# How is the holding force of this electromagnet so large?

I am designing a system that uses the magnetic force to fix an object in its place. I plan to connect an I-core to my object and fix it to a fixed C-core by letting a magnetic flux flow through both cores. By using $$E = \frac{B^2V}{2\mu_0\mu_r}$$ for the magnetic energy, with $$\B\$$ the flux density and $$\V\$$ the volume of the magentic core, my derivation for the magnetic force would be: I drew an E-core, but it doesn't really matter for the derivation. The holding force is therefore given by: $$F = \frac{B^2A}{2\mu_0\mu_r}$$

While searching for a magnetic core, I talked with several manufacturers, and they all said that the maximal saturation flux density of state-of-the-art materials is around $$\B_{sat}=1.5-2 \text{ T} \$$, which limits the holding force. I presumed that in order to increase the holding force, I must use a larger core (increasing $$\A\$$), but then I saw this core from Amazon, which has a holding force of around $$\1.2 \text{ kN}\$$. If we assume $$\B=2 \text{ T} \$$, $$\\mu_r=3000\$$, and solve the equation for the holding force for $$\A\$$, we get $$\A=2.2 \text{ m}^2 \$$, which is much larger than the area of the electromagnet I saw on Amazon. What am I missing? It would have worked if the holding force term didn't contain $$\\mu_r\$$, which reduces the area to $$\A=7.52 \text{ cm}^2 \$$. Indeed, Wikipedia says that the holding force is given without $$\ \mu_r \$$, but I don't understand why it doesn't include in the magnetic energy term.

• @Arad You need to use $ and not  or . This is an issue for this site. You just have to memorize it. The  does work but it occupies the entire line and doesn't work within a sentence. Just FYI. Commented Oct 10, 2023 at 17:58 • there is a problem with editing on multiple SE sites ... the sites will probably go down for maintenance soon Commented Oct 10, 2023 at 18:12 • @Arad$\mu_r\\$ isn't in the equation. So you worked that out, already. Commented Oct 10, 2023 at 18:21
• Can you state what the parameters are in your equation (first one... I recognize the denominator of course but, E, B and V are still confusing me) Commented Oct 10, 2023 at 18:32
• FYI, no need to go through a lengthy derivation: B^2 / (2 mu) is an energy density, which is also a pressure, the Maxwell stress. Multiply by area and you have F simply. Commented Oct 10, 2023 at 20:27

The holding force doesn't contain the $$\\mu_r\$$ term, because when you try to open the gap, you're computing the increase in magnetic energy in the gap, to differentiate with respect to distance, and the gap is vacuum, or at least air, which is much the same thing for magnetism.

You do use the $$\\mu_r\$$ term to compute the ampere.turns/m you need, the H field, to get the flux through the core in the first place, as it's the core that 'shortens the magnetic distance' round the core by a factor of $$\\mu_r\$$ compared to that in air.

It's worth having a handy conversion ratio to hand, for evaluating these door locks. If you have a 1 T field in the gap when closed, there is a pressure of 400 kPa, 4 atmospheres, 4 Bar, keeping them together. The pressure increases as the square of the field, so 16 Bar if you manage to get 2 T.

• Thanks! you helped me find my way to the solution, I edit the main post
My answer is inspired by the answer of Neil_UK. I (the OP) understand now where my problem was. The derivative I (the OP) calculated was for the magnetic energy in the core, but it's actually not the right term to calculate the energy. The total energy is given by: $$E = \frac{B^2V_\text{core}}{2\mu_0\mu_r} + \frac{B^2V_\text{gap}}{2\mu_0}$$ Where $$\E_\text{gap} >> E_\text{core} \$$. Now $$\\text{d}V_\text{core} = 0\$$ since it's a property of the core. Now if we continue the derivation like above (With $$\ V = V_\text{gap} \$$) we get the right term for the holding force: $$F = \frac{B^2A}{2\mu_0}$$