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 Question:
We are given a magnet's physical properties (mass, density) and its total magnetic moment. We must calculate its "Intensity of Magnetisation" ($I$ or $M$).

Step 2: Detailed Explanation:
The Intensity of Magnetisation ($I$) is defined as the magnetic moment ($M$) per unit volume ($V$) of the material.
$I = \frac{M}{V}$
We are given the magnetic moment:
$M = 2.4 \text{ Am}^2$
We need to find the volume ($V$) of the iron bar. We know:
$\text{Density } (\rho) = \frac{\text{Mass } (m)}{\text{Volume } (V)} \implies V = \frac{m}{\rho}$
Given physical parameters:
Mass ($m$) = $66 \text{ g} = 66 \times 10^{-3} \text{ kg}$ (Must convert to kg to match density!)
Density ($\rho$) = $7700 \text{ kg/m}^3$
Calculate Volume ($V$):
$V = \frac{66 \times 10^{-3}}{7700}$
$V = \frac{66}{77} \times 10^{-5}$
Simplify the fraction (divide by 11):
$V = \frac{6}{7} \times 10^{-5} \text{ m}^3$
Now, calculate the Intensity of Magnetisation ($I$):
$I = \frac{M}{V}$
$I = \frac{2.4}{\frac{6}{7} \times 10^{-5}}$
Multiply by the reciprocal:
$I = \frac{2.4 \times 7}{6} \times 10^5$
$I = 0.4 \times 7 \times 10^5$
$I = 2.8 \times 10^5 \text{ A/m (or Am}^{-1})$

Step 3: Final Answer:
The intensity of magnetisation is $2.8 \times 10^5$, matching option (b).