At a certain place , horizontal component of earth's field is `sqrt3` times the vertical component of earth's field . The angle of dip at this place is

Options

A. `30^(@)`
B. `45^(@)`
C. `60^(@)`
D. `90^(@)`

Answer

To find the angle of dip at a place where the horizontal component of Earth's magnetic field is \(\sqrt{3}\) times the vertical component, we can follow these steps:

### Step-by-Step Solution:

1. **Define the Components**:
Let the horizontal component of the Earth's magnetic field be \(B_H\) and the vertical component be \(B_V\). According to the problem, we have:
\[ B_H = \sqrt{3} \cdot B_V \]

2. **Understand the Angle of Dip**:
The angle of dip \(\theta\) is defined as the angle between the horizontal component of the magnetic field and the resultant magnetic field. The tangent of the angle of dip can be expressed as:
\[ \tan(\theta) = \frac{B_V}{B_H} \]

3. **Substitute the Relationship**:
Substitute the expression for \(B_H\) from step 1 into the equation for \(\tan(\theta)\):
\[ \tan(\theta) = \frac{B_V}{\sqrt{3} \cdot B_V} \]
This simplifies to:
\[ \tan(\theta) = \frac{1}{\sqrt{3}} \]

4. **Find the Angle**:
To find \(\theta\), we take the inverse tangent:
\[ \theta = \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) \]
From trigonometric values, we know that:
\[ \tan(30^\circ) = \frac{1}{\sqrt{3}} \]
Therefore:
\[ \theta = 30^\circ \]

5. **Conclusion**:
The angle of dip at this place is \(30^\circ\).