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marcusl
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Not necessarily. We can use a switch to connect the inductor to a resistance and let the current decay exponentially as field energy dissipates in the resistance.tech99 said:
Not necessarily. We can use a switch to connect the inductor to a resistance and let the current decay exponentially as field energy dissipates in the resistance.tech99 said:I don't think "allowed to decay" is quite accurate. To stop the current, we have to apply a force, a reverse voltage, just like stopping a flywheel, and the energy stored in the inertia of the system does work against our force by producing movement or current.
Sorry, no part of this post makes sense to me. I described the process by which the creation of a field requires work to be done from an external source, and the sense in which that work is stored as energy in the field. Please rephrase if you still have questions.Puglife said:Doesn't the heat dissipation, as well as the current into a resistor equaling the current out of a resistor, or in this case wire, conserve it's energy? How does the production if an electromagnetic field, continuously follow the conservation of energy. What is the energy loss in the circuit, that physically causes the electromagnetic field to form? where does the energy to do the initial work to create the field come from, because isn't the entirety of the energy drop from resistance dissipated through heat?
It can be summarised like this:Puglife said:ok, in that case, where does the energy physically come from, in order to produce the magnetic field, if their is no energy loss in terms of electricity in the circuit? does it not apply to conservation of energy for some reason?
So the energy that started to build the field was the power of the circuit? So in the case of the super conductor, the power of the magnetic field would be the full vi because the power I'd not being dissipated in any other way?NascentOxygen said:It can be summarised like this:
to build the field requires energy, so we apply a voltage and the inductor draws an ever increasing current,
to maintain the field requires a fixed current but [ideally] no voltage,
to collapse the field requires that energy be taken from the inductor, it can be reabsorbed by the source or electronically switched into another circuit.
Where there is no resistance there are no i2.R losses, and the energy returned when the field decays equals the energy that was earlier supplied in building it.
ok, but none the less, does the energy from the em field come from the power? (energy over time)marcusl said:Power and energy are different: power is a change in energy per unit time.
It came from the circuit's input power.Puglife said:So the energy that started to build the field was the power of the circuit?
Because v and i are not both constant, it's not very informative to speak of their product, vi or power, that's why I explained the cycle in terms of energy. Conservation of energy when none is lost as heat, means that averaged over timeSo in the case of the super conductor, the power of the magnetic field would be the full vi because the power I'd not being dissipated in any other way?
ok, so as per my original question, i know that the energy form the magnetic field stays their, but how does it stay their without a constant energy being input into it?NascentOxygen said:It came from the circuit's input power.
Because v and i are not both constant, it's not very informative to speak of their product, vi or power, that's why I explained the cycle in terms of energy. Conservation of energy when none is lost as heat, means that averaged over time
Electrical Energy in = Electrical Energy out
The current is being held fixed and this maintains the field. No work is needed for this as the ideal inductor is lossless.Puglife said:ok, so as per my original question, i know that the energy form the magnetic field stays their, but how does it stay their without a constant energy being input into it?
Puglife said:but how does it stay their without a constant energy being input into it?
DaveThis does come up time to time and it isn't a matter of semantics. Back EMF simply is not the spike that you get from a coil when the field collapses.
You charge a coil - lenz's law describes the counter current or back emf that opposes the forward current and resists the forward current's ability to bring the coil's charge up...
Once the coil is charged and you disconnect power, the spike you get back is the "inductive spike" or "transient spike."
You can see Lenz's law here:
http://www.energeticforum.com/redirect-to/?redirect=http%3A%2F%2Fwww.mwit.ac.th%2F%7EPhysicslab%2Fhbase%2Felectric%2Ffarlaw.html%23c2
It is at the bottom.
Look at this nice simple answer:
WikiAnswers - What is the formula for transient spike computation in an inductive load
"E=I x R. The inductive spike occurs as the circuit is opened. The collapsing magnetic field causes the inductor to become the source of the circuit. For example consider a circuit consisting of a 10 volt battery, a 10 mh inductor, and a 10 ohm resistor all in series. With the switch closed, 1 amp will flow through the circuit (after 5 mS). The 5 mS is the time it takes the current to rise from 0 to 1 amp. This is given by the formual TC=L/R where TC is the time constant in seconds, L is the inductance in henries, and R is the resistance in ohms. It takes 5 time constants for the current to reach the maximum current which is determined by I=E/R (Ohm's Law). The delay is caused by the counter EMF generated in the coil as flux lines cut through adjacent turns of the inductor. After 5 time constants, the current is at 1 amp. When we open the switch, it will take 5 time constants for the current to drop to 0 amps. However, this will not be 5 mS because the resistance is now much larger do to the opening switch contacts. The voltage across the switch contacts will be whatever is necessary to maintain the current flow for the 5 time constants. After one time constant, the current will have dropped to 32% of the maximum current or in this case, 0.32 amps. If the resistance of the switch gap is 1 megohm, the the voltage will be 320,000 volts. More than enough to ionize the air and create a conductive path. If we assume an average resistance of 1 megohm, it will take 50 nS for the current to drop to 0. Of course during this time, the switch contact gets zapped. Placing a diode across the inductor such that the diode is reverse biased with the switch closed will give the current an alternate path as the polarity of the inductor reverses when the magnetic field collapses and the inductor becomes the source. This lowers the voltage from 1,000,000 volts to 0.7 volts. The downside is that the time it takes for the current decrease increases bo the ratio of 1,000,000/0.7. In a relay, this may cause the relay to "chatter" when opening. Adding a zener diode in series anode to anode with the spike suppressing diode will alleivate most chattering problems. A 34.3 volt zener will raise the voltage from 0,7v to 35v and shorten the time by a factor of 50 (35/.0.7). "
davenn said:it doesn't ! ... as soon as the external power source is removed the magnetic field starts to collapse
As it collapses it cuts through the windings of the inductor. This generates a voltage (EMF) across the inductor
with a current flowing in the opposite direction to the original externally supplied current
Dave
oh, ok, that makes total sense in that case, so if i have any coil of wire, it will somewhat resist changes in current, just because it takes work to decrease the current in it, which is supplied by the resistance?NascentOxygen said:The current is being held fixed and this maintains the field. No work is needed for this as the ideal inductor is lossless.
Just as no work is involved in keeping a rubber band stretched---you just stretch it over two fixed pegs spaced well apart on the wall.
(If using a lossless inductor you can just connect the two ends of the winding together and let the current go round and round forever.)
It takes work to increase the current in an ideal inductor, and the system returns that energy when you allow the current to decrease. If there is any resistance present it does its usual job of wasting some electrical energy as heat, according to i2.RPuglife said:oh, ok, that makes total sense in that case, so if i have any coil of wire, it will somewhat resist changes in current, just because it takes work to decrease the current in it, which is supplied by the resistance?
my2cts said:Not if you believe Feynman, as stated above.
Do you have the theory to prove this?
The "alternative theory" is an arxiv paper referenced by my2cts in post #57. The paper is here:
http://arxiv.org/abs/physics/0106078
I have only skimmed it and have not checked out the journal it was published in. But at a minimum I would say it is not appropriate for the thread it was posted in.
As for the issue actually being discussed in the thread, it looks to me like anorlunda is saying "the energy stored in the inductor has to be coming from the external circuit", which is true (and obvious from energy conservation), and my2cts is misinterpreting that as meaning "the energy stored in the inductor has to be coming through the wires", which, according to the understanding Doc Al describes (which is also my understanding of the proper application of Maxwell electrodynamics to this case), is not true. But anorlunda is not claiming that it is.