Dipole moment of earth is order of the

Options

A. `10^(15)` J/T
B. `10^(22) ` J/T
C. `10^(-15) `J/T
D. `10^(-22)` J/T

Answer

To find the dipole moment of the Earth, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Concept**:
The Earth can be modeled as a bar magnet, and its magnetic field (B) can be expressed in terms of its dipole moment (M) and the distance (d) from the center of the dipole.

2. **Using the Formula**:
The formula for the magnetic field due to a magnetic dipole is:
\[ B = \frac{\mu_0}{4\pi} \cdot \frac{M}{d^3} \]
where:
- \( B \) is the magnetic field (given as \( 4 \times 10^{-5} \) Tesla),
- \( \mu_0 \) is the permeability of free space (\( 4\pi \times 10^{-7} \, \text{T m/A} \)),
- \( M \) is the magnetic dipole moment,
- \( d \) is the diameter of the Earth (given as \( 6.4 \times 10^6 \) meters).

3. **Rearranging the Formula**:
To find the dipole moment \( M \), we rearrange the formula:
\[ M = \frac{B \cdot 4\pi d^3}{\mu_0} \]

4. **Substituting Values**:
Substitute the known values into the equation:
\[ M = \frac{(4 \times 10^{-5}) \cdot (4\pi) \cdot (6.4 \times 10^6)^3}{4\pi \times 10^{-7}} \]

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

6. **Calculating \( d^3 \)**:
Calculate \( (6.4 \times 10^6)^3 \):
\[ (6.4 \times 10^6)^3 = 6.4^3 \times 10^{18} \approx 262.144 \times 10^{18} \approx 2.62144 \times 10^{20} \]

7. **Final Calculation**:
Now plug this back into the equation for \( M \):
\[ M = (4 \times 10^{-5}) \cdot (2.62144 \times 10^{20}) \cdot 10^{7} \]
\[ M = 4 \times 2.62144 \times 10^{22} = 10.48576 \times 10^{22} \approx 2.62 \times 10^{22} \, \text{Joule per Tesla} \]

8. **Conclusion**:
The order of the dipole moment of the Earth is approximately \( 10^{22} \).

### Final Answer:
The dipole moment of the Earth is of the order of \( 10^{22} \, \text{Joule per Tesla} \).

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