A long solenoid has 500 turns per meter and carries a current of 4...

The fractional increase in the magnetic field inside the solenoid is equal to the magnetic susceptibility $\chi$ of the material.

Therefore, $\chi = 2 \times 10^{-5}$.

The magnetic intensity $H$ inside the solenoid is given by $H = nI$, where $n$ is the number of turns per unit length and $I$ is the current.

$H = 500 \times 4 = 2000\text{ A/m}$.

The magnetization $M$ of the material is related to the magnetic intensity by the relation $M = \chi H$.

$M = (2 \times 10^{-5}) \times 2000 = 0.04\text{ A/m}$.

Answer: $0.04\text{ A/m}$