If a distance of 40 cm at an axial position of a dipole, the ''magnetic potential'' (analogous to electric potential) is `2.4xx10^(-5)"J A m"^(-1)` then the magnetic moment of the dipole is

Options

A. `28.6Am^(2)`
B. `32.2Am^(2)`
C. `38.4Am^(2)`
D. None of these

Answer

To find the magnetic moment of the dipole, we can use the formula for the magnetic potential \( V \) at an axial position of a magnetic dipole:

\[ V = \frac{\mu_0 \cdot m \cdot \cos(\theta)}{4 \pi r^2} \]

Where:
- \( V \) is the magnetic potential,
- \( \mu_0 \) is the permeability of free space (\( 4\pi \times 10^{-7} \, \text{T m/A} \)),
- \( m \) is the magnetic moment,
- \( r \) is the distance from the dipole,
- \( \theta \) is the angle between the dipole moment and the line joining the dipole to the point where the potential is measured.

Since we are at the axial position, \( \theta = 0^\circ \) and \( \cos(0) = 1 \).

### Step-by-step Solution:

1. **Identify the given values:**
- Magnetic potential, \( V = 2.4 \times 10^{-5} \, \text{J A m}^{-1} \)
- Distance, \( r = 40 \, \text{cm} = 0.4 \, \text{m} \)

2. **Substitute the known values into the formula:**
\[ 2.4 \times 10^{-5} = \frac{(4\pi \times 10^{-7}) \cdot m}{4\pi \cdot (0.4)^2} \]

3. **Simplify the equation:**
- The \( 4\pi \) terms cancel out:
\[ 2.4 \times 10^{-5} = \frac{10^{-7} \cdot m}{(0.4)^2} \]

4. **Calculate \( (0.4)^2 \):**
\[ (0.4)^2 = 0.16 \]

5. **Rearranging the equation to solve for \( m \):**
\[ 2.4 \times 10^{-5} \cdot 0.16 = 10^{-7} \cdot m \]

6. **Calculate \( 2.4 \times 10^{-5} \cdot 0.16 \):**
\[ 2.4 \times 0.16 = 0.384 \quad \Rightarrow \quad 0.384 \times 10^{-5} = 3.84 \times 10^{-6} \]

7. **Now, solve for \( m \):**
\[ m = \frac{3.84 \times 10^{-6}}{10^{-7}} = 0.384 \, \text{A m}^2 \]

8. **Final answer:**
\[ m \approx 0.4 \, \text{A m}^2 \]

### Conclusion:
The magnetic moment of the dipole is approximately \( 0.4 \, \text{A m}^2 \).