The horizontal component of the earth's magnetic field at any place is `0.36 xx 10^(-4) Wb m^(-2)` If the angle of dip at that place is `60 ^@` then the value of the vertical component of earth's magnetic field will be ( in `Wb m^(-2))`

Options

A. `0.12xx10^(-4)`
B. `0.24xx10^(-4)`
C. `0.40xx10^(-4)`
D. `0.62xx10^(-4)`

Answer

To find the vertical component of the Earth's magnetic field given the horizontal component and the angle of dip, we can follow these steps:

### Step 1: Identify the given values
- Horizontal component of the Earth's magnetic field, \( B_H = 0.36 \times 10^{-4} \, \text{Wb/m}^2 \)
- Angle of dip, \( \theta = 60^\circ \)

### Step 2: Use the relationship between components and angle of dip
The relationship between the horizontal component \( B_H \), the vertical component \( B_V \), and the angle of dip \( \theta \) is given by:
\[ \tan(\theta) = \frac{B_V}{B_H} \]

### Step 3: Rearranging the equation to find \( B_V \)
From the equation above, we can express the vertical component \( B_V \) as:
\[ B_V = B_H \cdot \tan(\theta) \]

### Step 4: Calculate \( \tan(60^\circ) \)
We know that:
\[ \tan(60^\circ) = \sqrt{3} \approx 1.732 \]

### Step 5: Substitute the values into the equation
Now we can substitute the values of \( B_H \) and \( \tan(60^\circ) \) into the equation:
\[ B_V = 0.36 \times 10^{-4} \, \text{Wb/m}^2 \cdot 1.732 \]

### Step 6: Perform the multiplication
Calculating the above expression:
\[ B_V = 0.36 \times 1.732 \times 10^{-4} \, \text{Wb/m}^2 \]
\[ B_V \approx 0.62292 \times 10^{-4} \, \text{Wb/m}^2 \]

### Step 7: Round the result
Rounding to two decimal places, we get:
\[ B_V \approx 0.62 \times 10^{-4} \, \text{Wb/m}^2 \]

### Final Answer
The vertical component of the Earth's magnetic field is approximately:
\[ B_V \approx 0.62 \times 10^{-4} \, \text{Wb/m}^2 \]

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