- #1

crick

- 43

- 4

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?

$$\mathrm{emf}=-\frac{d \Phi}{dt}=-\frac{d \Phi_{dS}}{dt} \tag{1}$$

(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)##?