Chapter 31 IQ Questions - 31.2.1 Moving rod 8; Lenz law I D...

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Unformatted text preview: 31.2.1 Moving rod 8; Lenz law I D A metal rod AD is sliding to the right, along two N metal railings, with a constant speed v. ADzL. The resistance is R. Magnetic field B is constant. It points into the paper. Determine the direction of BLnail (induced magnetic field). | 1 2| 3 | 4 | in ‘n'ght‘ out ‘ left ‘ direction of BM Hint: Lenz law says that Bind, the induced magnetic field vector, opposes the change of the flux defined by the loop. Extra: Show that duMdt: BVL. Explanation: As the rod is sliding to the right, the ingoing—flux, defined by the expanding loop ABCDA, increases. To prevent the increase of this ingoing—flux, BM must point outward. Extra: The growing ingoing—flux defined by the loop is given by ¢:BA:BxL. So the rate of change of the flux dwdt= d(BxL)fdt 2 EV]... 31.2.3 Moving rod & power A metal rod AD is sliding to the right along the ]| railings. The resistance of the loop is R. B is into the paper. Direction of induced current i is shown. Denote Fan to be external force which keeps AD movin with constant v. direction of Fm _) [r <— J, Hint: Let the magnetic force on AD be FB=iLXB. To keep v constant, we must have Fm+ FB=0. Extra: Show that the mechanical power applied to the rod is given by Pmech= Fm°v=izR. (Recall: smd=BvL) Explanation: FBziLBsineziLBzFem. Direction of FB is: FB Direction of FExt is opposite to F3. Thus it is to the right. Extra: Pmech = Fm-vziLBwiemd = 'ZR. This shows that the mechanical energy, which is transferred to the electrical energy, equals to the dissipative energy. 31.2.2 Moving rod & induced current I D A metal rod AD is sliding to the right along the H railings. The resistance is R. The magnetic field B is into the paper. L=AD. Determine the direction of the induced current. 3 direction of imd clockwise 0 counterclockwise Hint: Lenz law— Bind tends to maintain the original flux. dd 0’ , 5' Extra: Based on slimf—E=—:II(M, [Ami- (20’. Show that im: BVLIR. Explanation: As the loop expands, the flux pointing into the paper increases. So Bind is pointing “ou ”. The light hand rule #3 implies that it is counterclockwise. Extra: From the expressions given above, we get the magnitude of the induced emf lewd: BVL. By the Ohm’s law, imam, IR 2 BvaR. 31.2.4 Rotating Bar cutting B—lines '\ Consider a metal bar 0A which is rotating in the plane L to the ma tic field B. Which end has a hiher otential? Hint: The magnetic force due to B asserted on q, a positive charge on the rod, is given by F=qva. This force “pushes” q from a low potential point to a higher potential point. Extra: Show IVD-Vfil, the magnitude of the potential difference is (0BR212, where RzOA and 0) is the angular frequency of rotation. m: When q is pushed by a displacement Ar, the corresponding potential difference is given by AV: FAr/q = quArfqszrAr. Explanation: By inspection on the direction of the vector—cross—product, the magnetic force on q, i.e. on a positive charge on the rod, should be directed radially inward. So a positive charge is being “pushed” by the magnetic field from A to O. This implies that VD>VA. Extra: From the hint, emf! dv=IwBrd1=wBR2/2, where the range of integration was taken from r:O to r:R. 31.2.5 A falling rod R D C A metal rod AB is falling downward along two || vertical metal railings. ABzL. The resistance is R. Magnetic field B is constant. It points out of the paper. Determine the direction of i11m (induced current). direction of iind Hint: Lenz law says that Bind, the induced magnetic field vector, opposes the change of the flux defined by the loop. Extra: If the rod weighs mg, how can one express its terminal velocity in terms of R, m, g, B and L? Explanation: As the rod is falling, the outgoing—flux, defined by the expanding loop ABCDA, increases. To oppose the increase of this flux, Bind must point inward. The right hand rule implies that iiml is from D to C. Extra: When the terminal velocity is reached, the falling speed is constant and the net force on the rod is zero. At this point the magnetic force is balanced by the weight, so imdLB=mg, where imd=BvL/R. These two expressions enable one to solve for v. 31.4.1 Varying flux & Induced emf / ----- \ [/"flx ;‘\\ / x x x l’x x x x \ XXXXX . -4‘ xxxxx/ \x x/ \z x x/ \‘-_—o’ B1att1 Bzattz Given: 1=1m. AI; t1: 0 sec, B1=1T. At t2: 2 sec, B2=2T. Find: Induced emf and, involts. 1 2 | 3 4 and 11: 11/2 11: 11/ 2 (in volts) (ii: of clock- clock- counter- counter- emd wise wise clockwise clockwise BzA—BIA Hint: g. dzlflz .Use Lenz law m d! Extra: Show the induced electric field Ema: (1/4)V/n1. Explanation: Based on the formula given in the hint, the magnitude of induced emf |8de= (2-1)TE/(2-0) =Tt/2volts. Direction: Bind opposes the increase of flux Within the circular loop. So Bind is out. RHR #3 implies that and is counterclockwise. Extra: and = 27:1“ Emd = all or Ema =(l/4r)=(1/4)volts/m. Verify the units: volts/m = N/C. 31.3.1 Levitation of a magnet Consider levitation of magnet by a superconducting surface. The north pole is pointing downward. The direction of induced current as viewed from above is: 1 2 3 clockwise 0 counterclockwise Hint: The loop of induced current should correspond to an equivalent induced dipole which repels the magnet. Extra: 1f the polarity of susupending magnet is reversed, what is the direction of induced current? Will the magnet fall? Explanation: The flux at the superconducting surface due to the floating magnet is pointing downward. The induced currrent loop prevents the penetration of the downward flux, so the induced dipole points upward, or the current is counterclockwise. Extra: When the polarity of suspending magnet is reversed, Lenz law now implies that the induced current should also reverse its direction. Again no penetration of the magnetic flux lines is possible, or the magnet remains being levitated. 