Please look for flaws in my reasoning, any help would be appreciated.(adsbygoogle = window.adsbygoogle || []).push({});

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.

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# Lenz's Law, 2nd opinion wanted

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