Question

A long solenoid has $500$ turns per $\text{metre}$ and carries a current of $0.5\text{A}$. It has a soft iron core of ${\mu }_r=999$. The core is heated beyond the curie temperature $T_c$. Then
$\left[\right(\chi {)}_{para}\approx {10}^{-5}\text{and}(\chi {)}_{Ferro}\approx {10}^3]$

• A. Both $\overrightarrow{H}$ and $\overrightarrow{B}$ in the solenoid are nearly unchanged.
• B. The magnetization in the core reverses direction
• C. Magnetization will decrease
• D. The $\overrightarrow{H}$ in the solenoid is nearly unchanged, but the $\overrightarrow{B}$ changes.

Answer 1

Answer: (C), (D)

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

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

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

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

$\overrightarrow{H}=n\overrightarrow{I}$

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, options $\left(C\right)$ and $\left(D\right)$ are the correct answer.