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I Electromagnetic induction

  1. May 14, 2016 #1
    I have a trivial question about electrodynamics.

    If you have a very long coil, a long solenoid. Keep the current constant and you will have no [itex]\vec{B}[/itex] outside (magnetostatics).
    Let's write down the Maxwell equations:

    \nabla\cdot\vec{B} &= &0 \\
    \nabla\times\vec{E} &= &-\frac{\partial\vec{B}}{\partial t} \\
    \nabla\times\vec{B} &= &\frac{\vec{j}}{\epsilon_0 c^2}

    For the stationary case the second equation equals to zero.
    If we slowly vary [itex]\vec{j}(t)[/itex] over time we have still a very weak field [itex]\vec{B}[/itex] outside the solenoid, say it is more or less 0.
    The inner of the solenoid has a changing field [itex]\vec{B}[/itex]. This means that the second equation is not zero. Which means we get an [itex]\vec{E}[/itex] which works against the change - self induction, so we get a reactance from the basic solenoid.

    If now another solenoid is wrapped around the basic solenoid, why does it feel a pretty strong induction?
    Is it because [itex]\vec{B}(t)[/itex] is weak but [itex]\frac{\partial\vec{B}}{\partial t}[/itex] is strong?
    Why if the magnetic field outside is more or less zero the change of the flux [itex]\vec{B}\cdot\vec{A}[/itex] is detected strongly?

  2. jcsd
  3. May 14, 2016 #2

    Charles Link

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    Using Faraday's law in integral form, each loop(turn) of the outer solenoid has EMF ## \mathscr{E}=-d \Phi/dt ## where ## \Phi ## is the complete magnetic flux over the area of each outer loop including that from the inside of the inner solenoid. The ## B ## inside of the inner solenoid also gets included in this flux. The EMF's from each of the loops of the solenoid will add in series to give the complete induced voltage of the solenoid.
    Last edited: May 14, 2016
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