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An aluminum ring of radius 5.00 cm and resistance 3.00 × 10–4 ? is placed on top of a long air-core solenoid with 1 000 turns per meter and radius 3.00 cm, as sh...

Question

An aluminum ring of radius 5.00 cm and resistance 3.00 × 10–4 ? is placed on top of a long air-core solenoid with 1 000 turns per meter and radius 3.00 cm, as shown in Figure. Over the area of the end of the solenoid, assume that the axial component of the field produced by the solenoid is half as strong as at the center of the solenoid. Assume the solenoid produces negligible field outside its cross-sectional area. The current in the solenoid is increasing at a rate of 270 A/s. What is the magn

An aluminum ring of radius 5.00 cm and resistance 3.00 × 10–4 ? is placed on top of a long air-core solenoid with 1 000 turns per meter and radius 3.00 cm, as shown in Figure. Over the area of the end of the solenoid, assume that the axial component of the field produced by the solenoid is half as strong as at the center of the solenoid. Assume the solenoid produces negligible field outside its cross-sectional area. The current in the solenoid is increasing at a rate of 270 A/s. What is the magnitude of the magnetic field produced by the induced current in the ring?



Answers

An aluminum ring of radius 5.00 cm and resistance 3.00 × 10–4 ? is placed on top of a long air-core solenoid with 1 000 turns per meter and radius 3.00 cm, as shown in Figure. Over the area of the end of the solenoid, assume that the axial component of the field produced by the solenoid is half as strong as at the center of the solenoid. Assume the solenoid produces negligible field outside its cross-sectional area. The current in the solenoid is increasing at a rate of 270 A/s. What is the magnitude of the magnetic field produced by the induced current in the ring?

Okay, So finally managed to open PMF. Can we go to absolute negative? Divided over DT? I'd be Here is the magnetic flux t Here's the time there will have team Manitou off the, um a t c equals d five d over DT. And we know the magnetic flux can be able to be a over to beast my night if you ace area. So now in his b A over to substitute for five b So have many to you off the math is equal to d o D C. Test be over to, which is you go to a over to test the or DDB and when Tony man, any fuel for this Illinois can be with the Museo tonight. You zero is the probability Constant only see number terms and eyes occurrence. So now he used museum on my substitute will be so half Uh oh, absolutely. Yeah MF you see a over to a test Todt test museum on I which is people to a over to test new zero and test the i o d d and we know area can go past where as the radius off These so annoying So no, we can do some real regiment here eventually will be magnet to off the m f s equals pi r square over to test Museo and and intensity I over DJ It's not. It's determined occurrence the Indians currents can be go to the Manitou off the MF over resistance which is equal to pi r Square over to our chance Museo and testy I over DT And we know the radius off the S. Illinois is given as three centimeter, which is Joe pointing Joe three meter and the resistance is given as three times 10 to the power of the name for Omega. And we know from Billy Constant coming with four point has 10 to the power off next seven. Tessa has meat over in here, and the number returns is came as 1000 tons per meter and we know the rate of occurrence is 270. NPR per second now has plugging these values spend to the equation to determine the currents which is equal to, um, I times 0.3 meter to the bar to over two times three times 10 to the power off before omega and then more Bubba um you zero in which is well by times 10 to the power of the next seven. Tesla has meter around here in times 1000 a meter, sometimes 270 MPO per second. And this will give us India currents. Is you goto one point by 99 and here, which is approximately equal to 1.60 I m here. So now it's determining the my nephew. So the my night, a few full party Kevin expressed in such a way things This is the question 40 magnetic view off the rain, which is he was a museum. I would to our I want here is the radius of the ring. So we know the radios off the ring is came as five centimeter, which is your 50.5 meter. And we know India's currents in this case is one point. Find an A. Therefore, when plugging these values back into the equation to determine e my nephew, which is you go to full by times 10 to the power over 97. Tessler has a meter, but I'm here times 1.599 a. M. Pierre over two times 0.1 Peter. And it will give us the man you feel is you goto 2.9 turns 10 to the power of ah, big by Tesla. If we can worry too. Uh, micro Testa me buying it, if you can, you go to 20 0.9 Michael Tessa things, um, Long micro Tesla. You see, people tend to a power off the next six. Tessa. Now let's take a look at Parsi so you could take a photograph. You in town at the direction over the a magnetic view is point a little writes, And when his point all the rice, the occurrence is increasing. And we know that the induce my interview is always opposed the original and that if you therefore the direction every minute My magnetic few produced by the Indians currents should be pulling toward the left since is always going against the original view. And these are the answers for this question.

