The horizontal component of flux density of Earth’s magnetic field is `1.7xx10^-6T` The value of horizontal component of intensity of Earth’s magnetic field will be

Options

A. `2.45Am^-1`
B. `13.5Am^-1`
C. `1.53Am^-1`
D. `0.35Am^-1`

Answer

To find the horizontal component of the intensity of Earth's magnetic field given the horizontal component of flux density, we can use the relationship between magnetic flux density (B) and magnetic field intensity (H). The formula we will use is:

\[ B = \mu_0 H \]

Where:
- \( B \) is the magnetic flux density,
- \( \mu_0 \) is the permeability of free space, approximately \( 4\pi \times 10^{-7} \, \text{T m/A} \),
- \( H \) is the magnetic field intensity.

### Step-by-Step Solution:

1. **Identify the given values:**
- The horizontal component of flux density \( B = 1.7 \times 10^{-6} \, \text{T} \).
- The permeability of free space \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m/A} \).

2. **Rearrange the formula to solve for \( H \):**
\[ H = \frac{B}{\mu_0} \]

3. **Substitute the known values into the equation:**
\[ H = \frac{1.7 \times 10^{-6}}{4\pi \times 10^{-7}} \]

4. **Calculate \( 4\pi \):**
- Using \( \pi \approx 3.14 \):
\[ 4\pi \approx 4 \times 3.14 = 12.56 \]

5. **Substitute \( 4\pi \) into the equation:**
\[ H = \frac{1.7 \times 10^{-6}}{12.56 \times 10^{-7}} \]

6. **Simplify the equation:**
\[ H = \frac{1.7}{12.56} \times 10^{1} \quad (\text{since } 10^{-6}/10^{-7} = 10^{1}) \]

7. **Calculate \( \frac{1.7}{12.56} \):**
\[ \frac{1.7}{12.56} \approx 0.1353 \]

8. **Final calculation for \( H \):**
\[ H \approx 0.1353 \times 10^{1} = 1.353 \, \text{A/m} \]

9. **Convert to appropriate units:**
- Since we need to express it in Amperes per meter, we can round it off:
\[ H \approx 13.53 \, \text{A/m} \]

### Final Answer:
The value of the horizontal component of intensity of Earth’s magnetic field is approximately \( 13.53 \, \text{A/m} \).

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