2012 Semifinal Exam Part B 16 The flux through the loop will then be PB= Hom sin do where 0 is the angle between a line along the axis of the loop and a line drawn between the monopole and any point on the rim of the loop. It is easy to see that 1 dΦB=24 n0 From Faraday's law,we have that a changing flux will induce a current I in the loop. dB =IR, dt where R is the resistance of the loop.We'll figure R out later. The induced current will create a magnetic field that will oppose the monopole motion.We need to use the law of Biot Savart to find that field.Along the axis of the loop,we have dB=uol di'xr 4r3 where ris a vector connecting the monopole with some point on the rim of the loop.Only compo- nents of B parallel to the axis of the loop will survive,so we can concern ourselves with dB Hol dl 4r r sin0 The integral is trivial;dl is around the circumference;nothing else changes,so B sind It is better to think in terms of b,the radius of the loop,than it is to deal with r,the distance from the rim of the loop to the monopole.In that case, sin=r b SO B=Ho! 26(sin0)3 The force on the monopole is then F=aB=gm明=觉m外h Note than multiplying through by dt gives an expression that is related to the change in momentum, o(sin) dp=4m2b 元dΦB=qm2D (sin 0)4do Using the provided integral, △p= -9m24023π 4bR 8 Copyright C2012 American Association of Physics Teachers2012 Semifinal Exam Part B 16 The flux through the loop will then be ΦB = Z B~ · dA~ = 1 2 µ0qm Z sin θ dθ where θ is the angle between a line along the axis of the loop and a line drawn between the monopole and any point on the rim of the loop. It is easy to see that dΦB = 1 2 µ0qm sin θ From Faraday’s law, we have that a changing flux will induce a current I in the loop. dΦB dt = IR, where R is the resistance of the loop. We’ll figure R out later. The induced current will create a magnetic field that will oppose the monopole motion. We need to use the law of Biot & Savart to find that field. Along the axis of the loop, we have dB~ = µ0I 4π d ~l × ~r r 3 , where ~r is a vector connecting the monopole with some point on the rim of the loop. Only compo￾nents of B~ parallel to the axis of the loop will survive, so we can concern ourselves with dB = µ0I 4π dl r 2 sin θ The integral is trivial; dl is around the circumference; nothing else changes, so B = µ0Ib 2r 2 sin θ It is better to think in terms of b, the radius of the loop, than it is to deal with r, the distance from the rim of the loop to the monopole. In that case, sin θ = b r , so B = µ0I 2b (sin θ) 3 The force on the monopole is then F = qmB = qm µ0I 2b (sin θ) 3 = qm µ0 2b (sin θ) 3 1 R dΦB dt Note than multiplying through by dt gives an expression that is related to the change in momentum, dp = qm µ0 2b (sin θ) 3 1 R dΦB = qm 2 µ0 2 4bR (sin θ) 4 dθ Using the provided integral, ∆p = qm 2µ0 2 4bR 3π 8 Copyright c 2012 American Association of Physics Teachers