A bar magnet of magnetic moment `10Am^(2)` has a cross sectional area of `2.5xx10^(-4)m^(2)` . If the intensity of magnetisation of the magnet is `10^(6)A//m`, then the length of magnet is

Options

A. `0.4m`
B. `0.04cm`
C. `0.04m`
D. `40 cm`

Explanation

To find the length of the bar magnet, we can use the relationship between the magnetic moment (M), intensity of magnetization (I), and the cross-sectional area (A) of the magnet. The formula for the magnetic moment is given by:

\[ M = I \times A \times 2l \]

Where:
- \( M \) is the magnetic moment,
- \( I \) is the intensity of magnetization,
- \( A \) is the cross-sectional area,
- \( l \) is the length of the magnet.

We can rearrange this formula to solve for the length \( l \):

\[ l = \frac{M}{I \times A \times 2} \]

Now, we can substitute the given values into this equation.

1. **Given Values**:
- Magnetic moment \( M = 10 \, \text{Am}^2 \)
- Intensity of magnetization \( I = 10^6 \, \text{A/m} \)
- Cross-sectional area \( A = 2.5 \times 10^{-4} \, \text{m}^2 \)

2. **Substituting the Values**:
\[ l = \frac{10}{(10^6) \times (2.5 \times 10^{-4}) \times 2} \]

3. **Calculating the Denominator**:
\[ (10^6) \times (2.5 \times 10^{-4}) \times 2 = 10^6 \times 2.5 \times 2 \times 10^{-4} = 5 \times 10^2 \]

4. **Calculating the Length**:
\[ l = \frac{10}{5 \times 10^2} = \frac{10}{500} = 0.02 \, \text{m} \]

Thus, the length of the bar magnet is:

\[ l = 0.02 \, \text{m} \quad \text{or} \quad 2 \, \text{cm} \]