Calculate the inductance of a solenóide wound on an iron core

[Teslas]

Hello friends of the forum! I'm here with a doubt! I have a solenoid of cross-sectional area 5cm and length 9cm with iron core of relative permeability (ur) 9,000, I wonder how to calculate this silenoid!

 

[tech99]

Hello friends of the forum! I'm here with a doubt! I have a solenoid of cross-sectional area 5cm and length 9cm with iron core of relative permeability (ur) 9,000, I wonder how to calculate this silenoid!

I think the difficulty is that the effective permeability of the core will be much less than 9,000 because the magnetic path is partly air. This problem comes up in connection with ferrite rod antennas, and there is an article here: http://g3rbj.co.uk/wp-content/uploads/2014/06/Web-The-Inductance-of-Ferrite-Rod-Antennas-issue-3.pdf

 

[Teslas]

I think the difficulty is that the effective permeability of the core will be much less than 9,000 because the magnetic path is partly air. This problem comes up in connection with ferrite rod antennas, and there is an article here: http://g3rbj.co.uk/wp-content/uploads/2014/06/Web-The-Inductance-of-Ferrite-Rod-Antennas-issue-3.pdf

My question is

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[Teslas]

My question is

 

[Teslas]

My question is

 

[tech99]

In the case of the air core, we see the formula with uo in it. But for the second case, iron core, we must also multiply by the effective mu of the magnetic path.

 

[Teslas]

[QUOTE = "tech99, post: 6127760, member: 477979"] No caso do núcleo de ar, vemos a fórmula com uo nela. Mas para o segundo caso, núcleo de ferro, devemos também multiplicar pelo mu efetivo do caminho magnético. [/ QUOTE]
Amigo, obrigada por me responder! Estou tentando fazer um eletroímã! E eu não conheço a bitola do fio, não sei a equação, alguém poderia colocar um exemplo com essa fórmula que eu coloquei? Yess permeability magnetc of iron core yesss!

 

[Teslas]

This question is magnéticFild ( B)

 

[Teslas]

This question

In the case of the air core, we see the formula with uo in it. But for the second case, iron core, we must also multiply by the effective mu of the magnetic path.

You are correct, we must multiply by the magnetic permeability of the vac (uo) together with the permeability of the magnetic nucleus (ur) in this image that I put, I am in doubt in this formula, B = k.uo.nl, this formula is missing divide by length (L) am I right?