Question

A toroid has a mean radius $R$ equal to $\frac{10}{\pi }\text{cm}$, and a total of $200\text{turns}$ of wire carrying a current of $1.0\text{A}$. An alloy ring at temperature $27{}^{\circ }\text{C}$ inside the toroid provides the core. The magnetization $I_1$ is $7.2\times {10}^{-3}{\text{Am}}^{-1}$ for the alloy at $27{}^{\circ }\text{C}$. If the temperature of the alloy ring is raised to $87{}^{\circ }\text{C}$, then find the magnetization $\left(I_2\right)$ and susceptibility $\left({\chi }_2\right)$?

• A. ${\chi }_1=6.2\times {10}^{-6}$
• B. $I_2=3\times {10}^{-3}{\text{Am}}^{-1}$
• C. ${\chi }_2=6\times {10}^{-6}$
• D. $I_2=6\times {10}^{-3}{\text{Am}}^{-1}$

Answer 1

Answer: (C), (D)

$I_2=6\times {10}^{-3}{\text{Am}}^{-1}$

Given:
The magnetic intensity $H$ in the core is,

$H=ni$

Where, the number of turns per unit length of the toroid is,

$n=\frac{N}{l}=\frac{200}{2\pi R}=\frac{200}{2\times \pi \times \frac{10}{\pi }\times {10}^{-2}}=1000$

$\therefore H=1000\times 1=1000{\text{Am}}^{-1}$

So, the susceptibility at temperature $(T_1=27+273=300\text{K})$ be,

${\chi }_1=\frac{I_1}{H}=\frac{7.2\times {10}^{-3}}{1000}=7.2\times {10}^{-6}$

According to the Curie's law

$\frac{{\chi }_2}{{\chi }_1}=\frac{T_1}{T_2}$

${\chi }_2=\frac{300\times 7.2\times {10}^{-6}}{360}[\because T_2=87+273=360\text{K}]$

$\therefore {\chi }_2=6\times {10}^{-6}$

Now, the magnetization intensity at $T_2=360\text{K}$ is,

$I_2={\chi }_2{H}_2$,

Since, $H$ of a material does not change with $T$, so $H_2=H_1=H$.

$\Rightarrow I_2=(6\times {10}^{-6})\times 1000=6\times {10}^{-3}{\text{Am}}^{-1}$

Hence, options $\left(\text{C}\right)$ and $\left(\text{D}\right)$ are correct.