At a certain place, the angle of dip is ` 60^(@)` and the horizontal component of the earth's magnetic field `(B_(H))` is `0.8xx10^(-4)` T. The earth's overall magnetic field is

Options

A. `1.5xx10^(-4)T`
B. `1.6xx10^(-3)T`
C. `1.5xx10^(-3)T`
D. `1.6xx10^(-4)T`

Answer

To find the Earth's overall magnetic field (B) given the angle of dip (δ) and the horizontal component of the Earth's magnetic field (B_H), we can use the relationship between these quantities.

### Step-by-Step Solution:

1. **Identify the given values:**
- Angle of dip (δ) = 60 degrees
- Horizontal component of the Earth's magnetic field (B_H) = \(0.8 \times 10^{-4}\) T

2. **Use the relationship between the components of the magnetic field:**
The relationship between the total magnetic field (B), the horizontal component (B_H), and the angle of dip (δ) is given by:
\[ B_H = B \cos(δ) \]
Rearranging this equation to find B gives:
\[ B = \frac{B_H}{\cos(δ)} \]

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

4. **Substitute the values into the equation:**
Now substitute B_H and cos(δ) into the equation for B:
\[ B = \frac{0.8 \times 10^{-4}}{\cos(60^\circ)} = \frac{0.8 \times 10^{-4}}{\frac{1}{2}} \]

5. **Simplify the equation:**
\[ B = 0.8 \times 10^{-4} \times 2 = 1.6 \times 10^{-4} \text{ T} \]

6. **Final Result:**
The Earth's overall magnetic field (B) is:
\[ B = 1.6 \times 10^{-4} \text{ T} \]