Points `A` and `B` are situated perpendicular to the axis of a `2cm` long bar magnet at large distances `X` and `3X` from its centre on opposite sides. The retio of the magnetic fields at `A` and `B` wil be approximately equal to

Options

A. `1:9`
B. `2:9`
C. `27:1`
D. `9:1`

Answer

To solve the problem, we need to find the ratio of the magnetic fields at points A and B, which are located at distances X and 3X from the center of a bar magnet, respectively.

### Step-by-Step Solution:

1. **Understanding the Magnetic Field of a Bar Magnet**:
The magnetic field (B) at a point on the axis of a bar magnet is given by the formula:
\[ B = \frac{\mu_0 m}{4\pi d^3} \]
where \( \mu_0 \) is the permeability of free space, \( m \) is the magnetic moment of the bar magnet, and \( d \) is the distance from the center of the magnet to the point where the magnetic field is being calculated.

2. **Identify Distances**:
- For point A, the distance from the center of the magnet is \( d_A = X \).
- For point B, the distance from the center of the magnet is \( d_B = 3X \).

3. **Calculate Magnetic Field at Point A**:
Using the formula for the magnetic field:
\[ B_A = \frac{\mu_0 m}{4\pi (X)^3} \]

4. **Calculate Magnetic Field at Point B**:
Similarly, for point B:
\[ B_B = \frac{\mu_0 m}{4\pi (3X)^3} \]
Simplifying this gives:
\[ B_B = \frac{\mu_0 m}{4\pi (27X^3)} = \frac{\mu_0 m}{108\pi X^3} \]

5. **Finding the Ratio of Magnetic Fields**:
Now, we need to find the ratio \( \frac{B_A}{B_B} \):
\[ \frac{B_A}{B_B} = \frac{\frac{\mu_0 m}{4\pi X^3}}{\frac{\mu_0 m}{108\pi X^3}} \]
The \( \mu_0 m \) and \( \pi X^3 \) cancel out:
\[ \frac{B_A}{B_B} = \frac{108}{4} = 27 \]

6. **Final Ratio**:
Thus, the ratio of the magnetic fields at points A and B is:
\[ B_A : B_B = 27 : 1 \]

### Conclusion:
The ratio of the magnetic fields at points A and B is approximately equal to \( 27 : 1 \).