Question

A bar of iron having magnetic moment 2.4 Am$^2$ weighs 66 g. If the density of the material of the bar is 7700 kg/m$^3$, the intensity of magnetisation in Am$^{-1}$ is ______.

• A. $1.4 \times 10^5$
• B. $2.8 \times 10^5$
• C. $1.4 \times 10^4$
• D. $2.8 \times 10^4$

Hint:

Don't reach for a calculator! Notice how $66$ and $77$ both perfectly share a factor of $11$, reducing to the trivial fraction $6/7$, which makes the final division by $6$ incredibly easy. Look for these hidden simplifications in CET papers!

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Intensity of Magnetization ($I$) is defined as Magnetic Moment per unit Volume ($M/V$).

Step 2: Formula Application:
Volume ($V$) = $\frac{\text{Mass}}{\text{Density}}$.
$I = \frac{\text{Magnetic Moment}}{\text{Mass} / \text{Density}} = \frac{M \times \rho}{m}$.

Step 3: Explanation:
Mass $m = 66$ g $= 66 \times 10^{-3}$ kg.
$\rho = 7700$ kg/m³.
$M = 2.4$ Am².
$I = \frac{2.4 \times 7700}{66 \times 10^{-3}} = \frac{2.4 \times 77 \times 10^2}{66 \times 10^{-3}} = \frac{2.4 \times 7}{6} \times 10^5$.
$I = 0.4 \times 7 \times 10^5 = 2.8 \times 10^5$ Am⁻¹.

Step 4: Final Answer:
The intensity of magnetization is $2.8 \times 10^5$ Am⁻¹.