So we are asked to find the induced current in the ring where the aluminum ring has a radius which I call our sub are for radius of the ring of 0.5 meters, right and a resistance of capital are of three times 10 to the minus four homes and is placed in a still annoyed, which has 1000 turns per meter. So in his equal, 1000 per meter, Um, in the smaller radius of the solenoid, ourselves s 0.3 centimeters. Okay, um and also tells us the current and the solenoid is increasing at a constant rate. So delta I delta t is 270 amperes per second. So how can we go ahead and find the current than from this well, from owns lock current, I is equal to the induced Ian meth Absalon over Resistance are okay. We only really care about the magnitude of the induced e m f. So let's go ahead and do magnitude of induced ium f here. Okay, so what is the magnitude of the induced Ian met? Because we know that resistance capital are so let's find that magnitude of the induced e M f well, the induce TMS magnitude is equal to the magnitude of the change of magnetic clucks belt If I divided by Delta t. Okay, so this is going to be equal to the magnitude of Delta. Um, what is the flux? Well, it's the magnetic field times the area, the magnitude of that value divided by Delta t Well, the area's not changing, right, But the magnetic field is so this is going to be equal to the area times the change of magnetic field dealt to be divided by Delta t. So then the question is, what's the change of magnetic field? Okay, let's go ahead and get a new page going here. So assuming that the magnetic field produced by the soul annoyed over the area at the end of the solenoid is 1/2 assed strong as the magnetic field at the center of the sill, Annoyed then, ah, dealt to be here. Let's go ahead. Don't to be is going to be equal to 1/2 the area or excuse me, the magnetic field in the solenoid call that Delta bi So, Bess. Okay, well, this is equal to 1/2 you not which is magnetic per me, a bit permeability of free space. It's four times 10 to the minus seven. You can look that up if you want to know more about that constant. It's a very commonly used, constant times the number of turns in the solenoid per meter times the change in current in this Illinois will call that Delta II. Okay, so now plugging these values back in to the expression for the induce TMS magnitude we find that the induced E m f is equal to you. Not so I'm pulling all the constants outfront times, end times the area of the solenoid, which is pie. Times are so this squared all divided by two multiplied by built I over delta t So I just plugged that value in for the change of magnetic field right into the equation. Well, we know everything in this in this expression, you know, it's just that constant. Four pi times 10 to the minus seven. We know the number of turns per meter pies, a constant we know are so best the radius of the solenoid and we know the change in current with respect to time, we were told that the change in car with respect to time is 270 peers per second. So plugging those values in this expression, we find that the magnitude of the induced IMF is 4.79 times 10 to the minus four. And of course the units are volts. So now going back to homes lock where the current is equal to magnitude of the induced E M f, which we just found divided by the resistance, which we were told it's a plugging those values into this expression. We find that this is equal to about 1.6 in the units for current are in Pierre's. That could be boxed in is the solution to our question.

In the given question, there is a solid night like this invests. The wire has been bounced in this manner. These are the number off terms off the wife passing tudo solid night And there is a ring. This is the rain passing near the end Left end off the solid night The radios off this ring This has been given us five centimeter or we can see my 0.0 into 10 days to the power minus two meter. This is stands off this coil Yes, 3.0 into temptation Par minus four Home In the solid night there are 1000 terms four meter the number of cars Body implant is given us how's end? Yes. We consider magnetic field near the center of the Sala. Annoyed as be not Then it just given asked The magnetic field near the end of the solenoid is half that at the center off the solid night. So it will be given us, Have you not and I Because the expression for magnetic were at the center of the still annoyed is given as new not And I there this I is the current passing through the solid night whose direction is as shown in the figure. So do do this direction The right end. Oh, this solid night will behave like not ball because this direction is and d clockwise and it's left. And when we have like South Pole, you know this is south and this is not We will use these directions later on. For the time being, the time rate of change of current through the passing through the solid night is given us 70 m beer, but a second. So the current induced in this coil having a singer done, maybe given asked the IMF in used divided by resistance off this coil I mean Asper Ferraris Laws off Commemorating induction. Yes, In news the energies given as minus defy by DT divided by this resistance so weakened by tips like minus deed by our and be a by ditty. So rearranging it we will get my escape by our area. Being constant can be taken out minus eight by odd and de by DT off expiration for this Be here. This huh? New not And I so half you not end. We'll also come out changing the form of the question like minus k you know, And by you are d I by DT which we know exactly here. So if I put all these values, I'll get in the first part of the game problem I get for area A by our sister and I knew the value Fight 3.14 So I'll get minus 3.14 radius radius off. This solid night is our to which is given as three centimeter or three in two tenders for minus two meter Because magnetic field will be confined only this region within this region like this so flux we'll begin by this area only means area off the solenoid. So in order to find the flux magnetic locks, we are using the area off the solid. Annoyed on Lee not the area off oil because there is no magnetic field outside the solenoid. It is also given in the question so that it will become three into 10 dished apart minus two. So the whole square then in place off you not now really off munity is four pi into 10 days to part minus seven test limiter. But I'm here, so yeah, it will be four an implicit by 3.14 You could standish to the bottom line of seven in 2000 for a number of plants. But in planned and for the Aib, I didn t know 70 and divided by two in tow are and our history into tended for minus four MP year should be a unit. So when I calculated how did you get minus 1.6 MP? Here is the answer for the first part So the current passing through the coil is minus 1.6 ampere in negative sign shows the direction of this current No, The second part. We have to find the magnetic field and the center of this coil. So the expression for magnetic field of a center is new, not die by two are one now the radius off the oil will be used here for you. Not again for pie tended bar minus seven currently unused 1.6. If you look only for the direction only for the magnet you know the direction and it will be too into five into 10 days apart, minus two The radius off the court, you know, already here so But I calculated it will come out to be 2.1 into 10 days apart, minus five. Tesla is the answer for second part. Now, in the third part, we have to find the direction of this magnetic field in news as the reason behind this magnetic field is this magnetic field off the solenoid which is coming out of the North Pole and then ending into the South Pole. So the direction off inducing magnetic you is left to right. So the direction of in news magnetic field will be right toe left as far lenses Law, direction off, Indy used magnetic field will be right. Who left? Thank you.

