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I was trying to understand faradays law intuitively. For this I need to understand if a stationary rod subjected to changing magnetic flux experience an emf, or difference in potentials $v_{1}$ and $v_{2}$ (in the picture). Is $v_{1}=v_{2}$? The velocity of an electron in the rod is almost $0$ because the rod is stationary, hence $\vec{F} =q\vec{V} \times \vec{B} =q\vec{0} \times \vec{B}=\vec{0}$, since there are no forces acting on an electron, the potential difference between $v_{1}$ and $v_{2}$ is zero. is that correct? lenz law but for a rod and not a loop

but if we connect four rods to form a loop, the current will flow as in the picture below enter image description here

What is my mistake? Is it because electrons always have velocity? Also in Faraday's law $$\oint _{C}\vec{E} \cdot d\vec{l} =-\frac{\partial }{\partial t}\int _{S}\vec{B} \cdot d\vec{a}$$ I would understand Faraday's Law better if the r.h.s. can be represented as a line integral instead of surface integral. Is there such integral form? Thank you!

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  • $\begingroup$ The right way to write the integral form is $\oint _{C}\vec{E} \cdot d\vec{l} =-\int _{S}\frac{\partial\vec{B} }{\partial t} \cdot d\vec{a} = - \frac{d}{dt}\int _{S}\vec{B}\cdot d\vec{a} -\oint _{C}\vec{v} \times \vec{B} \cdot d\vec{l}$ so that $\oint _{C} (\vec{E} + \vec{v} \times \vec{B}) \cdot d\vec{l} = - \frac{d}{dt}\int _{S}\vec{B}\cdot d\vec{a}$, being $\vec{v} $ the velocity of the contour $C$ $\endgroup$
    – basics
    Commented Nov 27, 2022 at 21:04

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For a stationary rod, the EMF is not caused by the magnetic force, but the induced electric force as a result of the changing magnetic field. There is no velocity dependance on the Electric force.

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