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Here is a transformer where we can see the eddy currents in the core:

enter image description here

I would like to estimate the voltage which induces the eddy currents. I know what voltage is applied to the primary and how much time this voltage is applied. The number of turns on the primary is 1,so I am able to determine what the variation of the flux is.

How can I estimate the voltage into the core as there is no winding, ie no turn?

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    \$\begingroup\$ When using Maxwell's equations to find induced emf from rate of change of flux, the loop imagined need not be a physical winding; it can be any imagined, closed loop. See the integral form labelled Faradays law. The eddis form in different sizes, so the voltage causing each eddy will be different. \$\endgroup\$
    – AJN
    Commented May 17, 2021 at 12:36
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    \$\begingroup\$ Thank you for your reply :) Not simple to estimate. If I consider the biggest section of my transformer and highest variation of the flux. I will be able to get the highest voltage ? \$\endgroup\$
    – Jess
    Commented May 17, 2021 at 13:43
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    \$\begingroup\$ But what do you want to do with that information ? The phrase "voltage into the core" is not clear to me. Perhaps you can detail why you need to calculate this value. \$\endgroup\$
    – AJN
    Commented May 17, 2021 at 14:02
  • \$\begingroup\$ Indeed it is not clear ... \$\endgroup\$
    – Jess
    Commented May 17, 2021 at 14:46

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I would like to estimate the voltage which induces the eddy currents.

Suppose the total flux through the core of the transformer is given by some function of time \$\Phi(t)\$. Suppose there are N equally sized laminations in the core. We can call the flux through any single lamination \$\phi(t)\$. Assuming that the flux is evenly distributed through the laminations,

$$\phi(t) = \frac{\Phi(t)}{N}$$

There current in a lamination will not be uniform, but will depend upon the loop for that current, and the amount of (changing) flux linked by that loop. The loop with the largest amount of linked flux will be the loop that follows the surface of the lamination. That loop will link all the flux in the lamination.

The emf of that loop will simply be

$$ V = -\frac{d\phi}{dt}$$

The average emf of the current loops in a lamination will obviously be lower.

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    \$\begingroup\$ A good answer to an unusual question. Usually, it is sufficient to know the expression to compute the losses generated in the form of heat. \$\endgroup\$ Commented May 17, 2021 at 15:45

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