A domain in ferromagnetic iron in the form of cube shaving `5 xx 10^(10)` atoms. If the side length of this domain is `1.5mu` and each atom has a dipole moment of `8 xx 10^(-24)Am^(2)`, then magnetisation of domain is

Options

A. `11.8xx10^(5)Am^(-1)`
B. `1.18xx10^(5)Am^(-1)`
C. `11.8xx10^(4)Am^(-1)`
D. `1.18xx10^(5)Am^(-1)`

Explanation

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

### Step 1: Calculate the Volume of the Domain
The volume \( V \) of a cube is given by the formula:
\[ V = \text{side}^3 \]
Given that the side length of the domain is \( 1.5 \, \mu m = 1.5 \times 10^{-6} \, m \), we can calculate the volume:
\[ V = (1.5 \times 10^{-6} \, m)^3 = 3.375 \times 10^{-18} \, m^3 \]

### Step 2: Number of Atoms in the Domain
The number of atoms \( N \) in the domain is given as:
\[ N = 5 \times 10^{10} \, \text{atoms} \]

### Step 3: Dipole Moment of Each Atom
The dipole moment \( m \) of each atom is given as:
\[ m = 8 \times 10^{-24} \, A \cdot m^2 \]

### Step 4: Calculate the Total Magnetic Moment
The total magnetic moment \( M_{max} \) of the domain can be calculated using the formula:
\[ M_{max} = N \cdot m \]
Substituting the values:
\[ M_{max} = (5 \times 10^{10}) \cdot (8 \times 10^{-24}) = 40 \times 10^{-14} \, A \cdot m^2 = 4 \times 10^{-13} \, A \cdot m^2 \]

### Step 5: Calculate the Magnetization
Magnetization \( M \) is defined as the total magnetic moment per unit volume:
\[ M = \frac{M_{max}}{V} \]
Substituting the values we calculated:
\[ M = \frac{4 \times 10^{-13} \, A \cdot m^2}{3.375 \times 10^{-18} \, m^3} \]
Calculating this gives:
\[ M \approx 1.18 \times 10^{5} \, A/m \]

### Final Answer
The magnetization of the domain is:
\[ M \approx 1.18 \times 10^{5} \, A/m \]

---