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I am doing an experiment where I am making a power transformer where I am keeping the cross sectional area of primary leg and secondary leg different. I don't want the primary leg to hold more than a certain amount of flux in it, I want to know a formula that describes the relationship between cross sectional area of primary leg, length of primary leg, number of turns of coil, voltage on coil (frequency as well if applicable) so that I can choose the primary leg with most optimum cross sectional area to hold given magnitude of inductance.

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    \$\begingroup\$ Sounds to me like a usual transformer design problem -- most will attempt to find the optimal cross-section for a given application. \$\endgroup\$ Commented Aug 18, 2022 at 14:44
  • \$\begingroup\$ You ask about area, but say the fluxes will be equal, so the voltages will be equal (as expected from turns ratio, less the leakage). What's the difference? Do you actually want different fluxes (a magnetic shunt is required) instead? \$\endgroup\$ Commented Aug 18, 2022 at 21:46
  • \$\begingroup\$ This makes little sense. If your primary cross sectional area is limited, making any other part of the core have a larger cross sectional area will only add cost and losses. \$\endgroup\$
    – winny
    Commented Aug 23, 2022 at 9:50

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I want to know a formula that describes the relationship between cross sectional area of primary leg, length of primary leg, number of turns of coil, voltage on coil (frequency as well if applicable)

Start with the voltage formulas associated with an inductor: -

$$V = L\dfrac{di}{dt}\hspace{1cm}\text{and}\hspace{1cm}V = N\dfrac{d\Phi}{dt}$$

So, if you equate them you get this: -

$$L\dfrac{di}{dt}= N\dfrac{d\Phi}{dt}$$

And, it follows that inductance per turn equals flux per amp: -

$$\dfrac{L}{N} = \dfrac{\Phi}{I}$$

Then you can then use this basic inductance formula: -

$$L = \dfrac{\mu_0\mu_R N^2 A}{\ell}$$

  • \$A\$ is the cross sectional area of the coil and
  • \$\ell\$ is the mean path round the transformer core that the flux travels.

So you can now make this equation: -

$$\dfrac{\mu_0\mu_R N A}{\ell} = \dfrac{\Phi}{I}$$

And you can convert flux (\$\Phi\$) to flux density (\$B\$) by multiplying by area hence: -

$$\dfrac{\mu_0\mu_R N }{\ell} = \dfrac{B}{I}$$

And this formula boils down a bit further: -

$$\mu_0\mu_R\cdot\dfrac{N\cdot I}{\ell} = B$$

Also known as \$B = \mu\cdot H\$ or...

Flux density = H-field multiplied by magnetic permeability (\$\mu_0\$ and \$\mu_R\$ combined).

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This looks like it should give you all the formulas you need other than knowing that the applied voltage and primary reactance dictate the current that flows for a given frequency i.e. \$X_L = 2\pi f L\$.

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