# Vertical circuit under a magnetic field

• carllacan
The electrical power is ##P = \epsilon I_{circuit} = \frac{\epsilon^2}{2R+2\beta l} = \frac{(\mu_0 I)^2}{2\pi R}\frac{ln^2(1+\frac{b}{a})}{2R+2\beta l}##. This must equal ##Fv_0##. So the force is ##F = \frac{(\mu_0 I)^2}{2\pi R}\frac{ln^2(1+\frac{b}{a})}{(2R+2\beta l)v_0}##. I think this is the same as you have, but you

## 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} ##

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carllacan said:
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.
Looks like you have the ##v## misplaced, but I think it's just a typo.

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} ##
Yes. Divide the induced emf by the total circuit resistance.

Thanks. Now to find the magnetic force on the moving segment I would need to integrate ##F = \int _a ^{a+b}\vec I_c \times \vec B(r)dr = -I_c\int _a ^{a+b}\vec e_x \times \vec e_z B(r)dr =-I_c\int _a ^{a+b}\vec e_y\frac{\mu_0 I_{wire}}{2\pi r}dr = \frac{\mu_0 i_c I_{wire}}{2\pi }ln(\frac{a+b}{a})\vec e_y ## right?

Yes, that looks right. (I'm not quite sure on the orientation of your coordinate axes. Therefore, I'm not sure how to interpret the direction of your force. In your next to last expression there is a negative sign that seems to disappear in the last expression.) There's a quicker way to get F from energy considerations. The electrical power consumed in the circuit must match the mechanical power input by the applied force F.

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