# How to show laminations minimizes eddy current?

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1. Jun 8, 2015

### kelvin490

In transformer, lamination is an effective method to minimize eddy current and thus energy loss. I just wonder how can we show that the energy loss due to eddy current in those laminates is added up to be less than that without lamination (suppose the total cross sectional areas are the same)?

2. Jun 8, 2015

### Hesch

There are also losses in iron due to magnetic hysterisis in iron:

I think the hysterisis losses will be the same, laminated iron or not. But anyway:

Connect the (unloaded) transformers to the mains supply. Measure their power consumptions by a watt-meter, or measure the temperature gradients of the cores.

Last edited: Jun 8, 2015
3. Jun 8, 2015

### Staff: Mentor

What is the Relevant Equation for calculating the current in a loop with a changing magnetic field piercing the loop? What part of that equation are you affecting when you divide up a piece of metal into laminations that are insulated from each other?

4. Jun 10, 2015

### kelvin490

I don't know...

5. Jun 10, 2015

### Staff: Mentor

6. Jun 10, 2015

### kelvin490

Path lengths are different, so the integral for one laminate is small. If I add it up, is it smaller than one large piece?

7. Jun 10, 2015

### Staff: Mentor

That may be one effect, but there is a more important one. Under Faraday's Law, when you have an area that has a changing flux, that induces a voltage around the loop area, which generates an eddy current in this case. Since the induced voltage and resulting eddy current depends on the _____ ...

8. Jun 10, 2015

### Hesch

@kelvin490: You must regard that laminated iron is mixed up with silicon. This silicon decreases the magnetic proporties a little, but the electrical proporties (conductivity ) are decreased much more, thereby lowering the Eddy current.

So be cautious analysing the results of the measurements.

Last edited: Jun 10, 2015
9. Jun 10, 2015

### Hesch

Solution to this problem:

Build a cube of laminated iron and make some coil that fits this cube. ( Saw a piece of the centerleg out of some transformer core ).

Now you induce a magnetic (AC) field parallel to the lamination, thereafter (moving the coil) induce the magnetic field perpendicular to the lamination.

Your measurements will be different ( I promise ). Think about why?

Is the magnetic field in a transformer parallel or perpendicular to the lamination?

Last edited: Jun 10, 2015
10. Jun 10, 2015

### kelvin490

Thanks for the answer. But I am not asking for experimental method to show they are different. Lamination has been used in transformers for years, why do I need experimental method to prove it?

What I ask is a theoretical approach.

11. Jun 10, 2015

### kelvin490

Area?

12. Jun 10, 2015

### Staff: Mentor

Yes!

The area ratios how with the circumference?

13. Jun 10, 2015

### kelvin490

But we need to show their energy loss added up to be smaller than that in a larger, undivided area.

14. Jun 10, 2015

### Staff: Mentor

So this is schoolwork?

And if so, why haven't you been studying Faraday's Law as part of the preparation for this homework question?

15. Jun 10, 2015

### kelvin490

It's not HW.

16. Jun 11, 2015

### Hesch

Last edited by a moderator: May 7, 2017
17. Jun 11, 2015

### kelvin490

Last edited by a moderator: May 7, 2017
18. Jun 11, 2015

### Staff: Mentor

BTW @kelvin490 -- Can you guess why higher frequency ferrite transformers do not have laminations?

19. Jun 11, 2015

### kelvin490

High reactance?

20. Jun 11, 2015

### Staff: Mentor

No, not reactance...

21. Jun 11, 2015

### kelvin490

low current.

22. Jun 11, 2015

### Staff: Mentor

Nope. What property of a magnetic material would minimize eddy currents?

23. Jun 11, 2015

### kelvin490

High permeability.

btw, could you help me on the Ampere's law problem in another thread? May be it is a stupid question but I have no idea how to solve that problem.

24. Jun 11, 2015

### Staff: Mentor

Nope. Try googling Ferrite Transformer Eddy Currents -- the first hit on the list for me is a wikipedia article that explains it.