# Electromotive force in constant flux

## Homework Statement

in book there is written that this turn will make electromotive force, but from faraday law we know that force generates only when flux is changing, what about this?
http://img821.imageshack.us/img821/747/schematu.jpg [Broken]
and they say that it generates electromotive force, but flux is constant, why?

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Flux is being "cut"

use -v cross B for each infinitely small unit length dl

The velocity goes with r d(theta)/dt

Yo dawg, consider each side of the rectangular loop.
The top horizontal side "cuts" no flux cause it ain't got no balls.
The right side cuts flux so there IS an EMF cause it's one badass playa.
The left side cuts flux too and there IS an EMF generated but it's like ya know consider the direction of the EMF innit.
They are like from different gangs, naam sayin'? They cancel each other out.

If you consider the scenario where the loop is entering or leaving the uniform field where just one of the rectangular sides are killing Flux in the name of Faraday, then there IS EMF. And you should also know that when entering and leaving the field, the EMFs cause a current in the loop which are in the opposite direction.

okay, thx, so just $$\mathcal E=-\frac{\mbox{d}\Phi}{\mbox{d}t}$$ for every small part of this thing?

this link dont work, is it paste good?

Consider the Lorenz force on each element dl of the rod (it is a conducting rod i.e. has "free" charges inside it):

FB = q(E + vB); each element moving in a constant speed, thus no net force on the element !

-> E = -vB; notice it is independent of the charge size !

if you path integrate the expression for the field generated by each element from 0 to L you will get the potential between to extreme points of the rod.

**Another way to solve this is to think the rod is connected to a wedge shaped circuit and use Ampere's Law to calculate the time change in magnetic flux (you need to make your area time depended)

Tell us how you did :)