When we define the field generate by EMF, why there is not negative sign in $\mathcal{E} = \oint \vec{E} \cdot d\vec{l}$?
Usually we talk about potential, there should be a negative sign, right?
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The conceptual problem here is that of EMF, $\mathcal{E}$ vs Electric Potential, V. They aren't really the same thing despite being measured in the same units. For instance the EMF is caused by an external agent that isn't the conservative electrostatic field, like say a chemical reaction in a battery or a solar cell. Work is done to cause a charge separation, this charge separation however produces an electrostatic field opposite in direction to the EMF that causes the separation (Example visualization, the EMF provides a positive field to separate negative and positive charges so that the positive ones move in the direction of the EMF to the right of the negative ones, for a given right pointing EMF). If you have drawn this situation you can now see that since the charge is separated, there now exists an electrostatic field caused by the charges themselves which points to the left equal in magnitude but opposite in direction to the applied EMF. Hence there is a sign difference between the definition of electric potential and EMF. This is kind of a simple visualization, there are more rigorous ways to prove it.
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$\begingroup$ Thanks for your answer. But can you direct me to a more rigorous proof? $\endgroup$ Commented Apr 11, 2014 at 19:11
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$\begingroup$ Yeah right now im doing homework at school once I find a good one I will link you or reference you to it $\endgroup$ Commented Apr 15, 2014 at 0:53