The horizontal and vertical components of earth's magnetic field at a place are 0.3G and 0.52G. The earth's magnetic field and the angle of dip are

Options

A. `0.3G and delta=30^(@)`
B. `0.4G and delta=40^(@)`
C. `0.5G and delta=50^(@)`
D. `0.6G and delta=60^(@)`

Explanation

To solve the problem, we need to find the Earth's magnetic field (B) and the angle of dip (δ) using the given horizontal (Bh) and vertical (Bv) components of the magnetic field.

### Step 1: Calculate the Earth's Magnetic Field (B)

The formula for the total magnetic field (B) is given by:

\[ B = \sqrt{B_h^2 + B_v^2} \]

Where:
- \(B_h\) is the horizontal component of the magnetic field.
- \(B_v\) is the vertical component of the magnetic field.

Given:
- \(B_h = 0.3 \, G\)
- \(B_v = 0.52 \, G\)

Substituting the values into the formula:

\[ B = \sqrt{(0.3)^2 + (0.52)^2} \]

Calculating the squares:

\[ B = \sqrt{0.09 + 0.2704} \]

\[ B = \sqrt{0.3604} \]

Now, calculating the square root:

\[ B \approx 0.6 \, G \]

### Step 2: Calculate the Angle of Dip (δ)

The angle of dip (δ) can be calculated using the formula:

\[ \tan(δ) = \frac{B_v}{B_h} \]

Substituting the values:

\[ \tan(δ) = \frac{0.52}{0.3} \]

Calculating the division:

\[ \tan(δ) \approx 1.7333 \]

Now, we need to find the angle whose tangent is approximately 1.73. We know that:

\[ \tan(60^\circ) = \sqrt{3} \approx 1.732 \]

Thus, we can conclude that:

\[ δ \approx 60^\circ \]

### Final Results

- The Earth's magnetic field (B) is approximately \(0.6 \, G\).
- The angle of dip (δ) is approximately \(60^\circ\).

### Summary of Results:
- Earth's Magnetic Field (B) = \(0.6 \, G\)
- Angle of Dip (δ) = \(60^\circ\)

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