Please look for flaws in my reasoning, any help would be appreciated. Question: HRW CH31 #28P(if you have the book handy) A long rectangular conducting loop of width L, resistance R, and mass m, is hung in a horizontal, uniform magnetic field B that is directed into the page and exists only above line aa. The loop is then dropped; during its fall, it accelerates until it reaches a certain terminal speed. Ignoring air resistance, what is this terminal speed? Answer: For the falling loop to reach a constant terminal speed the force of gravity pulling it downward must be cancelled by a Lorentz Force pullint it upward F(Lorentz)=mg eqn 1 To find F(L) we must first find the induce EMF, then the induced current and finally use this to determine F(L). Magnetic Flux:(MF)=Integral[B*dot*dA] Let x equal verticle length of loop in B field. (MF)=B*L*x EMF=d/dt*(MF)=d/dt*BLX=BL*dx/dt=BLv Now taking the derivative of both sides with respect to t: d/dt*EMF=dv/dt*BL=abL intergrate both sides with respect to t: Integral[d/dt*EMF*dt=aBL*Integral[dt] EMF=aBLt This sounds reasonable because EMF should be increasing with time since change of flux is increasing with time. Now in terms of current: i=EMF/R=abLt/R and Force: F=aB^2*L^2*t/R Subbing g for a and rewriting eqn 1 from way above: mg=gB^2*L^2*t/R solving for t: t=MR/(M^2*L^2) Using kinematics equation: V(terminal)=g*t Sound Plausible? Thanks for your time.