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I'm confusing with a problem.

Consider an ideal inductor with a shorted circuit and no resistance and having a current through it. we know this current is constant while the inductor has no resistance because of its formula

The magnetic field of this inductor produces a force on a nearby iron block and pull it.

this action moves the iron closer to inductor and increase its inductance . right ?

so after that we have an inductor with same current and a higher inductance . so it have more stored energy ? also we have a kinetic energy in iron block.

where is my mistake ? the inductor must lose energy .

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  • $\begingroup$ Why do you assume the current remains the same? $\endgroup$
    – Puk
    Commented Oct 25, 2020 at 8:20

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The energy stored in the magnetic field is $$E=\frac{1}{2}Li^2$$ where $L$ is the inductance and $i$ is the current. As the iron is pulled in, $L$ increases and part of this energy is converted to the kinetic energy of the iron. Both of these effects cause $i$ to be reduced. The total energy is still conserved.

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  • $\begingroup$ thank you , but how current is reduced ? is it right to say the current reduces when L increases ? is there any equation for it ? to relate the current to change in inductance ? $\endgroup$
    – MohammadAli Zeraatkar
    Commented Oct 25, 2020 at 9:16
  • $\begingroup$ The inductor energy equation above shows you how current will get reduced in response to increasing $L$. If you need to take kinetic energy into account, you need to set the initial inductor energy equal to the final inductor energy plus the kinetic energy, and solve for the final current. $\endgroup$
    – Puk
    Commented Oct 25, 2020 at 9:19
  • $\begingroup$ This is the inductor equation : V = L * (di/dt) . lets check it with a very small time step T : V = L * (i - last_i / T) . V is zero because of no resistance over the inductor so it results last_i = i . So we have a constant current over time $\endgroup$
    – MohammadAli Zeraatkar
    Commented Oct 25, 2020 at 12:14
  • $\begingroup$ $v=L\ di/dt$ doesn't apply when you have a time-varying inductance or time-varying external magnetic fields. $\endgroup$
    – Puk
    Commented Oct 25, 2020 at 14:57
  • $\begingroup$ Thank you , so my mistake was here . But why it does not apply ? $\endgroup$
    – MohammadAli Zeraatkar
    Commented Oct 25, 2020 at 17:55

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