At a place the earth's horizontal component of magnetic field is `0.36xx10^(-4) "Weber"//m^(2)`. If the angle of dip at that place is `60^(@)`, then the vertical component of earth's field at that place in `"Weber"//m^(2)` will be approxmately

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 at a place where the horizontal component is given and the angle of dip is known, we can use the following steps:

### Step-by-step Solution:

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

2. **Understand the Relationship**:
The relationship between the horizontal component \( B_H \), the vertical component \( B_V \), and the total magnetic field \( B \) is given by:
\[ B_H = B \cos(\delta) \]
\[ B_V = B \sin(\delta) \]

3. **Express the Vertical Component in Terms of Horizontal Component**:
From the above equations, we can express \( B_V \) in terms of \( B_H \):
\[ \frac{B_V}{B_H} = \tan(\delta) \]
Therefore,
\[ B_V = B_H \tan(\delta) \]

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

5. **Substitute the Values**:
Now substitute \( B_H \) and \( \tan(60^\circ) \) into the equation for \( B_V \):
\[ B_V = B_H \cdot \sqrt{3} = (0.36 \times 10^{-4}) \cdot \sqrt{3} \]

6. **Calculate the Final Value**:
Using \( \sqrt{3} \approx 1.732 \):
\[ B_V \approx 0.36 \times 10^{-4} \times 1.732 \approx 0.623 \times 10^{-4} \, \text{Weber/m}^2 \]

7. **Round the Result**:
Rounding this value gives us approximately:
\[ B_V \approx 0.62 \times 10^{-4} \, \text{Weber/m}^2 \]

### Final Answer:
The vertical component of the Earth's magnetic field at that place is approximately \( 0.62 \times 10^{-4} \, \text{Weber/m}^2 \).

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