The earth's magnetic field at the equator is approximately `0*4G`, Estimate the earth's dipole moment.

Options

A. `1.05xx10^(23)Am^(2)`
B. `2.05xx10^(23)Am^(2)`
C. `1.05xx10^(21)Am^(2)`
D. `2.05xx10^(21)Am^(2)`

Explanation

To estimate the Earth's dipole moment given that the magnetic field at the equator is approximately 0.4 G, we can follow these steps:

### Step-by-Step Solution:

1. **Convert Magnetic Field from Gauss to Tesla**:
- The magnetic field at the equator is given as 0.4 G.
- We know that \(1 \, \text{G} = 10^{-4} \, \text{T}\).
- Therefore, \(B = 0.4 \, \text{G} = 0.4 \times 10^{-4} \, \text{T} = 4 \times 10^{-5} \, \text{T}\).

2. **Use the Formula for the Magnetic Field due to a Dipole**:
- The magnetic field \(B\) at the equator due to a magnetic dipole is given by:
\[ B = \frac{\mu_0}{4\pi} \cdot \frac{m}{r^3} \]
- Here, \(m\) is the magnetic dipole moment, \(r\) is the radius of the Earth, and \(\mu_0\) is the permeability of free space, which is approximately \(4\pi \times 10^{-7} \, \text{T m/A}\).

3. **Rearranging the Formula to Solve for \(m\)**:
- Rearranging the formula to find \(m\):
\[ m = B \cdot \frac{4\pi r^3}{\mu_0} \]

4. **Substituting Known Values**:
- The radius of the Earth \(r\) is approximately \(6.4 \times 10^6 \, \text{m}\).
- Substitute \(B = 4 \times 10^{-5} \, \text{T}\), \(r = 6.4 \times 10^6 \, \text{m}\), and \(\mu_0 = 4\pi \times 10^{-7} \, \text{T m/A}\) into the equation:
\[ m = (4 \times 10^{-5}) \cdot \frac{4\pi (6.4 \times 10^6)^3}{4\pi \times 10^{-7}} \]

5. **Simplifying the Expression**:
- The \(4\pi\) cancels out:
\[ m = (4 \times 10^{-5}) \cdot \frac{(6.4 \times 10^6)^3}{10^{-7}} \]

6. **Calculating \(r^3\)**:
- Calculate \(r^3\):
\[ (6.4 \times 10^6)^3 = 2.62144 \times 10^{20} \, \text{m}^3 \]

7. **Final Calculation**:
- Now substitute \(r^3\) back into the equation:
\[ m = (4 \times 10^{-5}) \cdot \frac{2.62144 \times 10^{20}}{10^{-7}} = (4 \times 10^{-5}) \cdot (2.62144 \times 10^{27}) \]
- This gives:
\[ m = 1.048576 \times 10^{23} \, \text{A m}^2 \approx 1.05 \times 10^{23} \, \text{A m}^2 \]

8. **Conclusion**:
- The estimated Earth's dipole moment is approximately \(1.05 \times 10^{23} \, \text{A m}^2\).