I was completing a question for homework and I don't really understand the answer. The question says that when a core goes from being laminated to unlaminated the voltage across the secondary coil goes down and it asks to explain why. The only thing I could think of was heat loss due to eddy currents. However, although this was mentioned in the answer it also said the following:

eddy current create magnetic flux This flux opposes flux from primary coil Max flux from primary coil is reduced Max rate of change of flux reduced so emf across secondary is reduced.

Thinking about this I don't understand how the back emf works. Take this as an example. Consider a power station is putting out a certain amount of power - call it 100 (arbitrary units) then the power at the primary is 100. Let us ignore the heat loss from eddy currents which will mean that the power at the secondary is also 100. However, even if you could eliminate heat loss from eddy current (e.g. superconductor) surely they would still create a back emf which would reduce the flux from the primary and hence reduce the emf at the secondary which would mean the power is NOT 100. So where would the missing energy go?


You obviously understand the loss of energy if you have eddy currents in a core which has resistance - you get Ohmic $I^2R$ heating.

A changing current in the primary produces a changing magnetic field in the core.
This changing magnetic field produces a changing magnetic flux through the secondary coil.
The changing magnetic flux through the secondary coil induces an emf in the secondary coil - Faraday's Law.

The changing magnetic flux through the core also induces an emf in the core itself.
If the core was made of an insulator that induced emf produced in the core would not matter because there would not be an induced current.

If the core has resistance then an induced (eddy) current will flow.
Lenz's law tells you that the induced current will always try and oppose the change which is producing it.
This is how it happens.

If the magnetic field in the core is increasing then the induced (eddy) current will produce a magnetic field in the opposite direction to that of the increasing magnetic field.
This decreases the rate of change of magnetic field in the core which in turn decreases the rate of change of magnetic flux in the secondary coil which in turn reduces the induced emf in the secondary coil.

There is a similar result of a reduced induced emf in the secondary coil when the magnetic field in the core is decreasing.
Then the magnetic field due to the induced (eddy) current will be in the same direction as the decreasing magnetic field and thus reduce the rate at which that magnetic field is decreasing.

The larger the induced (eddy) current the larger the opposition to the change producing it and so the smaller is the induced emf in the secondary coil.

To reduce this effect one must reduce the magnitude of the eddy currents and this is done by increasing the electrical resistance of the core by laminating (layering) the core.
The core is made of thin sheets of iron separated by thin sheets of an insulating material.

enter image description here

This also at the same time reduces the Ohmic heating within the core.

| cite | improve this answer | |

You are getting confused between back e.m.f. and power.

"Power" can only be transferred when there is a voltage and a current, with some component of the current in phase with the voltage.

The power delivered at the secondary is a consequence of that fact that when you allow current to flow in the secondary, this creates a magnetic field in the core that allows a phase shift of the current in the primary.

When you don't have a laminated core, you get such currents flowing although the secondary might be open-circuit. The power delivered by the primary results in currents flowing: either in the secondary circuit, or in the core. If no currents flow, then no work is done and no power is delivered.

Note that a real transformer has a finite resistance R (and not just inductance), which means that power will be dissipated in the primary winding as well - the above is valid for an "ideal" primary.

Does that help?

| cite | improve this answer | |
  • $\begingroup$ Thanks for the comments but this still doesn't quite help. If you imagine the power from a power station is 100W and this is made up of 100V carrying 1A. It reaches a transformer so the primary is at 100V and 1A. The transformer is a step up transformer which is designed to double the voltage. I want to assume there is no energy loss from heating so I use a superconducting core. I should get 200V and 0.5A at the secondary. But even with no heat loss you still get a supressed magnetic flux from the eddy currents which would reduce primary to less than 100V but surely this reduces the power? $\endgroup$ – user37250 Feb 14 '16 at 23:51
  • $\begingroup$ The current won't be 1A unless there is something that takes the 100 W of power... if you connect a 100 V AC signal to an "open" transformer, you may find that you only draw a few W due to losses in the transformer. If your core is superconducting, it will suppress the generation of any magnetic flux inside the core - meaning that the transformer will appear to have no inductance, or that it looks like a short circuit on the primary side. The short circuited core is indistinguishable from a short circuited secondary. $\endgroup$ – Floris Feb 15 '16 at 0:39
  • $\begingroup$ thanks again. if you have a sufficient load how can the ratio of turns to voltge be the same (i.e. the transformer equation)? if you have back emfs from eddy currents and secondary coil wouldn't the primary voltage always get supressed? $\endgroup$ – user37250 Feb 15 '16 at 0:49
  • $\begingroup$ Yes eddy currents look like a "parasitic secondary" and this means you need a slightly higher number of turns in the secondary to get the voltage transformation you expect. The difference between an ideal, and a real transformer... $\endgroup$ – Floris Feb 15 '16 at 12:53
  • $\begingroup$ Thank you. From what I have been reading it says that heating from eddy currents gives a power loss but suppression of the primary flux from the eddy currents is not a form of energy loss - is this right? Could you explain how the secondary voltage is reduced without a power loss to the system? $\endgroup$ – user37250 Feb 15 '16 at 13:29

There are two ways eddy currents affect the transformer action. One is by dissipating energy, and the laminations cause longer pathlengths for the induced currents, which raises the resistance, lowering that dissipation. But the other, is that they shield a large part of the core from being magnetized. Your iron core is magnetized first in its outer layer, then the magnetization progresses to the center. While eddy currents actually surround most of the core's cross section, they keep it from magnetizing (and your transformer has much less induction, as though it had only an air core).

| cite | improve this answer | |
  • $\begingroup$ Thanks. The second point you made about the eddy currents keeping the core from being magnetised - what affect does this have on the primary and secondary voltages because if it reduces primary flux then it would lower secondary voltage so how is energy conserved here as I thought that back emf was not a form of energy loss? $\endgroup$ – user37250 Feb 15 '16 at 10:11
  • $\begingroup$ While eddy currents cause part of the core to remain unmagnetized, they do this only for a short time (and that means it is important only at higher frequencies than the designer intended- the core is sized for its intended use). At those high frequencies, the inductance of the primary is less than its low-frequency value, and the back-emf is not strictly proportionate to the low-frequency inductance, but to the lesser high-frequency inductance. This mainly means that one-frequency measurement of an inductor doesn't tell the whole story. $\endgroup$ – Whit3rd Mar 10 '16 at 8:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.