The dipole moment of a short bar magnet is `1.25 A-m^(2)`. The magnetic field on its axis at a distance of 0.5 metre from the centre of the magnet is

Options

A. `1xx10^(-4)NA^(-1)m^(-1)`
B. `2xx10^(-6)NA^(-1)m^(-1)`
C. `4xx10^(-2)NA^(-1)m^(-1)`
D. `6.64xx10^(-8)NA^(-1)m^(-1)`

Explanation

To find the magnetic field on the axis of a short bar magnet at a given distance from its center, we can use the formula for the magnetic field \( B \) at a distance \( r \) from the center of a magnetic dipole:

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

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

Given:
- The dipole moment \( m = 1.25 \, \text{A m}^2 \)
- The distance \( r = 0.5 \, \text{m} \)

### Step-by-Step Solution:

1. **Identify the constants and values:**
- \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m/A} \)
- \( m = 1.25 \, \text{A m}^2 \)
- \( r = 0.5 \, \text{m} \)

2. **Substitute the values into the formula:**
\[ B = \frac{4\pi \times 10^{-7}}{4\pi} \cdot \frac{2 \times 1.25}{(0.5)^3} \]

3. **Simplify the equation:**
The \( 4\pi \) in the numerator and denominator cancels out:
\[ B = 10^{-7} \cdot \frac{2 \times 1.25}{(0.5)^3} \]

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

5. **Substitute \( (0.5)^3 \) back into the equation:**
\[ B = 10^{-7} \cdot \frac{2 \times 1.25}{0.125} \]

6. **Calculate \( \frac{2 \times 1.25}{0.125} \):**
\[ \frac{2.5}{0.125} = 20 \]

7. **Now substitute this back into the equation:**
\[ B = 10^{-7} \cdot 20 \]

8. **Final calculation:**
\[ B = 2 \times 10^{-6} \, \text{T} \quad \text{(or equivalently, } 2 \times 10^{-6} \, \text{N/A m)} \]

### Final Answer:
The magnetic field on the axis of the magnet at a distance of 0.5 meters from its center is \( 2 \times 10^{-6} \, \text{T} \).