A compass needle made of pure iron (with density `8000 kg//m^3`) has a length 5 cm, width 1.0 mm and thickness 0.50 mm. The magnitude of magnetic dipole moment of an iron is `mu_"Fe"=2 xx 10^(-23)` J/T. If magnetisation of needle is equivalent to the alignment of 10% of the atoms in the needle, what is the magnitude of the neddle's magnetic dipole moment `mu` ? (mass of iron per mole `=0.05 kg//"mole", N_A=6xx10^23`)

Options

A. `2.4xx10^(-3)` J/T
B. `4.8xx10^(-3) `J/T
C. `7.2xx10^(-3)` J/T
D. `9.6xx10^(-8)` J/T

Answer

To find the magnitude of the needle's magnetic dipole moment, we can follow these steps:

### Step 1: Calculate the Volume of the Needle
The volume \( V \) of the needle can be calculated using the formula:
\[ V = \text{length} \times \text{width} \times \text{thickness} \]
Given:
- Length = 5 cm = 0.05 m
- Width = 1 mm = 0.001 m
- Thickness = 0.5 mm = 0.0005 m

Substituting the values:
\[ V = 0.05 \, \text{m} \times 0.001 \, \text{m} \times 0.0005 \, \text{m} = 2.5 \times 10^{-8} \, \text{m}^3 \]

### Step 2: Calculate the Mass of the Needle
Using the density \( \rho \) and volume \( V \):
\[ \text{mass} = \rho \times V \]
Given:
- Density \( \rho = 8000 \, \text{kg/m}^3 \)

Substituting the values:
\[ \text{mass} = 8000 \, \text{kg/m}^3 \times 2.5 \times 10^{-8} \, \text{m}^3 = 2.0 \times 10^{-4} \, \text{kg} \]

### Step 3: Calculate the Number of Moles of Iron
Using the mass of iron per mole:
\[ n = \frac{\text{mass}}{\text{mass of iron per mole}} \]
Given:
- Mass of iron per mole = 0.05 kg/mole

Substituting the values:
\[ n = \frac{2.0 \times 10^{-4} \, \text{kg}}{0.05 \, \text{kg/mole}} = 4.0 \times 10^{-3} \, \text{moles} \]

### Step 4: Calculate the Total Number of Iron Atoms
Using Avogadro's number \( N_A \):
\[ \text{Total number of atoms} = n \times N_A \]
Given:
- \( N_A = 6 \times 10^{23} \, \text{atoms/mole} \)

Substituting the values:
\[ \text{Total number of atoms} = 4.0 \times 10^{-3} \, \text{moles} \times 6 \times 10^{23} \, \text{atoms/mole} = 2.4 \times 10^{21} \, \text{atoms} \]

### Step 5: Calculate the Number of Atoms Contributing to Magnetization
Since only 10% of the atoms contribute to the magnetization:
\[ \text{Contributing atoms} = 0.1 \times \text{Total number of atoms} \]
Substituting the values:
\[ \text{Contributing atoms} = 0.1 \times 2.4 \times 10^{21} = 2.4 \times 10^{20} \, \text{atoms} \]

### Step 6: Calculate the Magnetic Dipole Moment of the Needle
Using the magnetic dipole moment of iron:
\[ \mu = \text{Contributing atoms} \times \mu_{\text{Fe}} \]
Given:
- \( \mu_{\text{Fe}} = 2 \times 10^{-23} \, \text{J/T} \)

Substituting the values:
\[ \mu = 2.4 \times 10^{20} \times 2 \times 10^{-23} = 4.8 \times 10^{-3} \, \text{J/T} \]

### Final Answer
The magnitude of the needle's magnetic dipole moment \( \mu \) is:
\[ \mu = 4.8 \times 10^{-3} \, \text{J/T} \]

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