In the magnetic meridian of a certain place, the horizontal component of the earth's magnetic field is `0*26G` and dip angle is `60^@`. What is the magnetic field of earth at this location?

Options

A. 0.50G
B. 0.52G
C. 0.54G
D. 0.56G

Explanation

To find the magnetic field of the Earth at a certain location given the horizontal component and the dip angle, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values:**
- Horizontal component of the Earth's magnetic field, \( H = 0.26 \, G \)
- Dip angle, \( \delta = 60^\circ \)

2. **Use the Relationship Between Components:**
The relationship between the horizontal component \( H \), the total magnetic field \( R \), and the dip angle \( \delta \) is given by:
\[ H = R \cos \delta \]
Rearranging this formula to solve for \( R \):
\[ R = \frac{H}{\cos \delta} \]

3. **Calculate \( \cos \delta \):**
Since \( \delta = 60^\circ \):
\[ \cos 60^\circ = \frac{1}{2} \]

4. **Substitute the Values:**
Substitute \( H \) and \( \cos \delta \) into the equation:
\[ R = \frac{0.26 \, G}{\frac{1}{2}} = 0.26 \, G \times 2 \]

5. **Calculate \( R \):**
\[ R = 0.52 \, G \]

### Final Answer:
The magnetic field of the Earth at this location is \( 0.52 \, G \).

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