# I Flux through rim created by moving circuit and induced EMF

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1. Jan 8, 2017

### crick

In Purcell- Electricty and Magnetism book, in the chapter on electromagnetic induction, I found the following explanation regarding the magnetic flux through a circuit of area $S$.

Consider the circuit in figure, moving in a time $dt$ in a magnetic field $B$, constant in time (but not uniform in space).

The flux through the surface of the circuit $S$ at time $t$ is
$$\Phi(t)=\int_{S} B \cdot da$$
And I'm totally ok with that. Nevertheless it is said that the flux at time $t+dt$ is
$$\Phi(t+dt)=\int_{S+dS} B \cdot da=\Phi(t) +\int_{dS} B \cdot da$$
i.e. $\Phi(t+dt)$ is given by the previous flux, plus the flux through the "rim" $dS$ (let's call this $\Phi_{dS}$).
Here is the problem: I think that stating
$$-\Phi(t+dt)+\Phi(t)+\Phi_{dS}=0$$
is against the law $\nabla \cdot B=0$. Infact, considering the volume enclosed in the circuit surface at time $t$ and $t+dt$ togheter with the surface of the "rim", then $\Phi(t+dt)$ and $\Phi_{dS}$ are outgoing fluxes from the volume, while $\Phi(t)$ is an ingoing flux in the volume.
Therefore, if the previous relation is true, than the total flux outgoing from the volume would be
$$+\Phi(t+dt)-\Phi(t)+\Phi_{dS}= 2 \Phi_{dS} \neq 0$$
Is this correct or is there something missing?

In the book, from this proof it is derived that the emf induced in the circuit is
$$\mathrm{emf}=-\frac{d \Phi}{dt}=-\frac{d \Phi_{dS}}{dt} \tag{1}$$
(indicating with $\Phi$ the flux through the circuit surface $S$ and with $\Phi_{dS}$ the flux through the rim).
But, as I also see on another textbook, I think it should be
$$\mathrm{emf}=-\frac{d \Phi}{dt}=+\frac{d \Phi_{dS}}{dt} \tag{2}$$
Which of the two is the correct one, $(1)$ or $(2)$?