Formula for Electromotive force?

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The discussion centers on the formula for electromotive force (emf) in circuits, specifically the contributions from electric fields and magnetic fields. It highlights the distinction between conservative and non-conservative electric fields, noting that only the term v x B contributes to emk when the magnetic field is constant. The confusion arises from two examples: one with a static circuit and another with a spinning metal ring, both involving constant magnetic fields. The key point is that in both cases, the electric field remains conservative due to the constancy of the magnetic field, leading to a zero contribution from the electric field in the emk calculation. Understanding these concepts is crucial for applying Faraday's law in practical scenarios.
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Homework Statement
difficulty understanding the formula for emk.
Relevant Equations
formulas for emk.
So the formula for emk for a circuit is
1760708819114.webp
where f_s is caused by non emf forces like a battery. E is the electricalfield inside the circuit? v is the velocity of the circuit and B is the external magnetic field. I have difficulty understanding E. There are two "types" of electrical field one caused by charges and one by variation in a external magnetic field. The first one is a conservative field so it does not contribute to emk but the second does.

I don't understand the following example
1760709178196.webp
1760709202535.webp
and
1760709215653.webp
.

Why is the only contribution to emk by v x B? Isn't there a variation in the magnetic field because the whole circuit is not in it so the flux through the circuit decreases. Shouldn't that cause an non-conservative E which contributes to the emk?

Another example
1760709692292.webp
. Here a metal ring spins with angular frequency w. The total velocity is 0 and although the magnetic field is constant the spinning of the metal ring causes a variation of the flux through the ring which causes a non conservative E that contributes to emk.

So what is the difference between the first and second example why is E conservative in the first but there is a non conservative E in the second case?



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The electric field that you are asking about comes from Faraday's law in integral form when you have no batteries and no moving parts, $$\oint_C\mathbf E\cdot d\mathbf l=-\frac{\partial}{\partial t}\int_S\mathbf B\cdot \mathbf{\hat n}~dA.$$ It is the non-conservative emf that you think is missing.
 
kuruman said:
The electric field that you are asking about comes from Faraday's law in integral form when you have no batteries and no moving parts, $$\oint_C\mathbf E\cdot d\mathbf l=-\frac{\partial}{\partial t}\int_S\mathbf B\cdot \mathbf{\hat n}~dA.$$ It is the non-conservative emf that you think is missing.
Sorry I'm completely lost are you talking about example 1 or 2?
 
I am talking about the ##\mathbf E## in the expression

Screen Shot 2025-10-17 at 10.50.30 AM.webp

that you provided. I assume it is the non-conservative electric field when the magnetic flux is changing with respect to time. How is it defined in the source you got it from? If my assumption is correct, this field is present in both examples.
 
I'm a bit rusty so forgive me -- but in both of those examples, the magnetic field ##\mathbf{B}## is a constant* (in the lab frame), so ##\partial \mathbf{B} / \partial t = 0##.

And this means that ##\mathbf{E}## is a conservative vector field in both of those examples, i.e. ##\int_C \mathbf{E} \cdot d\mathbf{l} = \int_S (\nabla \times \mathbf{E}) \cdot d\mathbf{S} = 0##.

As you pointed out, the EMF is defined by ##\mathcal{E} = \int_C (\mathbf{E} + \mathbf{v} \times \mathbf{B}) \cdot d\mathbf{l} = \int_C \mathbf{E} \cdot d\mathbf{l} + \int_C (\mathbf{v} \times \mathbf{B}) \cdot d\mathbf{l}##. The first term is zero, so you are just left with the integral of ##\mathbf{v} \times \mathbf{B}##.

*[N.B. this assumes the back-reaction due to the currents in the circuit are negligible. I believe this is an explicit assumption in these examples -- i.e. we are told to treat ##\mathbf{B}## is constant and ignore second order corrections. At least, this is the case with the majority of introductory electrodynamics problems].
 
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