Two short bar magnets with magnetic moments `400 ab– "A cm"^2 and 800 ab – A cm^2` are placed with their axis in the same straight line with similar poles facing each other and with their centres at 20 cm from each other. Then the force of repulsion is

Options

A. 12 dyne
B. 6 dyne
C. 800 dyne
D. 150 dyne

Answer

To solve the problem of finding the force of repulsion between two short bar magnets with given magnetic moments, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Magnetic moment of the first magnet, \( M_1 = 400 \, \text{A cm}^2 \)
- Magnetic moment of the second magnet, \( M_2 = 800 \, \text{A cm}^2 \)
- Distance between the centers of the magnets, \( r = 20 \, \text{cm} = 0.2 \, \text{m} \)

2. **Convert the Distance to Centimeters:**
- Since we are working in CGS units, we will convert the distance to centimeters.
- \( r = 20 \, \text{cm} \)

3. **Use the Formula for the Force of Repulsion:**
- The formula for the force of repulsion \( F \) between two magnetic dipoles is given by:
\[ F = \frac{3 \mu_0}{4\pi} \cdot \frac{2 M_1 M_2}{r^4} \]
- Here, \( \mu_0 \) is the permeability of free space, which in CGS units is \( 1 \).

4. **Substitute the Values into the Formula:**
- Substitute \( M_1 = 400 \), \( M_2 = 800 \), and \( r = 20 \):
\[ F = \frac{3 \cdot 1}{4\pi} \cdot \frac{2 \cdot 400 \cdot 800}{(20)^4} \]

5. **Calculate \( (20)^4 \):**
- \( (20)^4 = 160000 \)

6. **Calculate the Numerator:**
- \( 2 \cdot 400 \cdot 800 = 640000 \)

7. **Substitute and Simplify:**
- Now substitute back into the equation:
\[ F = \frac{3 \cdot 640000}{4\pi \cdot 160000} \]
- Simplifying gives:
\[ F = \frac{3 \cdot 4}{4\pi} = \frac{12}{\pi} \]

8. **Calculate the Final Value:**
- Using \( \pi \approx 3.14 \):
\[ F \approx \frac{12}{3.14} \approx 3.82 \, \text{dyne} \]

### Final Answer:
The force of repulsion between the two magnets is approximately \( 3.82 \, \text{dyne} \).