Question

A toroid has a mean radius $R$ equal to $\frac{20}{\pi }\text{cm}$ and a total of $400$ turns of wire carrying a current of $2.0\text{A}$. An aluminium ring at temperature $280\text{K}$ inside the toroid provides the core. At this temperature magnetisation of core is $4.8\times {10}^{-2}\frac{\text{A}}{\text{m}}$ . If the tempertaure of the aluuminium ring is raised to $320\text{K}$, what will be the magnetization?

• A. $2.4\times {10}^{-2}\frac{\text{A}}{\text{m}}$
• B. $4.2\times {10}^{-2}\frac{\text{A}}{\text{m}}$
• C. $2.1\times {10}^{-2}\frac{\text{A}}{\text{m}}$
• D. $3.4\times {10}^{-2}\frac{\text{A}}{\text{m}}$

Answer 1

The correct option is B $4.2\times {10}^{-2}\frac{\text{A}}{\text{m}}$

Given
$R=\frac{20}{\pi }\text{cm}=\frac{20}{\pi }\times {10}^{-2}\text{m}$
$N=400$
$\because$No. of turns per unit length
$\therefore n=\frac{N}{L}=\frac{N}{2\pi R}=\frac{400}{2\pi R}$

current $i=2\text{A}$

Intensity of magnetisation $I=4.8\times {10}^{-2}\frac{\text{A}}{\text{m}}$
$\because$ Magnetisation field is $H=ni$
$\Rightarrow H=\frac{400}{2\pi \times \frac{20}{\pi }\times {10}^{-2}}\times 2=2000{\text{Am}}^{-1}$
So the magnetic susceptibility
$X=\frac{I}{H}=\frac{4.8\times {10}^{-2}}{2000}=2.4\times {10}^{-5}$
By curie temp $X\propto \frac{1}{T}$
$\Rightarrow \frac{X_1}{X_2}=\frac{T_2}{T_1}$
$\Rightarrow X_2=\frac{T_1}{T_2}{X}_1$
$\Rightarrow X_2=\frac{280}{320}\times 2.4\times {10}^{-5}$
$\Rightarrow X_2=2.1\times {10}^{-5}$
Now the intensity of magnetisation
$I_2=X_{2}H$
$I_2=2.1\times {10}^{-5}\times 2000$
$I_2=4.2\times {10}^{-2}\frac{\text{A}}{\text{m}}$