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Q) Amctal strip that is 2.00 _ wide and 0.05 . 20T cmnt thick caries cumcnnedi 30A in a Is shown in Figure 26-41. The Hall region with uniforn magnetic field of vel...

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Q) Amctal strip that is 2.00 _ wide and 0.05 . 20T cmnt thick caries cumcnnedi 30A in a Is shown in Figure 26-41. The Hall region with uniforn magnetic field of velocity _ voltage elettor- cnsund Curr Zurt be 85 #V: Fxplain how arises. You Cun fows from kefi think of itas # right.hich dinections do the clectrons 6) Find the number_ flow? Calculate the drift spced of the frec electrons density 0f the frce clectrons strip. the strip_ In which dineclion (ic: nietl dils pushed? noimt tnc trort and .

Q) Amctal strip that is 2.00 _ wide and 0.05 . 20T cmnt thick caries cumcnnedi 30A in a Is shown in Figure 26-41. The Hall region with uniforn magnetic field of velocity _ voltage elettor- cnsund Curr Zurt be 85 #V: Fxplain how arises. You Cun fows from kefi think of itas # right. hich dinections do the clectrons 6) Find the number_ flow? Calculate the drift spced of the frec electrons density 0f the frce clectrons strip. the strip_ In which dineclion (ic: nietl dils pushed? noimt tnc trort and . point "hthc is at the bxatck: higher potential? Explain yOur answer Lnu carefully. Which nowadays thit mcgitive electrons the positive edge? Jntigm CM the cunent_ chalrge differemt? Again . explain }Our Curricre Wen nosilive how inshet carcfully. would B-field 2.OOcm



Answers

ssM Figure $29-81$ shows a wire segment of length $\Delta s=3.0 \mathrm{~cm}$, centered at the origin, carrying current $i=2.0 \mathrm{~A}$ in the positive $y$ direction (as part of some complete circuit). To calculate the magnitude of the magnetic field $\vec{B}$ produced by the segment at a point several meters from the origin, we can use $B=\left(\mu_{0} / 4 \pi\right) i \Delta s(\sin \theta) / r^{2}$ as the Biot-Savart law. This is because $r$ Figure 29-81 Problem 75 . and $\theta$ are essentially constant over the segment. Calculate $\vec{B}$ (in unit-vector notation) at the $(x, y, z)$ coordinates (a) $(0,0,5.0 \mathrm{~m})$, (b) $(0,6.0 \mathrm{~m}, 0)$, (c) $(7.0 \mathrm{~m}, 7.0 \mathrm{~m}, 0)$, and (d) $(-3.0 \mathrm{~m},-4.0 \mathrm{~m}, 0)$

In this problem, we are going to look at the whole effect and the whole effect occurs in usually thin films. Um and they are actually a way to measure material properties inside of the film. But here we have a thin film of silver and it's carrying a fairly large current to the right of 100 and 20 amps with a magnetic field 200.95 Tesla, which is pointing perpendicular to the film. And that's kind of what we need is that magnetic field pointing perpendicular, so the direction of motion of the charges. So what happens, we could use the right hand rule with uh I. L. Crosby and we would figure out that positive charges would have a tendency to gather on the front part of the film. And the opposite charge sign would have a tendency to gather on the other side. And what this does is it sets up an electric field inside the film that points in this case from the back to the front. And those charges will continue to get built up until equilibrium occurs. And the equilibrium condition is that the electric force on moving charge would be equal to the magnetic force on that same moving charge and the drift velocity goes along with the current. So are velocity of those charges is perpendicular to be. Um And the electric field is perpendicular to both the V and the B. Um And we can write then the condition for the balance as E equals. Let's get the right color here. E magnitude wise is equal to V drift times B. Um Now a reminder that the magnitude of E is also proportional to the um voltage difference between plates which were kind of thinking of the two sides of the film as being plates divided by their spacing. And we see that the spacing is Z. One. Yeah. Um And that delta V. Is known as the hall voltage. And it's usually the parameter along with current that you measure while you're doing your experiment on the material. So we know the magnetic field. But in order to find that hall voltage, what will need to do is first of all find our drift speed salt for the electric field. And then we can solve for the hall voltage. So let's kind of take that in steps are. First step is to find the drift speed which is something we can relate to the properties of the material. And of course then the electric field will pop out and finally the hall voltage will be determined um from the electric field. So there is a relationship drift speed is related to the current in a fairly simple way, solve rate that equation down that the drift speed of the carriers in the material. And since this is a metal, we know that they are the electrons. But that drift speed is determined by the current divided by the number density of the carriers, number per cubic meter times the charge per carrier. And then cross sectional area. And in our slab, what's presenting the cross section is um fairly easy to determine kind of color. That in with a sharp truce has the area. So that's just a product of the Y one, C one. And um we know the charge on an electron of course, and the density. So we can quickly write down the numbers that we need in this equation. And since I'll put everything in S. I. Units, I know that the drift speed will come out in meters per second. Yeah. Okay, let me just get all these numbers in here. And then the area is um C one times Y one. So I'll get those numbers in there as well. But they multiply out to be 2.71 times 10 to the minus six in terms of square meters. And when I put that together, not surprisingly, the drift velocity inside of things like metals and wires is surprisingly small. Um it's usually of the order of a fraction of a meter per second. So you don't want to rely on electrons whizzing from one place to another inside of a material to make your circuit work. It's usually um the electric field which travels at the speed of light inside the material that causes your circuit to turn on and respond to changes anyway. So that was number one and then number two is find the electric field. So that is um 4.73 times 10 to the minus three times the 0.95 Tesla all in Si units. And so our electric field turns out to be quite small. Let's see 4.49 times 10 to the minus three. And I'll put this in volts per meter. And then finally our hall voltage is the electric field times see once in terms of magnitude. Anyway, um, and of course the side with positive charges gathering on them is going to be the higher potential side and get everything, get everything in S. I. Units and you're voltage will come out vaults. So it's a really typically a very small voltage. So in order to pick up the hall effect and detect it, you really need um, some careful measuring devices that can measure of the order of micro volts and block out other signals that could come in and ruin the experiment.

