Is My Calculation for Terminal Speed of a Falling Conducting Loop Correct?

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The discussion centers on calculating the terminal speed of a falling conducting loop in a magnetic field. The initial reasoning involves equating the Lorentz force to the gravitational force to find the terminal speed. The calculation progresses through determining the induced EMF and current, ultimately leading to a force equation. However, a key flaw is identified: the assumption that acceleration remains constant, while the magnetic force increases with speed. The conversation highlights the importance of recognizing that terminal velocity implies changing acceleration, ultimately leading to a more nuanced understanding of the physics involved.
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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 canceled 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.
 
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Here, I think, is a problem:

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

(You could have saved yourself some time by simply claiming that v = at, which in effect is what you have done.)

Problem is you are assuming that a is constant. I think you will agree that this is the case here. The magnetic force exerted on the coil is increasing as v increases.

In fact, if you just remember that you are looking for a terminal velocity, it will be obvious that ai=g, and af=0.
 
Ahhh :rolleyes: , thanks for pointing that out; I sensed a bit of redundancy(sp?), but wasn't sure enough to simplify. Then again, I can use the practice showing relationships via Calculus.
 
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