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Solenoid and energy

  1. May 27, 2009 #1
    Hi, this isn't an actual question, but is a confusion inspired by a question I did (hence me not using the template). I'm having trouble with this example. I'm talking about a solenoid (with core, but doesn't really matter), with n turns per unit length, and a current [itex]I_0 t[/itex] going in. [itex] \vec{z}[/itex] is along the axis of the solenoid, and the current is along [itex]\vec{\theta}[/itex].



    Using Ampere we get:

    [itex]H=In\hat{z}[/itex]

    [itex]B=\mu In \hat{z}[/itex]

    Using Faraday we get:

    [itex]E=-0.5\mu n I_0 r \hat{\theta}[/itex]

    the Poynting Vector is:

    [itex]\vec{E} \times \vec{H}[/itex]

    Integrating over the surface of some volume inside the solenoid to find the power flowing out, we get:

    [itex]\int \vec{N}.\vec{dS} = -\pi \mu n^2 l (I_0)^2 r^2 t[/itex]


    Also, the rate of change of energy stored in the magnetic field comes out as:

    [itex]\pi \mu n^2 l (I_0)^2 r^2 t = \frac{dU}{dt}[/itex]

    Also, work done against field (for that volume):

    [itex] - \xi I = \pi \mu n^2 r^2 l (I_0)^2 t = \frac{dW}{dt} [/itex]

    These three things don't seem to match up to the energy continuity equation - what am I thinking wrong?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. May 27, 2009 #2

    Hao

    User Avatar

    If we consider the surface of the solenoid, the Poynting vector is directed inwards - this represents the transfer of energy to the magnetic field.

    Hence, it is no coincidence that the rate of change of magnetic field energy is equal to the negative of the surface integral of the Poynting vector.

    If the consider a surface just outside the (infinite) solenoid, H = 0, and hence the Poynting vector is 0. Ie. There is no electromagnetic radiation whatsoever to consider.

    Hence,
    [itex]
    \int \vec{N}.\vec{dS} = -\pi \mu n^2 l (I_0)^2 r^2 t
    [/itex] represents the same thing as [itex] \pi \mu n^2 l (I_0)^2 r^2 t = \frac{dU}{dt}
    [/itex], which is equal to [itex]
    - \xi I = \pi \mu n^2 r^2 l (I_0)^2 t = \frac{dW}{dt}
    [/itex].

    It is not necessary to add both the surface integral and magnetic field energy terms together.

    On the other hand, if there was electromagnetic radiation, considering the Poynting vector alone would give the right answer for conservation of energy, as the Energy Flux integral would necessarily include the [itex]
    \frac{dU}{dt}
    [/itex] term.

    I hope this answers the question.
     
    Last edited: May 27, 2009
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