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I Magnetic Energy = 1/2LI^2?

  1. Nov 14, 2017 at 4:12 PM #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 at 7:51 PM
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  3. Nov 14, 2017 at 7:20 PM #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 at 1:03 PM #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 at 2:08 PM #4
    Yes
     
  6. Nov 17, 2017 at 9:09 AM #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 at 9:32 AM #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 at 9:56 AM #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 at 1:06 PM #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 at 5:13 AM #9
    I didn't understand. Can you elaborate more ?
     
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