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. `1:09`
B. `2:09`
C. `27:1`
D. `9:1`

Explanation

To solve the problem, we need to determine 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 1: Understand the magnetic field due to a bar magnet
The magnetic field \( B \) at a point along the perpendicular bisector of a bar magnet can be expressed as:
\[ B = \frac{\mu_0}{4\pi} \cdot \frac{2m}{r^3} \]
where:
- \( \mu_0 \) is the permeability of free space,
- \( m \) is the magnetic moment of the bar magnet,
- \( r \) is the distance from the center of the magnet to the point where the field is being calculated.

### Step 2: Calculate the magnetic field at point A
Point A is at a distance \( X \) from the center of the magnet. Therefore, the magnetic field at point A, \( B_A \), can be calculated as:
\[ B_A = \frac{\mu_0}{4\pi} \cdot \frac{2m}{X^3} \]

### Step 3: Calculate the magnetic field at point B
Point B is at a distance \( 3X \) from the center of the magnet. Therefore, the magnetic field at point B, \( B_B \), can be calculated as:
\[ B_B = \frac{\mu_0}{4\pi} \cdot \frac{2m}{(3X)^3} \]
This simplifies to:
\[ B_B = \frac{\mu_0}{4\pi} \cdot \frac{2m}{27X^3} \]

### Step 4: Find the ratio of the magnetic fields at A and B
To find the ratio \( \frac{B_A}{B_B} \), we substitute the expressions for \( B_A \) and \( B_B \):
\[ \frac{B_A}{B_B} = \frac{\frac{\mu_0}{4\pi} \cdot \frac{2m}{X^3}}{\frac{\mu_0}{4\pi} \cdot \frac{2m}{27X^3}} \]
The \( \frac{\mu_0}{4\pi} \) and \( 2m \) terms cancel out:
\[ \frac{B_A}{B_B} = \frac{27X^3}{X^3} = 27 \]

### Conclusion
The ratio of the magnetic fields at points A and B is:
\[ \frac{B_A}{B_B} = 27 \]