Two short bar magnets of length `1cm` each have magnetic moments `1.20 Am^(2) and 1.00 Am^(2)` respectively. They are placed on a horizontal table parallel to each other with their `N` poles pointing towards the south. They have a common magnetic equator and are separted by a distance of `20.0 cm`. The value of the resultant horizontal magnetic induction at the mid - point `O` of the line joining their centres is close to (Horizontal component of earths magnetic induction is `3.6 xx 10.5 Wh//m^(2)`

Options

A. `2.56xx10^(-4)Wb//m^(2)`
B. `3.50xx10^(-4)Wb//m^(2)`
C. `5.80xx10^(-4)Wb//m^(2)`
D. `3.6xx10^(-4)Wb//m^(2)`

Answer

To solve the problem, we need to calculate the resultant magnetic induction at the midpoint \( O \) between the two bar magnets, taking into account the magnetic fields produced by each magnet and the Earth's magnetic induction.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Length of each magnet, \( L = 1 \, \text{cm} = 0.01 \, \text{m} \)
- Magnetic moment of the first magnet, \( m_1 = 1.20 \, \text{Am}^2 \)
- Magnetic moment of the second magnet, \( m_2 = 1.00 \, \text{Am}^2 \)
- Distance between the magnets, \( d = 20.0 \, \text{cm} = 0.20 \, \text{m} \)
- Horizontal component of Earth's magnetic induction, \( B_E = 3.6 \times 10^{-5} \, \text{Wb/m}^2 \)

2. **Calculate the Magnetic Field at Point O:**
The magnetic field \( B \) due to a bar magnet at a point on its equator is given by the formula:
\[ B = \frac{\mu_0}{4\pi} \cdot \frac{m}{d^3} \]
where \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m/A} \).

3. **Calculate the Magnetic Field due to the First Magnet (B1):**
- The distance from the first magnet to point \( O \) is \( d_1 = \frac{d}{2} = 0.10 \, \text{m} \).
\[ B_1 = \frac{4\pi \times 10^{-7}}{4\pi} \cdot \frac{1.20}{(0.10)^3} = 10^{-7} \cdot \frac{1.20}{0.001} = 1.20 \times 10^{-4} \, \text{Wb/m}^2 \]

4. **Calculate the Magnetic Field due to the Second Magnet (B2):**
- The distance from the second magnet to point \( O \) is also \( d_2 = 0.10 \, \text{m} \).
\[ B_2 = \frac{4\pi \times 10^{-7}}{4\pi} \cdot \frac{1.00}{(0.10)^3} = 10^{-7} \cdot \frac{1.00}{0.001} = 1.00 \times 10^{-4} \, \text{Wb/m}^2 \]

5. **Calculate the Total Magnetic Field at Point O (B_total):**
Since both magnets are aligned such that their fields at point \( O \) add up:
\[ B_{\text{total}} = B_1 + B_2 + B_E \]
\[ B_{\text{total}} = 1.20 \times 10^{-4} + 1.00 \times 10^{-4} + 3.6 \times 10^{-5} \]
Convert \( B_E \) to the same power of ten:
\[ B_E = 0.36 \times 10^{-4} \]
Now, add all three:
\[ B_{\text{total}} = (1.20 + 1.00 + 0.36) \times 10^{-4} = 2.56 \times 10^{-4} \, \text{Wb/m}^2 \]

### Final Result:
The resultant horizontal magnetic induction at point \( O \) is:
\[ B_{\text{total}} \approx 2.56 \times 10^{-4} \, \text{Wb/m}^2 \]