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I know that for an inductor having self inductance $L$ energy stored in its steady state when a current $I$ has been established is given by $U = \frac{LI^2}{2}$.

But after this current has been established, if we suddenly cut the wires attaching the inductor to the potential source or short the circuit, what happens to the energy ?

It must not be stored anymore as $\frac{LI^2}{2}$ as there can be no $I$, could not have decayed as heat because we cut off the wires and did not have any circuit which may have allowed for reverse flow of current.

I have one thought that it might have gone as EM radiation but I am not sure.

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2 Answers

The place where you cut the wire acts as a temporary Capacitor where a huge potential difference is formed. This potential difference causes an intense electric field to develop, which is where the energy is initially stored. If the potential difference developed exceeds the dielectric breakdown voltage of the intervening medium, the charges are lost as a spark discharge which dissipates energy as EM waves and heat.

But usually the current is never cut down this abruptly, providing enough time for the energy to dissipate as the normal safe resistive heating. If not, then the energy will be lost by the aforementioned discharge which is very intense and might damage the equipment under consideration. Hence the use of a parallel capacitor with a large inductor, which allows slow dissipation of energy as LC oscillations (EM waves) and normal resistive heating.

EDIT
The said capacitance ceases to exist only if a spark discharge dissipates the gathered charge or, the instantaneous back emf is slowly reduced by resistive heating (the circuit is not cut-off). (i.e. if we assume the cessation of current occurred instantaneously, the developed field would exceed the breakdown field leading to a spark, or if we assume that the change is slow enough so that no spark is developed, then the finite time it takes for the current to die down, the resistances of the circuit would dissipate the energy in that time). The sudden stopping of the current is only an ideal occurrence and does not occur in practice.

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Barely the development of high electric field does not assert that all the energy has transfered from magnetic to electric field. At best you can treat it as an LC circuit for an instant, it does not explain when just after this instance you can not consider said capacitance. –  rijul gupta Feb 1 at 12:18
    
@rijulgupta see the edit in my answer. Ask if you need any further clarifications. –  Satwik Pasani Feb 1 at 13:19
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Satwik's answer is correct but I want to add a practical example. When we switch off some electric device(say an DC motor) there is a spark is produced ,this is where the energy is getting lost. This spark is so high that can be seen even with naked eyes at the switch.


PS: Do this experiment in vaccum then there be a part of energy dissipated as heat and the remaining will store in setting up an $\vec E$ in the wire and will cause the electrons to accumulate at the surface.
The capacitive energy will be $\dfrac{1}{2}CV^2$ ( $C$ is the stray capacitance of wire) and the remaining as heat energy produced by the movement of electrons from the body of conductor to the surface at a high current in a short time. Energy will not emit as EM waves in vaccum.
EM waves only generates due to corona discharge if there is some medium.

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@rijul dissconection of current means megnatic field has to decrease instantly which causes an Electric field of high intensity this is what swastic's answer tells at the core. –  Anupam Feb 1 at 12:46
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