# Power Loss Due to An Eddy Current

• I
• BlackMelon
In summary: Finally, In summary, the power loss due to eddy current can be calculated using the formula P = (π^2 * f^2 * B_max^2 * τ^2 * h * L) / (6 * ρ) where τ represents the thickness of the laminated sheet and the integration bounds are from 0 to τ/2.
BlackMelon
TL;DR Summary
Finding power loss due to eddy current
Reference website: https://www.electricalvolt.com/2019/08/eddy-current-loss-formula/?expand_article=1
Hi there!

Recently, I am studying this kind of power loss from the following link:
https://www.electricalvolt.com/2019/08/eddy-current-loss-formula/?expand_article=1
Just to summarize an idea,

Supposed that we got a material, which is penetrated by a magnetic flux. The material will generate the eddy current to oppose the change of the flux.
We divide this material into portions.
We treat each of the portion as a one-turn coil, having I_eddy flowing through.
Use the Faraday's Law to find the induced voltage (E) in each portion.
Use R = rho*(length)/(area) to find the resistance of each portion (that I_eddy flows through)
The power loss of each portion is dP = E^2/R
Integrate dP over all the portions to get "P: The power loss due to eddy current."

From this equation in the link, I am curious why the bounds of the integration is 0 to T/2. Should it be -T/2 to +T/2 instead?

Best Regards,
BlackMelon

Last edited:
Your diagram is hard to read as posted, although not impossible. I don't see any "T", or, "τ", which is what I think you meant. Don't make it hard for people to help you, they may just give up and move on. Please communicate clearly, we are not clairvoyant.

Hi All,
https://www.mediafire.com/file/tftzo99pddsvzhj/EddyCurrentLoss.png/file

According to the diagram in this link:
https://www.electricalvolt.com/2019/08/eddy-current-loss-formula/?expand_article=1
T and τ are the same. It is said "Let the length, height and thickness of the laminated sheet is L,h and𝞃 respectively". However on the diagram, the thickness is represented by T, instead of 𝞃. The dotted line represents the reference position, where x = 0.

The interval of the integration is ## [0, \frac \tau 2] ## because the interval ## [-\frac \tau 2, 0] ## is already included into the integration through the formula for the area which is ## A = 2 h x = 2 \cdot h x ## and through the formula for the resistance which is ## R = \rho \cdot \frac {2 h + 4 x} {L dx} = 2 \cdot \rho \cdot \frac {h + 2 x} {L dx} ##.

If the interval of the integration is ## [-\frac \tau 2, \frac \tau 2] ## the formula for the area will be ## A = h x ## , the formula for the resistance will be ## R = \rho \cdot \frac {h + 2 x} {L dx} ##

and there will be next:

## \Phi (t) = B (t) \cdot A = B_{max} \cdot \sin (\omega t) \cdot h x ##

## E = \frac {\sqrt 2} {2} \cdot B_{max} \cdot 2 \pi f \cdot h x ##
## E = \sqrt 2 \cdot B_{max} \cdot \pi f \cdot h x ##

## dP = \frac {E^2} {R} ##
## dP = E^2 \cdot \frac {L dx} {\rho \cdot (h + 2 x)} ##
## dP = E^2 \cdot \frac {L dx} {\rho \cdot h} ##
## dP = (\sqrt 2 \cdot B_{max} \cdot \pi f \cdot h x)^2 \cdot \frac {L dx} {\rho \cdot h} ##
## dP = 2 \cdot B_{max}^2 \cdot \pi^2 f^2 \cdot h^2 x^2 \cdot \frac {L dx} {\rho \cdot h} ##
## dP = 2 \cdot B_{max}^2 \cdot \pi^2 f^2 \cdot h x^2 \cdot \frac {L dx} {\rho} ##

## P_{eddy} = \frac {2 \pi^2 \cdot B_{max}^2 \cdot f^2 \cdot h L} {\rho} \cdot \int_{-\frac \tau 2}^{\frac \tau 2} x^2 \, dx ##
## P_{eddy} = \frac{2 \pi^2 \cdot B_{max}^2 \cdot f^2 \cdot h L} {\rho} \cdot (\int_{-\frac \tau 2}^{0} x^2 \, dx + \int_{0}^{\frac \tau 2} x^2 \, dx) ##
## \int_{-\frac \tau 2}^{0} x^2 \, dx = \int_{0}^{\frac \tau 2} x^2 \, dx ##
## P_{eddy} = \frac{2 \pi^2 \cdot B_{max}^2 \cdot f^2 \cdot h L} {\rho} \cdot (\int_{0}^{\frac \tau 2} x^2 \, dx + \int_{0}^{\frac \tau 2} x^2 \, dx) ##
## P_{eddy} = \frac{2 \pi^2 \cdot B_{max}^2 \cdot f^2 \cdot h L} {\rho} \cdot 2 \int_{0}^{\frac \tau 2} x^2 \, dx ##
## P_{eddy} = \frac {4 \pi^2 \cdot B_{max}^2 \cdot f^2 \cdot h L} {\rho} \cdot \int_{0}^{\frac \tau 2} x^2 \, dx ##
## P_{eddy} = \frac{\pi^2 \cdot f^2 \cdot B_{max}^2 \tau^2 } {6 \rho}\cdot (h L \tau) ##
.

## What is an eddy current?

An eddy current is a loop of electrical current induced within a conductor by a changing magnetic field. These currents flow in closed loops within the conductor, perpendicular to the magnetic field, and can cause significant energy losses in the form of heat.

## How does an eddy current cause power loss?

Eddy currents generate power loss primarily through the resistive heating of the conductor. When these currents flow through the material, they encounter electrical resistance, which converts some of the electrical energy into heat. This is known as eddy current loss and can reduce the efficiency of electrical devices like transformers and motors.

## What factors influence the magnitude of eddy current losses?

The magnitude of eddy current losses is influenced by several factors: the electrical conductivity of the material, the magnetic flux density, the frequency of the changing magnetic field, and the thickness of the conductor. High conductivity, high flux density, high frequency, and greater thickness generally increase eddy current losses.

## How can eddy current losses be minimized?

Eddy current losses can be minimized by using materials with lower electrical conductivity, laminating the core of transformers and motors to reduce the path of the currents, employing higher resistivity materials, and designing components to operate at lower frequencies. These methods help to disrupt the formation of large eddy currents and reduce associated power losses.

## Why are laminated cores used in transformers and electric motors?

Laminated cores are used in transformers and electric motors to minimize eddy current losses. By dividing the core into thin, insulated layers, the path available for eddy currents is restricted, which reduces their magnitude and the associated power loss. This improves the overall efficiency of the device.

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