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 very small bar magnet at an end-on position, we can use the formula for the magnetic field \( B \) at a distance \( r \) from a magnetic dipole moment \( m \):

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

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

### Step-by-Step Solution:

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

2. **Substitute the values into the formula**:
\[ B = \frac{\mu_0}{4\pi} \cdot \frac{2m}{r^3} \]

3. **Insert the value of \( \mu_0 \)**:
\[ B = \frac{4\pi \times 10^{-7}}{4\pi} \cdot \frac{2 \times 2.5}{(0.5)^3} \]
The \( 4\pi \) cancels out:
\[ B = 10^{-7} \cdot \frac{2 \times 2.5}{(0.5)^3} \]

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. **Substitute this value back into the equation**:
\[ B = 10^{-7} \cdot 40 = 4 \times 10^{-6} \, \text{T} \]

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

### Final Answer:
The magnetic field intensity at the end-on position at a distance of \( 0.5 \, \text{m} \) is \( 4 \, \mu\text{T} \).

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