Some context: I am trying to get the equation of motion for a dipole magnet falling through copper pipe. To proceed I need to calculate the EMF. We can do this by using Faraday's law,
$$\oint_{\partial \Sigma} \textbf{E}\cdot d\textbf{l} = – \int_{\Sigma}\frac{\partial\textbf{B}}{\partial t}\cdot d\textbf{S},$$
which can be simplified further,
$$U_{ind} = – \frac{d}{dt}\int_{\Sigma}\textbf{B}\cdot d\textbf{S}$$
$$U_{ind} = – \frac{d\Phi_{m}}{dt} \; .$$ But if I want to use that equation I have to calculate couple of really tedious integrals. So I found some online source where some dude used this suspicious equation $$U_{ind} = \int (\textbf{v} \times \textbf{B}) \cdot d\textbf{l}\, .$$
It was said that the above equation can be always used.
My question: Is this equation equivalent to the ones above? Why or why not? If yes, could you derive it from the one above. If not, could you derive it with proper assumption that were made to obtain it.
Best Answer
This other equation is for so-called motional emf, which is the part of the total emf induced in a conductor when this conductor moves in external magnetic field.
In the frame of the conducting pipe, the pipe does not move, so in this frame, the formula gives motional emf equal to zero — $\mathbf v$ is zero. There is also the induced emf, due to induced electric field. In this frame, all the emf is due to induced emf, and the familiar Faraday's law formula for a stationary loop applies.
In the frame of the magnet, the pipe is moving in the field of the stationary magnet, with variable velocity. So the formula does not give zero this time. It is the familiar Faraday formula that gives zero induced emf here. So the roles of the two mechanisms of EMF get reversed.
In both frames one gets the same value for the total emf, so in a sense, it does not matter which frame is chosen. However, one frame may be better for calculations.