Suspicious EMF equation

Question

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.

Answer 1

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 following equation can be always used.

My question: Is this equation equivalent to the ones above? Why or why not?

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.

Answer 2

You need to be precise by what you mean with $v$. The formula is true for the Lorentz induction when $v$ is the velocity of the wire. Writing $\Phi = \int B\cdot dS$, you have two contributions: one from the varying electric field (Neumann induction) and one from the movement of the loop (Lorentz induction). The formula applies for Lorentz induction only:

$$ \frac{d\Phi}{dt} = \int \frac{\partial B}{\partial t}\cdot dS +\int (\nabla \cdot B) v\cdot dS+ \oint (B\times v)\cdot dl \\ = -\oint (E+v\times B)\cdot dl $$

The first line explicits the time derivative (valid in general), the first term being von Neumann contribution and the second, third terms being Lorentz contribution. The second line is obtained by applying Maxwell's equations. You recognise the last expression as the emf, the force that the particles experience in the wire per unit charge.

You shouldn't use this though for your problem, since the loop is stationary on the wire.

Hope this helps.

Answer 3

The equation $$\Delta U_{ind} = U_{ind}(b) - U_{ind}(a) = \int_a^b (\vec{v} \times \vec{B}) \cdot d\vec{l} \; ,$$ is not suspicious. Remember the Lorentz force on a charge $q$ is $$ F=q(\vec{E}+\vec{v}\times\vec{B}).$$ So the integrand is that portion of the electric field that is generated by the magnetic field, contributing to the Lorentz force. Integrating over an electric field along a curve yields the voltage difference between the end points.

Caveat: The electric field created by changing magnetic fields is not conservative. So I'd expect the equation to hold only for a static magnetic field and possibly even only for a straight movement.