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?
 A: 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.
A: Another issue is that this equation seems to assume same permeability through the whole space. But solenoid's core is not closed, not a loop, magnetic flux has to go out in empty space on the outside of the coil. So effective permeability will be (much?) less than permeability of the core material.