In this problem, we apply a bio say words law by your say a word more to complete the main T field due to current caring segments at various locations. Bye. Or say what law says b it x Y Z is equal to mu nod divided by four pi which are a constant times. A current times Delta is cross our unit director divided by our square which is a constant mule I divide before by, um, Delta is cross our director derived by our three. We rotate the unit director on care into our vector divided by our so our camp is a call to our matter developed by its magnitude. Then our Delta X s story is a delta s main chute times in the Jacob direction and our factories excite plus wah jee plus z k the cross product use us Del Delta is cross on physical thio. Delta is Z I can't minus X k kept then for the what a off the problem. But a the field on nosy exes Between zombie 0005 meter it will be four by times 10 to the ball with minus seven, divided by four by five square power through You're too this above. We have two MPR current times three times dental power minus two meter downs. A five meter in I kept direction, which gives us 2.4 times dental power minus 10 Tesla in high cave direction. Similarly, for part B um man tick, he'll be at 06 meter and zero will be zero since Izzy Musical The X, which is 1/4 0 For her part, see off the problem. Similarly, be seven meter seven meters zero we can find by substituting a values a magnetic field or for minus 4.3 times dental bar minus 11. Tesla in KK of direction in body for the points minus three. Me to remind us for. And zero. We get the man tick field all from 1.4 times 10 to the power minus 10 Tesler in K kept direction.

Hi. And this given problem there is a uniform magnetic field which is directed into the plane of paper. Is shown by the process. Within this magnetic field there is a metallic Lord. Yeah which is being rotated about its. And see that role is named as cd frequency of rotation of the road. That is given S. F. Is equal to 5.0 revolutions or second end of this role between terminals C. And T. That is given us. L. Is equal to it is centimeter. Or also write it By zero point it zero peter is involved magnetic field. This is having a magnet europe 0.30 Tesla. So the rotational emotional IMF induced cruz. The ends of this roar that is given by an expression of B. L. S square into omega. And for omega we know that is given us to buy earth answering this too. Finally we get this expression to be bye into the earth. It will be into early square. Well finally plugging in all the known values here for five. This is 3.14 in two frequency Which was five r/s. Was supplied by a magnetic were Which is 0.0. Tesla Multiplied by the square of the length of the rolls. 2.8 meter. The whole sweat. Finally the CMF in news here comes out to be 3.0 hold rocks in it. Which is the answer for this. Given problem here. Thank you

So here we have a section off. This is very buzz on. We know that the drift velocity is related to the current density by the secretion on current density is given by how current over area are physical to live one times. So using this equation, we find the current density and used this value ofthe current density. In this, a question to find dying city on that comes out to before 0.7 times. Stand to the part, finest relatable a seconds. Now, for the second part, we know that in order for the night force to the zero, the electric field is to remind we have to determine the electric field our side. In order for the net force to this Edo, the electric force should be equal toe the magnetic force in magnitude on the directions off These forces should be opposite to it. So we like this. Two forces electric forces charged the magnitude of charges times the magnitude ofthe electric field on the magnet. E force has magnitude cue, more dying slowly drift velocity times the magnetic field, which only has a component along my direction. Andi, simplify desecration and we find the electric field to be just value and solve this and we get the magnitude ofthe electricity before 0.5 times 10 to the five minus three horse. It's going us now for their direction. You know that the drift velocity off the electron is in the opposite direction to the kind I by right and rule beget the cross product ofthe V N B to be along the negative city action. So, as you see here, are positive seeing is downward. So because Lee is along the negative direction on DH way have the magnetic force to charge times this cross presidencies charges there for the electron. The magnetic force is in the positive direction on from here. We can say that the electric for should be along negatives in direction from this equation over here on Since jars is negative, the electric force electricity, it should be opposite to the ineptly force. So the league trophy should be along posited city election. Now the whole enough is the potential difference between the two ages off the strip at sequel to Seo Ahn think what do you want? So using the business from Park being where we found the magnitude of the electric field, we can find this image, which is gonna be quite oh, the magnitude ofthe electric field that everyone on that gifts that you have to be Mike revolt against a 53 dyin standing by when Six World


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