# Solenoid with iron core

If I use a solenoid of $$h = 0.1\, \mathrm{m}$$, with a current of $$I = 30\, \mathrm{A}$$ , with $$62$$ turns and an iron core of permeability $$\mu = 0.25$$, then it will give me a magnetic field of:

$$B = \frac{\mu \times I \times N}{h} = 4650 \, \mathrm{T}.$$

That makes no sense. Did I just break the Guinness record with a home experiment?

• It's really the correct permeability according to Wiki: en.wikipedia.org/wiki/… Commented Jan 19, 2022 at 4:06
• Permeability you took here might be incorrect or it might be the relative Permeability. I think it's much more easy to understand that physically it's not possible for a solenoid of length jusr 0.1 meter and second of all the equation which you've used is only for the solenoid with the condition where radius of solenoid is << length of solenoid and you haven't specified the radius of the solenoid as well so it is difficult to conclude anything, But hey!!. And This is not possible for a normal solenoid to give you this massive amount of magnetic field strenght... Commented Jan 19, 2022 at 4:58

Your value of the permeability is not realistic. The real magnetic permeability of iron is complicated and depends on the geometry, the crystal structure, the purity, and the history of a particular sample. A typical value of the relative permeability for good everyday iron would be $$\mu_{r}\sim 1000$$. This makes the absolute permeability roughly $$\mu=\mu_{r}\mu_{0}=\mu_{r}\left(4\pi\times 10^{-7}\right) \,\mathrm{H}\cdot\mathrm{m}^{-1}\sim10^{-3}\,\mathrm{H}\cdot\mathrm{m}^{-1}.$$ For exceptionally pure and ideally machined iron, it is indeed possible to achieve permeabilities hundreds of time higher. However, it takes very little impurity to spoil this; the value you mention was for 99.95-percent-pure iron in the absence of reactive oxygen. Moreover, it is probably not really possible to maintain that effective permeability for large samples like your $$10\,\mathrm{cm}$$ example.