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I Flux through rim created by moving circuit and induced EMF

  1. Jan 8, 2017 #1
    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).
    6666666666666666666666666.png
    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)##?
     
  2. jcsd
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