A long solenoid has 1000 turns per meter and carries a current of 1A. It has a soft iron core of ${\mu }_r={10}^3$. The core is heated beyond curie temperature $T_c$

The $\overrightarrow{H}$ will remain unchanged, but the $\overrightarrow{B}$ will be high when the temperature is less than the curie temperature.

When temperature of the iron core of solenoid (which is ferromagnetic materials) is raised beyond curie temperature, then soft iron core behaves as para magnetic material.

Causing $\overrightarrow{B}$ to decrease drastically. 

$\therefore \frac{{\left({\chi }_m\right)}_{\text{Ferro }}}{{\left({\chi }_m\right)}_{\text{Para }}}=\frac{{10}^3}{{10}^{-5}}={10}^8$