If `B_H=4xx10^(-5)T and B_(V)=2xx10^(-5)T,` then the Earth's total field (in T) at the place is

Options

A. `6xx10^(-5)T`
B. `2sqrt(5)xx10^(-5)T`
C. `4xx10^(-5)T`
D. `3xx10^(-5)T`

Answer

To find the Earth's total magnetic field at the given location, we will use the horizontal and vertical components of the magnetic field. The total magnetic field \( B \) can be calculated using the Pythagorean theorem since the two components are perpendicular to each other.

### Step-by-Step Solution:

1. **Identify the Components**:
- Given:
- Horizontal component \( B_H = 4 \times 10^{-5} \, \text{T} \)
- Vertical component \( B_V = 2 \times 10^{-5} \, \text{T} \)

2. **Use the Pythagorean Theorem**:
- Since \( B_H \) and \( B_V \) are perpendicular, we can find the total magnetic field \( B \) using:
\[ B = \sqrt{B_H^2 + B_V^2} \]

3. **Calculate \( B_H^2 \) and \( B_V^2 \)**:
- Calculate \( B_H^2 \):
\[ B_H^2 = (4 \times 10^{-5})^2 = 16 \times 10^{-10} \, \text{T}^2 \]
- Calculate \( B_V^2 \):
\[ B_V^2 = (2 \times 10^{-5})^2 = 4 \times 10^{-10} \, \text{T}^2 \]

4. **Add the Squares**:
- Now, add the two results:
\[ B_H^2 + B_V^2 = 16 \times 10^{-10} + 4 \times 10^{-10} = 20 \times 10^{-10} \, \text{T}^2 \]

5. **Take the Square Root**:
- Now, take the square root to find \( B \):
\[ B = \sqrt{20 \times 10^{-10}} = \sqrt{20} \times 10^{-5} \, \text{T} \]

6. **Simplify \( \sqrt{20} \)**:
- We can express \( \sqrt{20} \) as:
\[ \sqrt{20} = \sqrt{4 \times 5} = 2\sqrt{5} \]
- Therefore:
\[ B = 2\sqrt{5} \times 10^{-5} \, \text{T} \]

### Final Answer:
The Earth's total magnetic field at the place is:
\[ B = 2\sqrt{5} \times 10^{-5} \, \text{T} \]