The vertical component of the earth’s magnetic field at a place is `0.24sqrt3xx10^-4`T Find out the value of horizontal component of the earth’s magnetic field, if angle of dip at that place is `30^@`.

Options

A. `1.32xx10^-5T`
B. `2.16xx10^-6T`
C. `0.91xx10^-4T`
D. `0.72xx10^-4T`

Answer

To find the horizontal component of the Earth's magnetic field given the vertical component and the angle of dip, we can use the following relationship:

1. **Understanding the Components**:
The Earth's magnetic field can be resolved into two components: the horizontal component (B_H) and the vertical component (B_V). The angle of dip (δ) is the angle between the Earth's magnetic field and the horizontal plane.

2. **Using the Relationship**:
The relationship between the vertical and horizontal components of the Earth's magnetic field is given by:
\[ \tan(\delta) = \frac{B_V}{B_H} \]
Rearranging this gives:
\[ B_H = \frac{B_V}{\tan(\delta)} \]

3. **Given Values**:
- Vertical component, \( B_V = 0.24 \sqrt{3} \times 10^{-4} \, \text{T} \)
- Angle of dip, \( \delta = 30^\circ \)

4. **Calculating \(\tan(30^\circ)\)**:
We know that:
\[ \tan(30^\circ) = \frac{1}{\sqrt{3}} \]

5. **Substituting Values**:
Now we can substitute the values into the equation for \( B_H \):
\[ B_H = \frac{0.24 \sqrt{3} \times 10^{-4}}{\tan(30^\circ)} = \frac{0.24 \sqrt{3} \times 10^{-4}}{\frac{1}{\sqrt{3}}} \]

6. **Simplifying**:
This simplifies to:
\[ B_H = 0.24 \sqrt{3} \times 10^{-4} \times \sqrt{3} = 0.24 \times 3 \times 10^{-4} = 0.72 \times 10^{-4} \, \text{T} \]

7. **Final Result**:
Therefore, the horizontal component of the Earth's magnetic field is:
\[ B_H = 0.72 \times 10^{-4} \, \text{T} \]