A bar mangnet has length 3 cm cross sectional area 2 `cm^(2)` and magnetic moment 3 `Am^(2)` the intensity of magnetisation of the bar magnet is

Options

A. `2xx10^(5) Am^(-1)`
B. `3xx10^(5) Am^(-1)`
C. `4xx10^(5) Am^(-1)`
D. `5xx10^(5) Am^(-1)`

Answer

To solve the problem of finding the intensity of magnetization of the bar magnet, we will follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Length of the bar magnet (L) = 3 cm
- Cross-sectional area (A) = 2 cm²
- Magnetic moment (μ) = 3 Am²

2. **Convert Units:**
- Convert the length from centimeters to meters:
\[ L = 3 \text{ cm} = 3 \times 10^{-2} \text{ m} \]
- Convert the cross-sectional area from cm² to m²:
\[ A = 2 \text{ cm}^2 = 2 \times 10^{-4} \text{ m}^2 \]

3. **Calculate the Volume (V) of the Bar Magnet:**
- The volume of the bar magnet can be calculated using the formula:
\[ V = A \times L \]
- Substituting the values:
\[ V = (2 \times 10^{-4} \text{ m}^2) \times (3 \times 10^{-2} \text{ m}) = 6 \times 10^{-6} \text{ m}^3 \]

4. **Use the Formula for Magnetic Moment:**
- The magnetic moment (μ) is related to the intensity of magnetization (I) and the volume (V) by the formula:
\[ μ = I \times V \]
- Rearranging the formula to find the intensity of magnetization (I):
\[ I = \frac{μ}{V} \]

5. **Substitute the Values:**
- Now substitute the values of μ and V:
\[ I = \frac{3 \text{ Am}^2}{6 \times 10^{-6} \text{ m}^3} = \frac{3}{6 \times 10^{-6}} = \frac{1}{2 \times 10^{-6}} = 5 \times 10^{5} \text{ A/m} \]

6. **Final Result:**
- The intensity of magnetization of the bar magnet is:
\[ I = 5 \times 10^{5} \text{ A/m} \]

### Summary:
The intensity of magnetization of the bar magnet is \( 5 \times 10^{5} \text{ A/m} \).