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## Homework Statement

An undefinedly long circuit is set vertically besides a current wire. The circuit has a conducting straight wire (the line AB) that can slide over it. The segments AB and CD have both resistance R, and the rest of the circuit has a lineal resistance ##\beta##.

How much force do we need to apply to the segment AB so that it descends with constante velocity ##v_0##?

https://photos-1.dropbox.com/t/2/AABqrUoB2O4vc616FYj-7SjoKBE3HfVok5fdrE-mQjBsEA/12/28182931/png/1024x768/3/1422032400/0/2/Screenshot%20from%202015-01-23%2015%3A34%3A30.png/CJOTuA0gASACIAMoASgC/SZ8-4VjvN-JRevQHpIByWWJh9zKoEmwluexW0MgzuiQ [Broken]

## Homework Equations

Magnetic field of a straight wire ## \vec B = \frac{\mu_0 I}{2\pi r} ##

Faraday's law of induction ##\epsilon = - \frac{d\Phi}{dt}##

## The Attempt at a Solution

The first things I've done is find the expression for the magnetic flux through the circuit. I'll call ##l## the distance between the points C and A. To do so I integrate the magnetic field created by the wire at a distance r from a to a+b, and then multiply it by the height of the closed circuit, which is l.

## \Phi = l\int _{a} ^{a+b} B_{wire}(R) dR = l\int \frac{\mu_0 I}{2\pi R} dR = \frac{\mu_0 I l}{2\pi} (ln|a+b| -ln|a|) =l \frac{\mu_0 I }{2\pi} ln|1+\frac{b}{a}|##

Now the EMF on the circuit is simply

## \epsilon = - v\frac{d\Phi}{dt} = -\frac{\mu_0 I }{2\pi} ln|1+\frac{b}{a}|##, where ##v## is the velocity at which l grows, or equivalently the falling velocity of the wire.

Next I need the current along the segment, and here is where I start running into problems, since I'm not sure of how to treat the EMF in a circuit. I know it is a voltage, but when I normally work with circuits voltage are defined between two points in a circuit. Between which two points should I put ##\epsilon##? Between any

*pair*of points? Or should I divide ##\epsilon## by the total resistance of the circuit and take the result as a current through it? That is ##I_{circuit}=\frac{V}{R_T}= \frac{\epsilon}{2R+2\beta l} ##

Thank you for your time.

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