Question

A bar magnet made of steel has a magnetic moment of \(2.5 \, \text{Am}^2\) and a mass of \(6.6 \times 10^{-2} \, \text{kg}\). If the density of steel is \(7.9 \times 10^3 \, \text{kg/m}^3\), the intensity of magnetization of the magnet is:

• A. \(I = 30 \times 10^5 \, \text{A/m}\)
• B. \(I = 30 \times 10^7 \, \text{A/m}\)
• C. \(I = 30 \times 10^6 \, \text{A/m}\)
• D. \(I = 30 \times 10^4 \, \text{A/m}\)

Hint:

Always use SI units when calculating volume and magnetization intensity. Check for consistency in mass and density units before substituting into the formula.

Answer 1

The Correct Option is C

The intensity of magnetization $I$ is given by:
\[I = \frac{M}{V}\]
where $M$ is the magnetic moment, and $V$ is the volume of the magnet. The volume $V$ is:
\[V = \frac{m}{\rho} = \frac{6.6 \times 10^{-2}}{7.9 \times 10^{3}} \approx 8.35 \times 10^{-6} \text{ m}^3\]
Substituting into the formula for $I$:
\[I = \frac{2.5}{8.35 \times 10^{-6}} \approx 30 \times 10^{6} \text{ A/m}\]