A magnetic of magnetic moment 215 `Am^(2)` placed on the magnetic meridian with its N-pole towards grographic north.The distance between the two neutral points is 12 cm.The horizontal component of earth's field at that place is

Options

A. 0.1 T
B. 0.025 T
C. 1T
D. 25T

Explanation

To solve the problem, we need to find the horizontal component of the Earth's magnetic field (B_H) using the given information about the magnetic moment (M) of the magnet and the distance between the two neutral points.

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have a magnet with a magnetic moment \( M = 215 \, Am^2 \).
- The magnet is placed in the magnetic meridian with its north pole facing geographic north.
- The distance between the two neutral points is given as \( 12 \, cm \).

2. **Finding the Distance to Each Neutral Point**:
- Since the distance between the two neutral points is \( 12 \, cm \), the distance from the center of the magnet to each neutral point is half of that:
\[ R = \frac{12 \, cm}{2} = 6 \, cm = 6 \times 10^{-2} \, m \]

3. **Using the Formula for the Magnetic Field on the Equatorial Line**:
- The magnetic field \( B \) at a distance \( R \) from a magnetic dipole on its equatorial line is given by:
\[ B = \frac{\mu_0}{4\pi} \cdot \frac{2M}{R^3} \]
- Where \( \mu_0 \) (the permeability of free space) is approximately \( 4\pi \times 10^{-7} \, Tm/A \).

4. **Substituting the Values into the Formula**:
- Substitute \( M = 215 \, Am^2 \) and \( R = 6 \times 10^{-2} \, m \):
\[ B = \frac{4\pi \times 10^{-7}}{4\pi} \cdot \frac{2 \times 215}{(6 \times 10^{-2})^3} \]
- This simplifies to:
\[ B = 10^{-7} \cdot \frac{2 \times 215}{(6 \times 10^{-2})^3} \]

5. **Calculating \( (6 \times 10^{-2})^3 \)**:
- Calculate \( (6 \times 10^{-2})^3 = 216 \times 10^{-6} = 2.16 \times 10^{-4} \).

6. **Final Calculation**:
- Now substituting back into the equation:
\[ B = 10^{-7} \cdot \frac{430}{2.16 \times 10^{-4}} = 10^{-7} \cdot 1990740.74 \approx 0.1 \, T \]

7. **Conclusion**:
- The horizontal component of the Earth's magnetic field \( B_H \) is approximately \( 0.1 \, T \).

### Final Answer:
The horizontal component of the Earth's magnetic field at that place is approximately \( 0.1 \, T \).