Question

The magnetic moment of a sample of mass \( 2 \, \text{g} \) is \( 8 \times 10^{-7} \, \text{A}\cdot\text{m}^2 \). If density \( \rho = 4 \, \text{g/cm}^3 \), then the magnetisation \( M \) of the sample is: ?

• A. \( 0.4 \, \text{A/m} \)
• B. \( 1.6 \, \text{A/m} \)
• C. \( 4.0 \, \text{A/m} \)
• D. \( 6.4 \, \text{A/m} \)

Hint:

Remember to convert all quantities into SI units before substituting into formulas. Density should be in \( \text{kg/m}^3 \), mass in \( \text{kg} \), and volume in \( \text{m}^3 \).

Answer 1

The Correct Option is B

Magnetisation \( M \) is defined as: \[ M = \frac{\text{Magnetic moment}}{\text{Volume}} \] We are given: - Magnetic moment \( \mu = 8 \times 10^{-7} \, \text{A}\cdot\text{m}^2 \) - Mass \( m = 2 \, \text{g} = 2 \times 10^{-3} \, \text{kg} \) - Density \( \rho = 4 \, \text{g/cm}^3 = 4 \times 10^3 \, \text{kg/m}^3 \) Step 1: Calculate volume of the sample Using \( \text{Density} = \frac{\text{Mass}}{\text{Volume}} \Rightarrow \text{Volume} = \frac{\text{Mass}}{\text{Density}} \) \[ V = \frac{2 \times 10^{-3}}{4 \times 10^3} = \frac{2}{4} \times 10^{-6} = 0.5 \times 10^{-6} = 5 \times 10^{-7} \, \text{m}^3 \] Step 2: Apply the formula for magnetisation \[ M = \frac{8 \times 10^{-7}}{5 \times 10^{-7}} = \frac{8}{5} = 1.6 \, \text{A/m} \] Final Answer: \[ \boxed{1.6 \, \text{A/m}} \]