The small magnets each of magnetic moment `10 A-m^(2)` are placed end-on position 0.1 m apart from their centres. The force acting between them is

Options

A. `0.6xx10^(7)N`
B. `0.06xx10^(7)N`
C. `0.6N`
D. `0.06N`

Explanation

To find the force acting between two small magnets placed end-on with a given magnetic moment and separation, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values**:
- Magnetic moment of each magnet, \( m_1 = m_2 = 10 \, \text{A-m}^2 \)
- Distance between the centers of the magnets, \( R = 0.1 \, \text{m} \)

2. **Use the Formula for Force Between Magnetic Dipoles**:
The force \( F \) between two magnetic dipoles can be calculated using the formula:
\[ F = \frac{\mu_0}{4\pi} \cdot \frac{6 \cdot m_1 \cdot m_2}{R^4} \]
where \( \mu_0 \) (the permeability of free space) is approximately \( 4\pi \times 10^{-7} \, \text{T m/A} \).

3. **Substitute the Values into the Formula**:
- Substitute \( m_1 = 10 \, \text{A-m}^2 \), \( m_2 = 10 \, \text{A-m}^2 \), and \( R = 0.1 \, \text{m} \) into the formula:
\[ F = \frac{4\pi \times 10^{-7}}{4\pi} \cdot \frac{6 \cdot 10 \cdot 10}{(0.1)^4} \]
- Simplifying this gives:
\[ F = 10^{-7} \cdot \frac{600}{0.0001} \]
\[ F = 10^{-7} \cdot 6 \times 10^6 \]
\[ F = 0.6 \, \text{N} \]

4. **Final Result**:
The force acting between the two magnets is:
\[ F = 0.6 \, \text{N} \]