if `B_(y) = 1/sqrt(3)B_(y)`, then find the angle of dip:

Options

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

Explanation

To find the angle of dip given that \( B_y = \frac{1}{\sqrt{3}} B_H \), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Components**:
- The magnetic field can be resolved into two components: the vertical component \( B_V \) and the horizontal component \( B_H \).
- In this case, we are given that the vertical component \( B_V \) is equal to \( B_H \) multiplied by \( \frac{1}{\sqrt{3}} \).

2. **Using the Angle of Dip Formula**:
- The angle of dip \( \delta \) is related to the vertical and horizontal components of the magnetic field by the formula:
\[ \tan(\delta) = \frac{B_V}{B_H} \]

3. **Substituting the Given Values**:
- From the problem, we know:
\[ B_V = \frac{1}{\sqrt{3}} B_H \]
- Substitute this into the angle of dip formula:
\[ \tan(\delta) = \frac{\frac{1}{\sqrt{3}} B_H}{B_H} \]

4. **Simplifying the Expression**:
- The \( B_H \) in the numerator and denominator cancels out:
\[ \tan(\delta) = \frac{1}{\sqrt{3}} \]

5. **Finding the Angle**:
- To find the angle \( \delta \), we take the arctangent (inverse tangent) of \( \frac{1}{\sqrt{3}} \):
\[ \delta = \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) \]

6. **Identifying the Angle**:
- We know that \( \tan(30^\circ) = \frac{1}{\sqrt{3}} \), hence:
\[ \delta = 30^\circ \]

### Final Answer:
The angle of dip is \( 30^\circ \).