The horizontal component of the earth's magnetic field at any place is `0.36 xx 10^(-4) Wb m^(-2)` If the angle of dip at that place is `60 ^@` then the value of the vertical component of earth's magnetic field will be ( in `Wb m^(-2))`

Options

A. `0.12xx10^(-4)`
B. `0.24xx10^(-4)`
C. `0.40xx10^(-4)`
D. `0.62xx10^(-4)`

Answer

To find the vertical component of the Earth's magnetic field given the horizontal component and the angle of dip, we can use the relationship between the horizontal and vertical components of the magnetic field and the angle of dip.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Horizontal component of the Earth's magnetic field, \( B_H = 0.36 \times 10^{-4} \, \text{Wb/m}^2 \)
- Angle of dip, \( \theta = 60^\circ \)

2. **Use the Relationship Between Components:**
The relationship between the vertical component \( B_V \), horizontal component \( B_H \), and the angle of dip \( \theta \) is given by:
\[ \tan(\theta) = \frac{B_V}{B_H} \]

3. **Rearrange the Formula:**
To find the vertical component \( B_V \), we can rearrange the formula:
\[ B_V = B_H \cdot \tan(\theta) \]

4. **Calculate \( \tan(60^\circ) \):**
We know that:
\[ \tan(60^\circ) = \sqrt{3} \]

5. **Substitute the Values:**
Now, substitute the values into the equation:
\[ B_V = 0.36 \times 10^{-4} \cdot \sqrt{3} \]

6. **Calculate \( B_V \):**
Using the approximate value \( \sqrt{3} \approx 1.732 \):
\[ B_V = 0.36 \times 10^{-4} \cdot 1.732 \approx 0.622 \times 10^{-4} \, \text{Wb/m}^2 \]

7. **Final Answer:**
Thus, the vertical component of the Earth's magnetic field is:
\[ B_V \approx 0.622 \times 10^{-4} \, \text{Wb/m}^2 \]

### Summary:
The vertical component of the Earth's magnetic field at that place is approximately \( 0.622 \times 10^{-4} \, \text{Wb/m}^2 \).

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