Force between two identical bar magnets whose centres are r metre apart is `4.8 N`, when their axes are in the same line. If separation is increased to `2r`, the force between them is reduced to

Options

A. `2.4 N`
B. `1.2 N`
C. `0.6 N`
D. `0.3 N`

Explanation

To solve the problem, we need to determine how the force between two identical bar magnets changes when the distance between them is increased from \( r \) to \( 2r \).

### Step-by-Step Solution:

1. **Understanding the Initial Force**:
The initial force \( F \) between the two magnets when they are \( r \) meters apart is given as \( 4.8 \, \text{N} \).

2. **Using the Formula for Force Between Magnetic Dipoles**:
The force \( F \) between two magnetic dipoles is given by the formula:
\[ F \propto \frac{1}{d^4} \]
where \( d \) is the distance between the centers of the two magnets.

3. **Calculating the New Distance**:
When the distance is increased to \( 2r \), we need to substitute \( d = 2r \) into the formula.

4. **Setting Up the Ratio of Forces**:
Let the new force be \( F' \). Using the proportionality:
\[ \frac{F'}{F} = \frac{r^4}{(2r)^4} \]
Simplifying the right side:
\[ \frac{F'}{F} = \frac{r^4}{16r^4} = \frac{1}{16} \]

5. **Finding the New Force**:
Now we can express \( F' \) in terms of \( F \):
\[ F' = F \cdot \frac{1}{16} \]
Substituting \( F = 4.8 \, \text{N} \):
\[ F' = 4.8 \cdot \frac{1}{16} = \frac{4.8}{16} = 0.3 \, \text{N} \]

### Final Answer:
The force between the two magnets when the distance is increased to \( 2r \) is \( 0.3 \, \text{N} \).

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