Question

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

• A. The $\overrightarrow{H}$ field in the solenoid is nearly unchanged, but the $\overrightarrow{B}$ field changes.
• B. Both $\overrightarrow{H}$ and $\overrightarrow{B}$ fields in the solenoid are nearly unchanged.
• C. The magnetization in the core reverses direction
• D. The magnetization in the core diminishes by a factor of nearly ${10}^6$

Answer 1

The correct option is A The $\overrightarrow{H}$ field in the solenoid is nearly unchanged, but the $\overrightarrow{B}$ field changes.

We know that, if a magnetic particle is heated beyond curie temperature ($T_c$), the particle gets converted from ferromagnetic to paramagnetic.

Now, $\overrightarrow{H}=\frac{\overrightarrow{B}}{{\mu }_0{\mu }_r}$

$\overrightarrow{H}=\frac{{\mu }_0{\mu }_r}{{\mu }_0{\mu }_r}nI(\because B={\mu }_0{\mu }_{r}nI)$

$\overrightarrow{H}=nI$

Clearly, $\overrightarrow{H}$ is independent of ${\mu }_r$

But, $\overrightarrow{B}\propto {\mu }_r$

So, as the temperature is increased beyond $T_c$, ${\mu }_r$ changes and hence $\overrightarrow{B}$ changes.

$(\chi {)}_{para}\approx {10}^{-5}$

$(\chi {)}_{Ferro}\approx {10}^3$

So, magnetization is given by,

$\frac{(\chi {)}_{Ferro}}{(\chi {)}_{para}}=\frac{{10}^3}{{10}^{-5}}={10}^8$

Hence, option $\left(A\right)$ is the correct answer.