The vertical component of earth's magnetic field at a place is `0.16sqrt(3) xx 10^(-4)` Tesla. What is the value of horizontal component of earth's magnetic field if the angle of dip at that place is `30^(@)` ?

Options

A. `2 xx 10^(-5) T`
B. `2.6 xx 10^(-4) T`
C. `0.48 xx 10^(-4) T`
D. `6 xx 10^(-3) T`

Explanation

To find the horizontal component of the Earth's magnetic field given the vertical component and the angle of dip, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Vertical component of the Earth's magnetic field, \( B_v = 0.16 \sqrt{3} \times 10^{-4} \) Tesla.
- Angle of dip, \( \delta = 30^\circ \).

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

3. **Rearranging the Formula to Find \( B_h \):**
To find the horizontal component, rearrange the formula:
\[ B_h = \frac{B_v}{\tan(\delta)} \]

4. **Substituting the Known Values:**
Substitute \( B_v \) and \( \tan(30^\circ) \) into the equation. We know that:
\[ \tan(30^\circ) = \frac{1}{\sqrt{3}} \]
Therefore:
\[ B_h = \frac{0.16 \sqrt{3} \times 10^{-4}}{\frac{1}{\sqrt{3}}} \]

5. **Simplifying the Expression:**
When you divide by \( \frac{1}{\sqrt{3}} \), it is equivalent to multiplying by \( \sqrt{3} \):
\[ B_h = 0.16 \sqrt{3} \times 10^{-4} \times \sqrt{3} \]
\[ B_h = 0.16 \times 3 \times 10^{-4} \]
\[ B_h = 0.48 \times 10^{-4} \text{ Tesla} \]

6. **Final Result:**
Hence, the horizontal component of the Earth's magnetic field is:
\[ B_h = 0.48 \times 10^{-4} \text{ Tesla} \]