Every iron atom in a ferromagnetic domain has a magnetic dipole moment equal to `9.27 xx 10^(-24) A m^(2)`. A ferromagetic domain in iron has the shape of a cube of side ` 1mum`. The maximum dipole moment occurs when all the dipoles are alligned. The molar mass of uiron is 55 g and its specific gravity is 7.9 .The magnetisation of the domain is .

Options

A. `8.0 xx 10^(5) A//m`
B. `8.0 xx 10^(8) A//m`
C. `8.0 xx 10^(11) A//m`
D. `8.0 xx 10^(14) A//m`

Answer

To find the magnetization of a ferromagnetic domain in iron, we can follow these steps:

### Step 1: Calculate the Volume of the Cube
Given that the side of the cube is \(1 \, \mu m\):
\[ \text{Volume} = \text{side}^3 = (1 \times 10^{-6} \, m)^3 = 1 \times 10^{-18} \, m^3 \]

### Step 2: Calculate the Density of Iron
The specific gravity of iron is given as \(7.9\). The density of iron can be calculated using the formula:
\[ \text{Density} = \text{Specific Gravity} \times \text{Density of Water} \]
Since the density of water is \(1000 \, kg/m^3\):
\[ \text{Density} = 7.9 \times 1000 \, kg/m^3 = 7900 \, kg/m^3 \]

### Step 3: Calculate the Mass of the Cube
Using the density and volume, we can find the mass:
\[ \text{Mass} = \text{Density} \times \text{Volume} = 7900 \, kg/m^3 \times 1 \times 10^{-18} \, m^3 = 7.9 \times 10^{-15} \, kg \]

### Step 4: Convert Mass to Grams
To convert the mass from kilograms to grams:
\[ \text{Mass} = 7.9 \times 10^{-15} \, kg \times 1000 \, g/kg = 7.9 \times 10^{-12} \, g \]

### Step 5: Calculate the Number of Moles of Iron
Using the molar mass of iron, which is \(55 \, g/mol\):
\[ \text{Number of Moles} = \frac{\text{Mass}}{\text{Molar Mass}} = \frac{7.9 \times 10^{-12} \, g}{55 \, g/mol} = 1.43636 \times 10^{-13} \, mol \]

### Step 6: Calculate the Number of Atoms in the Cube
Using Avogadro's number (\(6.022 \times 10^{23} \, atoms/mol\)):
\[ \text{Number of Atoms} = \text{Number of Moles} \times \text{Avogadro's Number} = 1.43636 \times 10^{-13} \, mol \times 6.022 \times 10^{23} \, atoms/mol \approx 8.66 \times 10^{10} \, atoms \]

### Step 7: Calculate the Total Magnetic Dipole Moment
Each iron atom has a magnetic dipole moment of \(9.27 \times 10^{-24} \, A \cdot m^2\):
\[ \text{Total Magnetic Dipole Moment} = \text{Number of Atoms} \times \text{Dipole Moment} = 8.66 \times 10^{10} \, atoms \times 9.27 \times 10^{-24} \, A \cdot m^2 \approx 8.04 \times 10^{-13} \, A \cdot m^2 \]

### Step 8: Calculate the Magnetization of the Domain
Magnetization \(M\) is defined as the total magnetic dipole moment per unit volume:
\[ M = \frac{\text{Total Magnetic Dipole Moment}}{\text{Volume}} = \frac{8.04 \times 10^{-13} \, A \cdot m^2}{1 \times 10^{-18} \, m^3} = 8.04 \times 10^{5} \, A/m \]

### Final Result
The magnetization of the domain is approximately:
\[ M \approx 8.04 \times 10^{5} \, A/m \]