Two equal magnetic poles placed 10 cm apart in air attract each other with a force of `0.4 xx 10^(-4)N`. What should be the distance of separation between them so that the force of attraction of `0.6 xx 10^(-4)N` ?

Answer

To solve the problem, we need to use the relationship between the force of attraction between two magnetic poles and the distance between them. The force of attraction \( F \) between two magnetic poles is inversely proportional to the square of the distance \( D \) between them. Mathematically, this can be expressed as:

\[ F \propto \frac{1}{D^2} \]

From this relationship, we can derive the equation:

\[ F_1 D_1^2 = F_2 D_2^2 \]

Where:
- \( F_1 \) is the initial force of attraction,
- \( D_1 \) is the initial distance,
- \( F_2 \) is the final force of attraction,
- \( D_2 \) is the final distance we need to find.

### Step-by-step Solution:

1. **Identify the Given Values**:
- Initial force \( F_1 = 0.4 \times 10^{-4} \, \text{N} \)
- Initial distance \( D_1 = 10 \, \text{cm} \)
- Final force \( F_2 = 0.6 \times 10^{-4} \, \text{N} \)
- Final distance \( D_2 \) (unknown)

2. **Set Up the Equation**:
Using the relationship \( F_1 D_1^2 = F_2 D_2^2 \):
\[ (0.4 \times 10^{-4}) \times (10^2) = (0.6 \times 10^{-4}) \times D_2^2 \]

3. **Substitute the Values**:
\[ (0.4 \times 10^{-4}) \times (100) = (0.6 \times 10^{-4}) \times D_2^2 \]
Simplifying:
\[ 0.4 \times 10^{-2} = 0.6 \times 10^{-4} \times D_2^2 \]

4. **Isolate \( D_2^2 \)**:
Divide both sides by \( 0.6 \times 10^{-4} \):
\[ D_2^2 = \frac{0.4 \times 10^{-2}}{0.6 \times 10^{-4}} \]

5. **Simplify the Right Side**:
\[ D_2^2 = \frac{0.4}{0.6} \times 10^{2} = \frac{2}{3} \times 10^{2} \]

6. **Take the Square Root**:
\[ D_2 = \sqrt{\frac{2}{3} \times 10^{2}} = \sqrt{\frac{2}{3}} \times 10 \]

7. **Calculate the Value**:
Using \( \sqrt{\frac{2}{3}} \approx 0.816 \):
\[ D_2 \approx 0.816 \times 10 \approx 8.16 \, \text{cm} \]

### Final Answer:
The distance of separation \( D_2 \) should be approximately **8.16 cm**.