The intensity of magnetisation of a bar magnet is `5xx10^(4)A-m^(-1)`. The magnetic length and the area of cross section of the magnet are 12cm and `1cm^(-2)` respectivley. The magnitude of magnetic moment of this bar magnet (in SI unit) is.

Options

A. 0.6
B. 1.3
C. 1.2
D. 2.4

Explanation

To find the magnetic moment of the bar magnet, we will use the relationship between intensity of magnetization (I), magnetic moment (M), and volume (V) of the magnet.

### Step-by-Step Solution:

**Step 1: Understand the relationship between intensity of magnetization and magnetic moment.**
The intensity of magnetization (I) is defined as:
\[ I = \frac{M}{V} \]
Where:
- \( M \) is the magnetic moment,
- \( V \) is the volume of the magnet.

**Step 2: Rearrange the formula to find magnetic moment.**
From the above equation, we can express the magnetic moment as:
\[ M = I \times V \]

**Step 3: Calculate the volume of the bar magnet.**
The volume \( V \) of the bar magnet can be calculated using the formula:
\[ V = \text{Area} \times \text{Length} \]
Given:
- Length = 12 cm = 0.12 m (converted to meters),
- Area = 1 cm² = \( 1 \times 10^{-4} \) m² (converted to square meters).

Now, substituting the values:
\[ V = (1 \times 10^{-4} \, \text{m}^2) \times (0.12 \, \text{m}) = 1.2 \times 10^{-5} \, \text{m}^3 \]

**Step 4: Substitute the values into the magnetic moment formula.**
Now, we know:
- Intensity of magnetization \( I = 5 \times 10^4 \, \text{A/m} \),
- Volume \( V = 1.2 \times 10^{-5} \, \text{m}^3 \).

Substituting these values into the magnetic moment formula:
\[ M = I \times V = (5 \times 10^4 \, \text{A/m}) \times (1.2 \times 10^{-5} \, \text{m}^3) \]
Calculating this gives:
\[ M = 5 \times 1.2 \times 10^{4 - 5} = 6 \times 10^{-1} \, \text{A m}^2 = 0.6 \, \text{A m}^2 \]

**Final Answer:**
The magnitude of the magnetic moment of the bar magnet is \( 0.6 \, \text{A m}^2 \).