# Transformer losses due to eddy currents. Why these losses are compensated by the transformer?

Here is my understanding of a transformer. Let's assume that we have the following ideal circuit, in which a coil is wrapped around a core and there are no core losses⎯excuse my crude drawing.

And there's a current flowing through the coil. Now, this current produces a magnetic field, as shown above. Now, because we have a ferromagnetic core, then, from my understanding, the flux density at points must be

$$B_1 > B_2 > B_3$$

Whereas the "outside" of the core must be small due to air's low magnetic permeability. The difference in flux density is a consequence of Ampere's law, the intensity of the magnetic field drops as you move away from the source which produced it $$\oint\, \vec{H}\cdot\, \vec{ds} = i$$

And $$B = \mu H$$

Now, since there is no secondary winding, a current must flow through the coil such that the time-varying magnetic flux produces the input voltage $$V_{in} = N\frac{d\phi}{dt}$$

But now, if we assume that the core is not ideal, then due to the eddy currents, that is currents that are induced in the core due to the time-varying magnetic field, there is another current component to the excitation current in order to compensate this loss to produce the flux needed that induces the input voltage $$i_{ex} = i_m + i_{eddy}$$

My question is, why that is? Given that this loss is local to the core, the magnetic field produced by the coil closes through the coil, so why is this extra current needed to account for the losses since it doesn't affect the flux through the primary coil? My understanding is clearly wrong, I would like a purely physical explanation, if possible.

• "as shown above", um, no, the magnetic field lines would mostly follow the shape of the core, it's µ being much larger than that of free space Commented Apr 22 at 11:58
• @MarcusMüller What makes you say that they "follow" the shape? I would imagine that the magnetic field is radial, and the flux density just happens to be higher in the regions of the core at points where lines pass through, as in the picture, and this would lead to a flux density gradient along the core Commented Apr 22 at 12:14
• I agree with Marcus. Those field lines are not what you will get with a decent core in place AND you can assume all the flux densities are equal. Commented Apr 22 at 12:46

You are talking about initial state. Eddy current oppose the flux produced i.e. oppose/decrease the magnitude of excitation voltage, which in turn decrease the current and flux, if not compensated during the dynamic (i.e. after zero time). This become relevant if flux needed to be constant or follow a desired curve in cases like motor.

The thing is that the induced currents had their own induced field. Anyway, your initial assumption wasn't wrong. I'll try to use some simulations to support my arguments, but your question involves plenty of phenomena that needs to be addressed separately.

What your drawing didn't capture is the high magnetic permeability of the core, which you can observe below in the following simulation result. You can see that the flux lines are confined within the core. (flux density is near 1.8T here)

However, if we increase the flux density towards saturation, the reluctance of the core path increases and there will be some 'rebel' flux lines taking the shortcut, similarly to your drawing. (now flux density is over 2T)

Anyway, the current drawn due to the induced currents will always be quite low because the core is laminated, which leads to very low effective conductivity. But if there are some induced currents, of course it will drawn current from the source. It's true even for the no-load condition where only the magnetization current will flow.

Alternatively, when some load is applied to the secondary, higher currents will flow and there will be some leakage flux between primary and secondary. This flux tends to impinge metallic structure and induced higher load losses. But it's important to understand that the induced currents will produce it's own field that will affect the whole flux distribution. I mean, you can use superposition to understand that the final flux distribution will be the sum of the source field plus the induced currents field.