The energy in a current carrying loop is given as

$E_{\mathrm{m}}=\frac{1}{2 \mu_{0}} \int B^{2} d \tau$

If I cut the loop abruptly the current and hence the magnetic field quickly fall to zero and hence the energy goes to zero as well, as is evident from the above equation.

But where does that energy go to?


First off, if the wire is "ideal" the answer is undefined. The ideal models provide no solution if you try to solve them. What actually happens is defined by the non-ideal aspects of the system.

The magnetic field collapsing causes back-EMF, which induces a voltage on the wire. In small voltage cases, this ends up turning the wire into an antenna, radiating the energy outward. This sudden change in voltages due to a sudden cutting of the wire can be thought of as voltage changes at all frequencies. The capacitences and inductances of the wire and its environment will decide exactly how that works out.

If the magnetic field is strong enough, the back-EMF voltage can get quite large. At some point, this becomes sufficient to arc across the cut ends of the wire, effectively re-creating the wire using ionized air to fill in the gap.

This effect is a big deal for engineers working with large motors. Switching off a large motor like this is basically equivalent to cutting the wire on a very large inductive loop. The result is an arc. In the smaller voltage cases, this slowly damages the switch until it stops functioning. In larger voltage cases, this can bridge the contacts of the switch, and the motor keeps running even though you turned it off!

To deal with this, you'll find we tend to put a large resistor in parallel with the motor. When the circuit is opened, and the magnetic field has to collapse, it can use the resistor to dissipate the energy.

An alternative solution is seen in the AC power world. Quite often we will cut the circuit at a 0 crossing for the voltage. At this point, it can't create any arc (because there's no voltage), and all "antenna" like effects are at nice multiples of 60Hz. This gives us time to physically separate the switch enough to avoid arcing, and give the system lots of room to dissipate the energy.

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  • $\begingroup$ Thank you, but why won't an ideal model give an answer? $\endgroup$ – Yasir Sadiq Oct 22 at 13:39
  • $\begingroup$ @YasirSadiq In an ideal wire model, if you cut the wire instantaniously, you end up with infinite voltages trying to maintain the invariants of the simple wires and simple inductors. Its the non-ideal aspects which keep the values finite. $\endgroup$ – Cort Ammon Oct 22 at 20:36

There will be a spark between the cut ends and the energy goes into the heat, light, and bang associated with the spark.

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  • $\begingroup$ Does the spark always occur? Or is it based on the assumption that the wire is sufficiently "sharp" to cause a dielectric breakdown? $\endgroup$ – Kenzo Tenma Oct 21 at 17:09
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    $\begingroup$ The current cannot stop suddenly. The faster it stops the bigger the generated emf. The emf will therefore be as big as it needs to jump the gap and cause a spark. $\endgroup$ – mike stone Oct 21 at 17:28
  • $\begingroup$ And radiation ? $\endgroup$ – Yasir Sadiq Oct 22 at 13:36
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    $\begingroup$ @Yasir Dasiq Yes indeed! I guess that light is form of radiation though.... :) $\endgroup$ – mike stone Oct 22 at 13:37
  • $\begingroup$ Oh yes, Thank you for correcting me again. I had other non visible spectrum in mind. Just because the charges are accelerating so they'll emit radiation. $\endgroup$ – Yasir Sadiq Oct 22 at 13:41

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