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Revaw0.850-ITI-Icng Faris pulledt2 Fghl siesdu 6.00 I!/ $ pappendicular uniam 0.Sl magnetic Fie d The bar rdes Metal railg conneciedtnrouoh {arirtor (Figure I} a0 t...

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Revaw 0.850-ITI-Icng Faris pulledt2 Fghl siesdu 6.00 I!/ $ pappendicular uniam 0.Sl magnetic Fie d The bar rdes Metal railg conneciedtnrouoh {arirtor (Figure I} a0 the apparatus makes compate circuit Igncre the resistance 0 the bar and Ihe rails Calculate tha magnitude_ the emf irducac tne circuuit Eyores: Vouransuer Wilh Inc appropriatc unils. Value Units Submit RequerAnaWer Part Find tNe direciion ofthe current induced circum by Using tha magnetic force charaes Momng Dar Faradav" Lanz'< awy clcoMisA Flgure Ccuntercicckwes Submit requebt Anbwer Part € Cuculale currurl Whcuegh Iha rosislor: ansie with the appropriate units Express Value Units



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A 1.50 -m-long metal bar is pulled to the right at a steady 5.0 $\mathrm{m} / \mathrm{s}$ perpendicular to a uniform, 0.750 - $\mathrm{T}$ magnetic field. The bar rides on parallel metal rails connected through a $25.0-\Omega$ resistor, as shown in $\mathrm{Fig}$ . 29.36 , so the apparatus makes a complete circuli. You can ignore the resistance of the bar and the rails. (a) Calculate the magnitude of the emf induced in the circuit. (b) Find the direction of the current induced in the circuit (i) using the magnetic force on the charges in the moving bar; (i) using Faraday's law; (iii) using Lenz's law. (c) Calculate the current through the resistor.

Hi. In the given problem 1.50 meter long metal bar. It's moving over the barrel reels. The land for this metal bar is one point 50 meter. The steed with which it is being pulled towards right is 510 Meet up or second, the magnetic fear. But then vehicular to the plane off paper and getting into it as here, showing the figure These are there to metallic rails. This iss a metallic bar. It is being pulled towards right. The speed we and these two rails are connected. You get a suspense up the magnetic field in this region misdirected into the plane off paper for 10. Vehicular like this. The magnitude of this magnetic field is 0.750 s LA and the assistance which is joining these two rails is 25.0 home. No. In the first part of the problem, we have to find the magnitude off the anything used in the circuit. And this Emma will be in views across the ends off this metallic but and expression for this elephant news. This is known as emotional unit and the expression is it is equal to the product off magnetic field, the speed and the length off this metallic rock. So when we put these values for magnetically, this is 0.750 s LA What a speed This is 510 meter per second and for length This is 1.50 meter and then we calculated the value comes out to be 5.62 by world, this is the enough induced in the circuit across the eggs off this moving metallic bar. No. In the second part of the problem, we have to find the direction off current induced in this metallic bar. A b so who find it? Use lemmings, right hand rule, whichever the rule we used a direction off the current induced in this metallic bar will come out to be saying so if we use Fleming citing rule according to which the stretch. But come the index finger and the middle finger off our right hand perpendicular to each other then and come gives the direction off motion of this bar. The index finger gives the direction of magnetic field which is in work on the middle finger, will give the direction of current induced in this metallic bar which will come out to be upward in this metallic bar, means from B to a so using this flaming site and rule the direction of current in this metallic bar is from B do. Yeah, so whichever the rules we apply here whether it is magnetic force on the charge in the moving bar or using Faraday's law or using lenses law using all the rules, the direction off currently come out to B from B to a Now, in the third part of the problem, we have to find this parents the magnitude of current flowing through this circuit and that magnitude of current will be given using form so according to which current is given by the ratio off Yemen by the resistance of the circuit. So it will come out to be 5.625 world to be invited by 25 home and this value off current will come out to be 0.22 fine Ambu This is the answer for the third part of the problem Thank you

