At a given place on earth's surface the horizontal component of earths magnetic field is `2xx10^(-5)T` and resultant magnetic field is `4xx10^(-5)T`. The angles of dip at this place is

Options

A. `30^(@)`
B. `60^(@)`
C. `90^(@)`
D. `44^(@)`

Explanation

To find the angle of dip (δ) at a given place on Earth's surface, we can use the relationship between the horizontal component of Earth's magnetic field (H), the resultant magnetic field (R), and the angle of dip. The steps to solve the problem are as follows:

### Step-by-Step Solution:

1. **Identify the given values:**
- Horizontal component of Earth's magnetic field (H) = \(2 \times 10^{-5} \, T\)
- Resultant magnetic field (R) = \(4 \times 10^{-5} \, T\)

2. **Use the relationship between H, R, and δ:**
The horizontal component (H) is related to the resultant magnetic field (R) and the angle of dip (δ) by the equation:
\[ H = R \cos(δ) \]

3. **Rearrange the equation to find cos(δ):**
From the equation, we can express cos(δ) as:
\[ \cos(δ) = \frac{H}{R} \]

4. **Substitute the known values into the equation:**
\[ \cos(δ) = \frac{2 \times 10^{-5}}{4 \times 10^{-5}} \]

5. **Calculate cos(δ):**
\[ \cos(δ) = \frac{2}{4} = \frac{1}{2} \]

6. **Determine the angle δ:**
We know that:
\[ \cos(60^\circ) = \frac{1}{2} \]
Therefore, we conclude that:
\[ δ = 60^\circ \]

### Final Answer:
The angle of dip at that place is \(60^\circ\).

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