31.4.3 Thee light bulbs with a pair in parallel A solenoid is producing the same steadily increasing magnetic flux through two circular circuits shown above. Case A: Two identical bulbs 1 and 2 are in series. Each has a resistance R. Their brightnesses (or their electric powers) are the same, i.e. P1=P2. Case B: Bulbs 2 and 3 are in parallel, and the bulbs 2 and 3 is in series with the bulb 1. Each bulb has a resistance R. We label the respective electric powers of the bulbs by Pl’, P2’ and P3’. Compare the power of bulb #1 for two cases. _ Extra: Show Pz’f P2249. Explanation: For case A, denote the current i, the loop equation is: s-2iR=0, or i=E/(2R). For case B, label currents through the bulbs by il’, i2’ and i3’ respectively. By symmetry i2’2i3’. Since il’:iz’+i3’, then il’:2i2’. This leads to the loop equation: E-il’R-iZ’R : E-(3f2)i1’R:0. Solving for il’ gives: il’=(2/3)£/‘R =(2/3)(2i) =(4/3)i. Here il’>i, so P1’> P1. Ans=3. Extra: Notice that i2: i: s/(2R) and i2’=i1’/2=213R. I {Tilt-M=[(%R>/(%R)12=3- 31.4.2 Light bulbs in series Case A Case B A solenoid is producing a same steadily increasing magnetic flux through two circular circuits shown above. Case A: Two identical bulbs 1 and 2 are in series. Each has a resistance R. Their brightness (or their electric power) is the same, i.e. Plsz. Case B: Three bulbs are in series, each with a resistance R. #3 is close to #2. Electric powers are labelled by P1 ’, Pf’ and P3’. Compare the powers of bulbs #1 and #2. Extra: Show PZ’: P224/9. Explanation: For case A, denote the current i. The loop equation gives £-2iR=0, or i=E/(2R). Correspondineg for case B, £=3i’R, or i’=E/(3R)=(2/3)i. Since i’<i, P1’< P1. Bulb 1 and bulb 2 are in series, so Pl’z Pz’. Ans=3. . i.2 Em%=fi=<r2wrag/(amig- 31 .4.4 Two light bulbs and a short wire A solenoid is producing the same steadily increasing magnetic flux through two circular circuits shown above. Case A: Two identical bulbs 1 and 2 are in series. Each has a resistance R. Their brightnesses (or their electric powers) are the same, i.e. P1=P2. Case B: There is a short CD which is to the right of the center line We label the respective electric powers of the bulbs by Pl’, and PZ’. Compare the power of bulb #1 for the two cases. Extra: If CD coincides with the center line compare P1’ with PI for this new situation. Explanation: For case A, denote i to be the loop- current, the loop equation is: £—2iR=0, or i=8/(2R). For case B, label currents through the bulbs by i1’ and iz’, and through the short by i’. The loop equations for C1 DC and C2DC are: E,’— il’R=0 and 82* iz’R=0 respectively. The emf is proportional to area of the enclosed flux. By inspection E,’> sl2> 82’. Thus i1’>i> iz’, or P1’> P1. Extra: For the case where CD is the center line, the identity of two areas of the enclosed flux implies that 81’: 52’. In other words, i1’= iz’:[ £I2]/‘R:i. So Pl’: P1. 31.4.5 Current along the short Case A Case B A solenoid is producing the same steadily increasing magnetic flux through two circular circuits shown above. Case A: Two identical bulbs 1 and 2 are in series. Each has a resistance R. The area of magnetic region is A. The induced emf is 8=-AdB/dt. The loop equation s-2iUR=0, implies that the induced current is ngS/(QR). Case B: There is a short CD. Given the area of the magnetic region to the left of CD be AL=2A/3 and to the right of CD be AR=A/3. The direction of the current flow along the short is (choose one): 1 2 frothoC fromCtoD Extra: Show the magnitude of the current flow along the Shell is Explanation: For case B, by inspection current through bulb #1 is larger than the current through #2, i.e. il’ >i2’ . At C, steady flow requires the current to go from D to C. Extra: For case B, the lefi loop (CIDC) has the induced emf 8L=-(2A/3)dB/dt =28/3, giving the current i1’=(28/3)/R =4ig/3. The right loop (i.e. D2CD) has the induced emf 8R=81 3, giving i2’=(8./ 3)/R=2i0/ 3. The node equation at C reads i1’= i2’+ iDC, or iDc= i1’- if: Zia/3. 31 .6.1 Damped Pendulum Consider the setup of a daifipEdpendulumThe magnetic field B is into the paper. Find the direction of the force due to B asserted on metal plate, as it leaves the region. 1 2 Dir of PB / / Extra: Find the direction of F3 as it reenter the region. Explanation: As the plate is leaving the rectangular region, Bind opposes the flux change in the plate. Bind has the same direction as that of B, iind is clockwise, FfiziimAL XB, which leads to the pulling back force. XXXXX XXXXXX /'swing up Extra: As the plate is swinging back, to maintain the no-flux status, Bind should be opposite to B, i.e. Bind is out of the paper. Thus the direction of magnetic force vector asserted on the plate is also reversed. ...
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This note was uploaded on 02/04/2012 for the course PHY 303L taught by Professor Turner during the Fall '08 term at University of Texas at Austin.

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Chapter 31 IQ Questions - 31.2.1 Moving rod 8; Lenz law I D...

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