All right. The human problem. We are having our Sahlin oId having a large number of times he has shown in the middle. These are their guns carrying a current and if shoot here. So if you look at this current from right and off this solenoid the direction of this current is counterclockwise and hence it will be in a clockwise direction. When we look it wrong, we left and then there is hurry at the end. Doctor, solenoid the radios off the ring, I think even as are too. This is our and radios off This so denied has bean given as our money Are you like this? Is the ideas off this soon enough? The number of terms for you at length in the solenoid is n system number. Well are but yeah. So the magnetic field off the solenoid at the ends off the solenoid Guinness Uh huh. New not Anna. As we know, The magnetic field near the center of the solenoid is we know tonight. So near the end, it will be half off this magnet feel things. How new? Not and I So the flux magnetic clubs linked through this solid I'd believe find the which will be given by it could be into area off that solid nine. And remember, this should be the area off solid night. So we will take it as it as the Madrid if reliance are confined Onley within this region. So if we have to find the flux linked through this coil also the flux in the only this be multiplied by eight. So if we come out to be hard, um, you know, deny not by five Are you slept? This is the total flux linked with the solenoid and as well as with this boy the time weight off change off magnetic. The hundreds of change of current in the solenoid has been given us Delta I by 10. 30. If this rate the current is increasing no. In the first part of the problem, we have to find the current in news in this. Why, as a result, off this time off rate of change of current industrial annoying, so do that. First of all, we will find the anything news in the oil which, using Faraday's laws off electromagnetic induction, will be you and ask the crime rate off change off magnetic Plus, it is linked to destroy. And we know this. So we become negative differentiation off this pipe is half, you know, and I multiplied by by car swept. So taking all other things being constant out off this durability we will get here. This is minus. Why are to set? Um, you know, in. And then this is 100 off changer electric current, which is nothing but delta. I buy the strategy just giving in the problem. So finally, the current inducing this larger coil the beginning losing homes law as, um, opinion was directed by the resistance. So finally it will come out to be minus new. Not, and by are two square invited by Wiesel are multiplied by help. I fighters 30. This is the expression or the current induced in this spoiled? No. In the second part of the problem, we have to find the magnetic field at the center off destroy so that magnetically induced at the center of the coin will be given by the expression knew Not I buy tow. Our and their R is the radius of this coil, which is Skinner's are one. So yes, magnetically will come out to be do not buy two or one multiplied by the expression for current induced which waas you know, and by divided by two are are two square Delta I 5 10 20 Here we are taking just the magnitude of this current. So they are ignoring the negative sign. Finally this magnetic feeling used represent off the oil will come out to be new, not square. And why are two square invited by four times off? Our resistance and urban really sorta bigger point were declared by I he expression for that magnetically at the center of the coil? No, In the third part we have to find the direction of this magnetically. So yes, in this solenoid this is some problem while so it will start getting like not full. And this is here. The current is clockwise So this stays off. The solenoid will be here like South Pole in which the magnetic reliance will enter. So the magnetic field lines are entering and the current is increasing, So these magnetic reliance will also increasing so we can see the reason behind the current in used in the oil is increasing magnetically lines in inward action. So this is in world increasing so their opposition, as for lenses law will be outward increasing. So the direction of currently in use in the oil should be such that So the magnetic really news at the center should be coming out off. It means we can see the directional magnetic field inducing the center off. The coil should be left. Then only it will be said, Toby, increasing in outward direction. And then these lines will be opposing these black, the lands magnetically. Thank you.


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