Ph1c Analytic Quiz 2 Solution

Transcription

1 Ph1 Analyti Quiz 2 olution Chefung Chan, pring 2007 Problem 1 (6 points total) A small loop of width w and height h falls with veloity v, under the influene of gravity, into a uniform magneti field B between the poles of a magnet. The loop has mass m and resistane R. Throughout the problem, you may assume that the field of the magnet is uniform between the pole faes and vanishes outside. For parts a,b, assume the loop has only partially entered the region between the magneti poles. (a) (2 points) Find the magnitude of the urrent that flows in the loop. What is its diretion? peify lokwise or ounterlokwise as seen from the +y axis (as in the above figure), and justify your answer. uppose the loop has fallen a distane z < h into the field region. If we take the normal to the area of the loop to point in the y diretion, Φ = B da = B( haty) (daŷ) = Bzw From Faraday s law, the indued EMF in the loop is given by ε = 1 dφ dt = Bwv I = V R = Bwv R 1

2 Using Lenz s Law, we find the diretion of the urrent to oppose the hanging flux through the loop. As the loop falls, the net flux beomes larger in the ŷ diretion sine z inreases with time. To oppose this, we need the indued flux through the loop. As the loop falls, the net flux beomes larger in the ŷ diretion sine z inreases with time. To oppose this, we need the indued flux to point in the +ŷ diretion. The right hand rule shows that the urrent must flow ounterlokwise (as seen from +ŷ). (b) (1 point) What is the net magneti fore on the loop? Reall the formula for the fore on a straight wire of length L in a uniform magneti field. F = L I B For the bottom edge of the loop, I = Iˆx and L = w. This gives rise to F = w (I( ˆx) B( ŷ)) F = wib ẑ = B2 w 2 v 2 R ẑ This is the fore opposes the fall of the loop. ine the top of the loop is not yet in the B field, it feels no fore. The vertial sides of the loop have I = ±Iẑ (depending on whih side), and L = z. These lead to equal and opposite ompressive fores on the loop in the ±ˆx diretions, but no net fore. () (2 points) While entering the field region, the loop reahes terminal veloity v t, where its net aeleration is zero. Find v t, negleting the field due to the urrents that flow in the loop itself. The motion of the loop is just given by F net,z = F g + F m = ma z At terminal veloity v = v t, a z = 0, and so F net,z v t = mg + B2 w 2 v t 2 R = 0 = mg2 R B 2 w 2 (d) (1 point) Now onsider the ase when the loop is entirely immersed in the field. Desribe the loop s motion, and explain the reason for its behavior. One the loop is entirely between the poles of the magnet, the flux through the loop is Φ = Bwh = onstant dφ dt = 0 2

3 Without a hanging flux, no EMF or urrent is indued and there is no magneti fore on the loop. Thus the loop falls normally under the influene of gravity, aelerating from its terminal veloity (found above) aording to v(t) = v t + gt starting from the time of full immersion in the field. Eventually when the loop reahes the bottom of the region of magneti field, it will be travelling faster. As the loop leaves the field, a similar effet will our where the hanging flux results in a magneti fore that again slows the deent of the loop. ine F m v, the magneti fore ould be quite strong initially, and the loop ould slow to lose to v t before fully leaving the field region. Problem 2 (4 points total) A toroidal oil with retangular ross setion has N evenly spaed, tightly paked turns. The inner radius of the oil is a, the outer radius is b, and its height is h. A separate long straight wire is plaed along the axis of the toroid, extending from z = L to z = L, where L is muh greater than a,b, or h. The ends of the straight wire are onneted by a semiirular ar of radius L, forming a wire loop. (a) (1 point) Find the mutual indutane M of these two loops There are two ways to ompute this (letting I flow in either loop), and both of whih give the same M by the reiproity theorem. Either ase an be done here, and the other is done for part b below. Consider the ase that a urrent flows in the torus. From the homework problem 6.14, we know that the field outside vanishes, and inside the torus the field is irumferential given by B torus = 2NI r ˆφ 3

4 Taking the semiirular surfae of the long wire loop, we see that it intersets the torus in a square ross-setion. This is the only region that generates flux sine B torus = 0 outside the torus. Φ = B da Φ = Φ = ( ) 2NI r ˆφ h/2 b h/2 Φ = 2NhI a ( ) daˆφ 2N Idrdz r From Faraday s Law and the definition of mutual indutae, ε = M di dt = 1 dφ dt M di ( 2Nh = ln b ) di dt 2 a dt M = 2Nh ln b 2 a (b) (2 points) how expliitly that M 12 = M 21. Explain why the B field of the semiirular ar of the wire loop an be ignored when alulating Φ 12, the flux through toroidal oil due to the wire loop. Consider the ase that a urrent I flows in the long wire, and it generates B wire = 2I r ˆφ where we have ignored the semiirular ar for now and treated the wire as infinitely long give L a,b,h. Idealizing the torus, we onsider it being made of N square loops. The total flux through the torus is just N times that of the flux through a single square loop. Φ = N B da Φ = N Φ = N ( ) 2I r ˆφ h/2 b h/2 Φ = 2NhI a ( ) daˆφ 2Idrdz r This is the same flux as before, and similarly we get the same value of M. M = 2Nh 2 4

5 Regarding the semiirular loop, using the Biot-avart Law, we an ompute db = I L 2dl ˆr 1 L 2 ine the length of the ar is πl L, the B field due to the semiirular ar is then 1 L. We know L a,b and thus B wire B ar. This justifies why the B field of the semiirular ar of the wire loop an be ignored when alulating the flux through the toroidal oil due to the wire loop. () (1 point) The axial wire is now displaed a distane d < a towards one side of the torus. What is the mutual indutane of the system now, M? There are two ways to ompute this problem, letting I flow in either loop. However, this time the ase of omputing the flux through the torus due to the off-enter wire is too diffiult. Instead, onsider I flowing in the torus, and ompute the flux through the semiirular surfae of the large loop. As in the above alulation, the flux omes only from the square ross-setion where the surfae intersets the torus. Furthermore, the flux is exatly the same as it was before sine B torus only depends on distane r from the torus s axis and not where the long wire is. Here we use the assumption that d < a so the intersetion is still the whole (b a) h square. Thus M = M, the same as it was in the above ases. 5