For the first item off. This question we use for it is law which states that induces electro motive Force is given by minus that derivative off the flux of magnetic field with respect to time. The area that we have to use them for these law is this area is the area off the circuit, and that area is changing because this part of the circuit is moving to the right. Then we can calculate the flux as follows. The flux off magnetic field is given by the integral off D A, which is an area element times the magnetic field. Be now notice the following the orientation off that surface A is given by your vector that is normal to the surface. A factor that is normal to the surface is a factor that is pointing either our ports or in words. We can choose the orientation of that factor. It is conventional to choose, eat as pointing outwards. So the vector e should be something like this. Then I'm going with this convention. The reform. You can see that A and B are untie parallel so we can write it as minus the integral off D A times be where now we are working with magnitudes The magnitude of the magnetic field is constant So we have miners b times the integral off the A which is just the area. So the flux is minus B times a The area off that rectangle is given by the product off its size. So the flux is my news. Be times l which is these side times is more l which is disorder site Now we can take the derivative respect Time to get the following Mine has been times l They are constant times the l divided by the teeth The LTTE is the reach off change off l which is the same as the velocity off this bar search results in minus b times l a times we there afford induce it e n f is minus minus beat LV which results in B times L A times. We therefore plugging the numbers we get 0.75 times 0.65 times five, which results in an e m F off 2.4375 which can be rounded to 2.4 votes. And this is the answer to the first item before proceeding. Let me clear my board. Okay. The second item has three super items in all of them. We have to do to remind what is the direction of the current, but using different methods, The first item would have to use the magnetic force that is acting on the current. So you let us suppose that there is a charge seating right here? Discharge is a charge Q. Then the magnetic force acting on these charge is given by cube times V cross beat. We can see that in the situation V is pointing to the right while B is pointing inside the board like this. Then using the right head who we conclude that the magnetic force we will be pushing the charge upwards. So the magnetic force pushes the charge in this direction, which means that they induce a current is a counterclockwise current like this. Now, in the second soup item, we have to use foreign Desslok Well, you noticed that the flux is negative and it's negative and decreasing the reform. We have the derivative off the flux off magnetic field beings. Moledet Zero before to compensate for that we need an e M F that is bigger than zero. It means that, according to the orientation that we had chosen that the electric current should run counter clockwise. So this is how you use for a these lot to solve these questions. Now for the last item, we have to use lenses Low Liz Low is very similar to far these law, but instead of treating the E. M F, it treats the magnetic field directly. So you you have to think about the same things. Something has to happen in order for the flex. Two becomes motor. Two things can happen. They are quite equivalent. From a general point of view, these two things are there. Should there must be a positive IMF more. On the other hand, some magnetic field should be induced in the opposite direction off the doctor on magnetic field in order to lower the floods off magnetic field lines. In order for this to happen, these inducing magnetic few to shoot points to an opposite direction with respect to the background magnetic field. So we have a background magnetic field pointing in wards, the reform, these reducing magnetic future point outwards and the magnetic field pointing our parts is induce it by a current that goes in this direction in that circuit so that outwards pointing magnetic field is endings that by a counterclockwise current, and this is how you can use lenses low to justify the answer for the second item. Finally, we have to calculate the current from the resistor. For that we can use orders. Law homes Low tells us that the voltage difference is equals to the resistance off that resistor times the current. So the current is the voltage divided by the resistance on these results. In chu 0.4375 divided by 25 noticed that I'm using the voters. We wrote approximations to get a more accurate results and then the results in a current that is zero 0.975 and this could be rounded to 0.0 98 which is 98. Milli amperes uses the answer to the last item

Think Surgery, Chapter 29 Problem 31. So, in this problem, yeah, they circuit with a resistor here, and these two ends of the circuit leased to wire. They're connected by a medal bar is called a little bit better. They're connected by a metal bar here, and this is all in a B field that is pointing into out of the page so he field out the page is the least l and they tell us it is moving towards spar is moving towards the left. Okay, so it's right. Simply saying sounds 036 meters is the length of the bar tells us the resistance of the resistor is 45 homes, and it tells us the magnitude of the Byfield is 0.65 kessler. Okay, so we want to figure out the direction of the induced current in the circuit. So to do that, we use of Lindsay's law and loses laws. The one that says the induced current produces a D field that opposes the change in magnetic flux. Okay, so so look at this flux here for a second. You know, the magnetic flux is that the field times the area So the Byfield was constant. Right? So this is constant. But as the philosophy moves to the left, this loop of our circuit is decreasing. Some things are flux and decreasing sorrow. Far flux is out of the page and decreasing. To oppose that, the induced magnetic field must be out of the page. Okay, so now we got to use the right hand rule and figure out which direction of a current in this loop produces a Byfield out of the page. And if you do the right hand, really? You see that this is going to be counterclockwise. Okay, Cool. So now let's figure out what the current value is because we're going to need that for the second part of the problem. So let's figure out the magnitude of this induced current. This is given by the path over the resistance, which we know the IMF has given my fair in his laws. The negative change didn't magnetic flux. So we're just worrying about magnitudes here so they could drop the drop the negative sign you, Khun, Simplify this too. B the tea. Since he's constantly and now we can even simplified. The peace of this becomes B over are the of tea. Okay, so it says we want to do this at the instant where it's moving 5.9 meters per second on the left so we can now plug everything in. So we have 0.65 Tesla's time, 0.36 meters times 5.9 meters per second, all over 45 homes. And this comes out to be a 030 seven stamps. Cool. So now we can figure out the second part of the problem, which is asking about the great ofwork. So the rate of work has also known as a power. So we can find this two different ways when we can do by the a mechanical work of thieves force times the velocity of this bar's moving but by energy conservation, we know that this is exactly equal to the thermal power loss which has given his eye square are in the circuit. So since we know high from the first part, we can easily calculate this of the power is given as 0.307 squared times 45 owns and this comes up to the 0.0 for two parts awesome. And again, since we have a kind of clockwise current, this is Mrs the magnitude of occurrence and this is the power dissipated.

So we can draw exactly what's going on here for party. We can say we can just simply draw. We have the bar, eh? Up here. Be down here. Going coming right out of the page. Just draw a circle first. Coming right out of the pages. The magnetic field. We have the positively charged particle here. Going straight up would be the force. And then going straight to the right would be the velocity of thie, Positively charged particle. So for party, it's asking us to find the current here. The magnetic force is towards the top of the bar. As we can see in the diagram eso here the induced current. I is from B to A through the bar and then we can say and counterclockwise through the circuit rather in the circuit. And then for part B, we can say that here the magnetic field is into the page so we can say that magnetic fields into the page. We know that the flux the magnetic flux is increasing as thie area of the circuit increases. And this is because the flux is equal to the 92 of the magnetic field multiplied by the area so flux is actually directly proportional to area. So if the area of the circuit increases the magnetic flux off, the magnetic flux also increases. We can say that here, the magnetic field of the induced current is therefore out of the page. And at this point, Ah, we can say that to produce this magnetic field in this direction, um, induced current must be again counterclockwise. So that would be your answer for part B. That is the end of the solution. Thank you for watching.


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