As we know that, the magnetic field intensity $\mathrm{H}=\mathrm{nI}$, where $\mathrm{n}=$ number of turns per metre of a solenoid and $\mathrm{I}=$ current.
And the magnetic induction
$$
\mathrm{B}=\mu_0 \mu_{\mathrm{r}} \mathrm{nI}
$$
So, for solenoid $\mathrm{H}=\mathrm{nI}$
$$
\mathrm{H}=1000 \times 1=1000 \mathrm{Am}
$$
( $\because$ given, $\mathrm{n}=1000)$
So, $\mathrm{H}$ is a constant, then it is nearly unchanged.
From(i),
$$
\begin{aligned}
\mathrm{B} &=\mu_0 \mu_{\mathrm{r}} \mathrm{nI} \\
&=\left(\mu_0 \mathrm{nI}\right) \mu_{\mathrm{r}}=\mathrm{k} \mu_{\mathrm{r}} \\
&(\therefore \text { where, } \mathrm{K}=\text { constant })
\end{aligned}
$$
So, $\quad \quad B \propto \mu_r$
Hence, we find that, B varies with the variation in $\mu_{\mathrm{r}}$. By curies law, when temperature of the iron core of solenoid or ferromagnetic material is raised beyond Curie temperature, then it behave as paramagnetic material, where,
Susceptibility of $\left(\chi_{\mathrm{m}}\right)_{\mathrm{Ferol}}=10^3$
Susceptibility of $\left(\chi_{\mathrm{m}}\right)_{\text {Para2 }}=10^{-5}$
So, $\frac{B_1}{\mathrm{~B}_2}=\frac{X_1}{X_2}=\frac{\left(\chi_m\right)_{\text {Fero1 }}}{\left(\chi_{\mathrm{m}}\right)_{\text {Para } 2}}=\frac{10^3}{10^{-5}}=10^8$,
$\mathrm{B}_1=10^8 \mathrm{~B}_2$
or, $\quad \mathrm{B}_2=10^{-8} \mathrm{~B}_1$