become side of the this world will become the 1.8 for eight, which will come out to be 1.36 centimeter. This is the site off the square, Lou. So experiment terribly become for into side means four into 1.36 which you will become five wine four, 376 Neater. So as the oil is having 50 terms So total land prop the wire used to make this a square. You we begin an ass 50 into a very meter. News 50 intell 5.4376 meter, which comes out to be to 71. Point it it or nearly so. 72 meter length off wire. We need to meet the square. Luke. Thank you.

So according to Friday's Ella off electromagnetic induction, we can write and USDA MF as negative Read up change off flex And according to the definition off flags we have flex equal toe dot product off beyond a That is equal toe be a because I know Kita s is the problem. We are given that plane of qualities perpendicular to the magnetic felines. So the bacteria area off pulling up oil will be parallel toe magnetic field. So here angle between area victor and bees. Zero degree. This gives us flags equal toe be times air because I know off zero degree on because I know zero is equal to one So we will get flags is equal to be okay Now we can write induce enough equal toe minus do you by DT many times. And here is the surface area off putting off oil which is Costin's so we can write induce him a physical toe minus and toe db upon duty. So this gives us induced a Memphis proportional toe. The negative rate of change off magnetic field are we can, right? This is in Memphis proportional do Then I get you slahv off the magnetic field. Now here. Graph is plotted between time and magnetic field from zero to Taiwan. Magnetic flux. Sorry. From zero to T one. Magnetic field increases here. Slope of magnetic field is positive. Now here we can write from Gino Toe Dear. Magnetic field increases. So here we have so love off the graph. Positive. But we have a Memphis. The negative of the Slav. So here Mm will be negative. And from the 31 slow off the grafitti Constant. So here we will get negative I am of so we will get Ah graph. Like this far a members is time now In the second case from he won 32 magnetic field is constant. So here's lobbies zero This gives us and you see um Africa all too zero So from t one to t two, we will have in Africa will toe zero No from Ditto to t three. Magnetic field decreases here again. Slope is constant and negative, so we will have induce him off. That will be positive


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