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I am trying to design an inductor. I have free hand of using any geometric form and ferromagnetic material.

It's my first time doing it. I studied magnetics design in college and I always thought that the cross section of a material can affect the saturation, flux density [T] and stored magnetic energy [H].

I was baffled when double checking my math to see that the cross section ONLY affects the magnetic energy (inductance) in the following conditions:

  • When I increase the ferromagnetic material cross section, the inductance [H] increases.
  • When I decrease the ferromagnetic material cross section, the inductanc3 [H] decreases

How can I, for a known DC magnetic field, use a ferromagnetic material of 1mm^2 cross section or a feromagnetic material of 100mm^2 and have the same flux density [T] through it?

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  • \$\begingroup\$ cross section of a ferromagnetic material with a known relative permeability \$\endgroup\$
    – i33SoDA
    Commented Nov 5, 2023 at 17:00
  • \$\begingroup\$ To obtain the same flux density for different cross area you can change number of turns. It is also possible change voltage or frequency if you can. Or core material. \$\endgroup\$ Commented Nov 5, 2023 at 18:41

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Why doesn't the cross section of a material affect flux density?

To answer this, a good place to start is understanding magnetic reluctance \$(\mathcal{R})\$.

\$\mathcal{R}\$ defines the total flux \$(\Phi)\$ produced from the applied magneto motive force \$(\mathcal{F})\$.

As a reminder, \$\mathcal{F} = N\cdot I\$ (number of turns × current). Hence, we get: -

$$\mathcal{R} = \dfrac{N\cdot I}{\Phi}\hspace{1cm} \text{or} \hspace{1cm} \Phi = \dfrac{N\cdot I}{\mathcal{R}}$$

So, reluctance is a kind of limiter to the flux based on the applied current just like ohmic resistance determines current flow when a voltage is applied. You can liken this relationship to the equivalent of ohm's law but for magnetics.

But, \$\mathcal{R}\$ also can be shown to equal this: -

$$\mathcal{R} = \dfrac{\ell}{\mu_0\cdot \mu_r\cdot A}$$

Where \$A\$ is the cross sectional area of the core, \$\ell\$ is the length of the path around the core in which the lines of flux take and, the two other terms define the magnetic permeability. Then, if you equate the two \$\mathcal{R}\$ formulas, you find this: -

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

Therefore, the total flux produced is proportional to area and, this means that flux density (\$B\$) remains constant. In other words, if area increases then flux increases but, the bigger area means that flux density \$(B)\$ remains constant.

If you keep \$A\$ constant, then flux (and flux density) both reduce with a longer mean length around the core. If you wanted to go further you can use the definition of inductance per turn: -

$$\dfrac{L}{N} = \dfrac{\Phi}{I}\hspace{1cm}\text{or}\hspace{1cm}L = \dfrac{\Phi\cdot N}{I}$$

Then you can use the \$\Phi\$ formula shown earlier to find inductance: -

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

I'm sure you might have come across that formula.

How can I, for a known DC magnetic field, use a ferromagnetic material of 1mm^2 cross section or a ferromagnetic material of 100mm^2 and have the same flux density [T] through it?

As you increase the cross sectional area of the core, you also increase the inductance and, because higher inductance produces more flux per amp, the flux density remains constant despite the area term.

Of course, if you were driving the inductor with a voltage at a fixed frequency then, as the inductance increased (due to cross sectional area increasing), so does the inductive reactance and, inevitably, this means there is less current flow. So now, we have a scenario where the flux density reduces when cross section area increases.

This is extremely important in switch mode power supply design.

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  • \$\begingroup\$ Great! Thank you! Makes perfectly sense! \$\endgroup\$
    – i33SoDA
    Commented Nov 6, 2023 at 7:19
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Well, your premise is "for a known magnetic field" (presumably B). That the presence of a magnetic conductor is not going to affect the "known" magnetic field is sort of an audacious assumption.

In a similar vein, the cross section of an electric conductor does not influence the current density for a known electric field. But that assumes that the electric field is not fazed by putting a conductor in, and that means a voltage source that is strong compared to the conductivity of the conductor you are going to place, and enough resistance that the conductor won't just evaporate due to the developing heat.

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