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I Magnetic Energy = 1/2LI^2?

  1. Nov 14, 2017 #1
    As we know, work done by an outside agency in creating some current ''I" in an inductor 'L' is 1/2LI2. Now this result is derived by quasistatic approximation if I am not wrong. Now, I am assuming that in non quasistatic (real) scenarios, the work done by outside agency would be different(If you actually calculate by using Poynting Vector, there would be some radiation flying off ......). Am I right till here ?

    My question is :Let W be the work done in charging inductor to current I in time T (at a rate of I/T amperes per second) without making qusi static approximations. Is W > 1/2LI2 ? Also, if T approaches infinity (i.e. the rate of increasing current is too darned slow), will W approach 1/2LI2 ?

    << Post edited by Mentor for language >>
     
    Last edited by a moderator: Nov 14, 2017
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  3. Nov 14, 2017 #2

    Charles Link

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    To help answer your question ## L=\frac{\Phi}{I} ## , and for a long cylinder ## B=n \mu_o I ##. The result is with ## E=(1/2)LI^2 ## and with ## \Phi=nL( BA) ## that ## E=(1/2) \frac{B^2}{\mu_o} V ##. This assumes nothing gets radiated away=otherwise, yes, there is power that is lost and not described by these basic equations. It is still correct that ## E=(1/2) LI^2 ##. That part of the energy is recoverable.
     
  4. Nov 16, 2017 #3
    Yes. Thanks. I think I get it now.

    Also, is my assumption about super-slow process true(will W= 1/2LI^2 in that limiting case where T approaches infinity)
     
  5. Nov 16, 2017 #4

    Khashishi

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    Yes
     
  6. Nov 17, 2017 #5
    One more question : then how can an lossless LC circuit exist ? It should lose energy by means of radiation since current in the circuit is definitely greater than zero(hence quasistatic approximation certainly doesnot hold)
     
  7. Nov 17, 2017 #6

    Charles Link

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    I don't know that anyone ever said that an LC circuit is completely lossless. Normally, LC circuits come with a quality factor ## Q ##, and the higher the ## Q ##, the smaller the losses. The losses can be both resistive and radiative, and it can be somewhat difficult to determine how much is resistive and how much is radiative.
     
  8. Nov 17, 2017 #7
    Thanks. "Lossless" word stuck with me and I guess I never cared so deeply about these questions until now. Maxwell equations has certainly enlightened me. I almost see the whole picture now. Thanks again :)
     
  9. Nov 17, 2017 #8

    tech99

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    It sounds as if you are charging the inductor from a DC source such as a battery.
    Any EM wave radiated would then have to be unidirectional. The usual view is that this is not possible.
    So it looks as if the radiated energy may be zero for this case.
     
  10. Nov 18, 2017 #9
    I didn't understand. Can you elaborate more ?
     
  11. Nov 18, 2017 #10

    tech99

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    In order to obtain EM radiation, we require an alternating current. You are using a DC supply so there is no alternating current.
    If the inductor is connected to AC, then there will be some radiated energy, depending on the dimensions and the size of the current.
     
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