A long solenoid has 1000 turns per meter and carries a current of 1 A. It has a soft iron core of ${\mathrm{\mu }}_{\mathrm{r}}=1000$. The core is heated beyond the Curie temperature, ${\mathrm{T}}_{\mathrm{C}}.$

(a) The H field in the solenoid is (nearly) unchanged but the B field decreases drastically.
(b) The H and B fields in the solenoid are nearly unchanged.
(c) The magnetization in the core reverses the direction.
(d) The magnetization in the core diminishes by a factor of about ${10}^{8}$.

1. (a, c)
2. (a, d)
3. (b, c)
4. (c, d)

(2)
Hint: The magnetisation field intensity does not depend on the material of the core but the magnetisation depends on the material of the core.
Step 1: The magnetic field intensity; $\stackrel{\to }{\mathrm{H}}=\mathrm{nI}$
Magnetisation, $\stackrel{\to }{\mathrm{B}}={\mathrm{\mu }}_{\mathrm{o}}{\mathrm{\mu }}_{\mathrm{r}}\mathrm{nI}$
So the magnetic field intensity remains unchanged and the magnetisation decreases.
Step 2: When the temperature of the iron core of solenoid (which is ferromagnetic materials) is raised beyond curie temperature, then soft iron core behaves as paramagnetic material.
${\left({\mathrm{\chi }}_{\mathrm{m}}\right)}_{\mathrm{ferro}}={10}^{5}$
${\left({\mathrm{\chi }}_{\mathrm{m}}\right)}_{\mathrm{para}}={10}^{-3}$
So the magnetization in the core diminishes by a factor of about ${10}^{8}$.