2012 Semifinal Exam Part B 17 as the monopole moves from one side to the other. If the monopole started from rest a distance H above the loop,then the initial energy of the system is mgH,and the initial momentum when passing through the loop (assuming there is no loop)is then po =mV2gH The monopole will lose Ap from the momentum on each pass through the loop,so the number of times it passes through the loop N is 9m24023xy -1 N= =mv2gH △p 4bR 8 or N= 32bRmv2gH 3rμ02qm2 Oh,we still need to do R.Since a<b,we can treat it as a long thin cylindrical wire,and R= p2πb Ta2 so we finally get N= 6462pmv2gH 3r1402qm2a2 Version 2 Instead of force,focus on the power dissipated in the loop,which is P= 2 R μ0gm 2 P= 62+x2)正 2πbp P=o29m2a23 1 2 dx 8p (b2+x2)3 dt The energy is lost from the particle: P=- dl m2 dt2' v P=-mv dt Combining these,.and sincev=器， dv Lo2qm2a23 1 dx dt 8pm (b2+x2)3dt Use the provided integral (or do a trig subsitution that gives the other provided integral),and continue. Copyright C2012 American Association of Physics Teachers2012 Semifinal Exam Part B 17 as the monopole moves from one side to the other. If the monopole started from rest a distance H above the loop, then the initial energy of the system is mgH, and the initial momentum when passing through the loop (assuming there is no loop) is then p0 = m p 2gH The monopole will lose ∆p from the momentum on each pass through the loop, so the number of times it passes through the loop N is N = p0 ∆p = m p 2gH  qm 2µ0 2 4bR 3π 8 −1 or N = 32bRm√ 2gH 3πµ0 2qm 2 Oh, we still need to do R. Since a  b, we can treat it as a long thin cylindrical wire, and R = ρ2πb πa2 so we finally get N = 64b 2ρm√ 2gH 3πµ0 2qm 2a 2 Version 2 Instead of force, focus on the power dissipated in the loop, which is P = E 2 R P =  1 2 µ0qm b 2 (b 2+x2) 3 2 dx dt 2 2πbρ πa2 P = µ0 2 qm 2a 2 b 3 8ρ 1 (b 2 + x 2) 3  dx dt 2 The energy is lost from the particle: P = − d dt 1 2 mv2 P = −mv dv dt Combining these, and since v = dx dt , dv dt = − µ0 2 qm 2a 2 b 3 8ρm 1 (b 2 + x 2) 3 dx dt Use the provided integral (or do a trig subsitution that gives the other provided integral), and continue. Copyright c 2012 American Association of Physics Teachers