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Your title (above) is problematic because, you cannot design a transformer core without knowing the number of turns. Forget about the secondary winding; that comes good after you design the primary and, whatever you might have learned or might have been taught, the primary winding is still just a plain ordinary inductor. So, the first part to designing a transformer is to forget that it's a transformer and, instead, design an inductor where that inductor also serves as the primary winding.

  Ideally, you would want the inductance to be high so that when you apply an AC voltage to it, the current is low. Remember I said we are designing an inductor first and foremost so, I didn't refer to that current as the magnetization current (although that is what it becomes when we add a secondary and think about it as a transformer). But from now, I will call it magnetization current because it is this current that magnetizes the core and creates magnetic flux in the core.

So, what determines that flux level and, how do we convert magnetization current into flux? The starting formula is what we call magneto motive force (MMF). It equals the number of turns multiplied by the current that flows in that coil.

This is only part of the story; we need to calculate the \$H\$ field (magnetic field strength) in the prospective core and, this is MMF divided by length of the magnetic field path that circulates in the core: -

$$H = \dfrac{MMF}{\ell}$$

A few words about the above. H is made smaller by having a longer magnetic field path. Smaller H leads to lower flux density and this is a good thing. However, a longer path length leads to a lower primary inductance and, this is not ideal given that we want high inductance to avoid a significant level of core saturation. So, generally, when we make the magnetic path longer we also increase the cross sectional area of the core so that the inductance isn't compromised too much.

Then we have a formula that most folk will have seen and relates to what is known as the BH curve. That curve is the amount of flux density within a core that exists due to a driven level of H-field. It relates the permeability of the core to B and H: -

But, it still all comes back to the number of turns on the core and how much inductance that creates. Most cores have an \$A_L\$ figure that tells you the inductance for 1 turn and, if you double the turns the inductance becomes 4 times greater hence, the working inductance of a given coil is: -

_{Note that I've used \$A_e\$ for the core cross sectional area.}

And how does this relate to the formulas shown above?

Start by understanding that V is the RMS value of the applied sinewave and that we are interested in knowing the peak magnetization current that flows due to V being applied across the primary coil of inductance L (a reactance of \$2\pi f\cdot L\$): -

Your title (above) is problematic because, you cannot design a transformer core without knowing the number of turns. Ignore the secondary winding for now; that naturally comes after you design the primary and, whatever you might have been taught, the primary winding is still just a plain ordinary inductor. 

So, the first part is to forget that it's a transformer and, instead, design an inductor where that inductor eventually becomes the primary winding. Ideally, you want inductance to be high so that when you apply an AC voltage, the current drawn is low. This current is what magnetizes the core and creates magnetic flux in the core.

So, what determines that flux level (and possibly saturation) and, how do we convert the magnetization current into flux? The starting formula is what we call magneto motive force (MMF). It equals the number of turns multiplied by the current that flows in that coil.

This is only part of the story; we need to calculate the H-field (magnetic field strength) in the prospective core. This is MMF divided by the length (\$\ell\$) of the magnetic field path that circulates in the core: -

$$H \text{ (magnetic field strength)}= \dfrac{MMF}{\ell}$$

A few words about the above...

The H-field is reduced by using a longer magnetic field path (\$\ell\$). A reduced H-field leads to a lower flux density (a good thing). However, a longer path length leads to a lower primary inductance and, this is not ideal given that we want high inductance to avoid saturation of the core. So, generally, when we make the magnetic path longer we also increase the cross sectional area of the core so that the inductance isn't compromised too much.

Then we have a formula that most folk will have seen and relates to what is known as the BH curve. That curve is the amount of flux density within a core that exists due to a driven level of H-field: -

Image from my basic website. 

It relates the permeability of the core to B and H: -

But, it still all comes back to the number of turns on the core and how much inductance that creates. Most cores have an \$A_L\$ figure (aka permeance) that tells you the inductance for 1 turn and, if you double the turns the inductance becomes 4 times greater hence, the working inductance of a given coil is: -

_{I've used \$A_e\$ for the core cross sectional area. The "e" stands for "effective" hence \$A_e\$ is "effective area".}

How does this relate to the formulas I've shown above?

Start by recognizing that V is the RMS value of the applied sinewave and that we are interested in knowing the peak magnetization current that flows due to V being applied across the primary coil of inductance L (a reactance of \$2\pi f\cdot L\$): -

And how does this relate to the formulas shown earlier in my answer?

$$I = \dfrac{\sqrt2\cdot V}{2\pi f\cdot L}$$

_{Note that I've used \$A_e\$ for the core cross sectional area.}

And how does this relate to the formulas shown above?

$$I = \dfrac{\sqrt2\cdot V}{2\pi f\cdot L}$$

_{The \$\sqrt2\$ comes from converting RMS current to peak current.}

But what about the formula at the top of the question: -

$$V = 4.44\cdot f\cdot N\cdot B \cdot A_e$$

And how does this relate to the formulas shown earlier in my answer?

Start by understanding that V is the RMS value of the applied sinewave and that we are interested in knowing the peak magnetization current that flows due to V being applied across the primary coil of inductance L (a reactance of \$2\pi f\cdot L\$): -

$$I = \dfrac{\sqrt2\cdot V}{2\pi f\cdot L}$$

We can multiply current by the number of turns (N) and divide by the path length (\$\ell\$) to get H then multiply by permeability (\$\mu\$) to get B: -

$$B = \dfrac{\sqrt2\cdot V\cdot N\cdot \mu}{2\pi f\cdot L\cdot \ell}$$

But, we also know that \$L = N^2\cdot A_L\$ hence: -

$$B = \dfrac{\sqrt2}{2\pi f}\cdot\dfrac{V\cdot N\cdot \mu}{\ell\cdot N^2\cdot A_L}$$

From inductor theory we know that \$A_L\$ is the core permeance (the inverse of core reluctance) and this equals \$\hspace{0.2cm}\mu\cdot\dfrac{A_e}{\ell}\$ so we can say: -

$$B = \dfrac{\sqrt2}{2\pi f}\cdot\dfrac{V\cdot \mu}{\ell\cdot N\cdot \mu\cdot\frac{A_e}{\ell}}$$

Rearranging and cancelling terms we get: -

$$V\hspace{0.5cm} = \hspace{0.5cm}\dfrac{2\pi}{\sqrt2}\cdot f\cdot N\cdot B \cdot A_e\hspace{0.5cm} \approx \hspace{0.5cm}4.443\cdot f\cdot N\cdot B \cdot A_e$$