The magnetic field intensity due to a very small bar magnet having magnetic dipole moment as `2.5 Am^2` at end on position at a distance of `0.5 m` in `mu T` is_______.

Explanation

To find the magnetic field intensity due to a small bar magnet at an end-on position, we can use the formula for the magnetic field intensity (B) at a distance (d) from a magnetic dipole moment (m):

\[ B = \frac{\mu_0}{4\pi} \cdot \frac{2m}{d^3} \]

Where:
- \( \mu_0 \) (the permeability of free space) is approximately \( 4\pi \times 10^{-7} \, \text{T m/A} \)
- \( m \) is the magnetic dipole moment
- \( d \) is the distance from the magnet

Given:
- \( m = 2.5 \, \text{A m}^2 \)
- \( d = 0.5 \, \text{m} \)

### Step-by-step Solution:

1. **Identify the values**:
- Magnetic dipole moment \( m = 2.5 \, \text{A m}^2 \)
- Distance \( d = 0.5 \, \text{m} \)

2. **Substitute the values into the formula**:
\[ B = \frac{4\pi \times 10^{-7}}{4\pi} \cdot \frac{2 \times 2.5}{(0.5)^3} \]
Here, \( \mu_0 \) cancels out with \( 4\pi \).

3. **Simplify the expression**:
\[ B = 10^{-7} \cdot \frac{2 \times 2.5}{(0.5)^3} \]

4. **Calculate \( (0.5)^3 \)**:
\[ (0.5)^3 = 0.125 \]

5. **Substitute \( (0.5)^3 \) back into the equation**:
\[ B = 10^{-7} \cdot \frac{5}{0.125} \]

6. **Calculate \( \frac{5}{0.125} \)**:
\[ \frac{5}{0.125} = 40 \]

7. **Now substitute back**:
\[ B = 10^{-7} \cdot 40 \]

8. **Convert to microtesla**:
\[ B = 4 \times 10^{-6} \, \text{T} = 4 \, \mu T \]

### Final Answer:
The magnetic field intensity at a distance of 0.5 m from the bar magnet is **4 microtesla (μT)**.