The angle of dip at a place is `37^(@)` and the vertical component of the earth's magnetic field is `6xx10^(-5) T`. What is the earth's magnetic field at this place?
`(tan 37^(@)=3/4)`

Options

A. `6xx10^(-4) T`
B. `3xx10^(-5) T`
C. `10^(-4) T`
D. `2xx10^(-5) T`

Explanation

To find the Earth's magnetic field at a place where the angle of dip is \(37^\circ\) and the vertical component of the Earth's magnetic field is \(6 \times 10^{-5} \, \text{T}\), we can follow these steps:

### Step 1: Understand the relationship between components of the magnetic field
The angle of dip (\(\delta\)) is related to the vertical component (\(V\)) and the horizontal component (\(H\)) of the Earth's magnetic field by the formula:
\[ \tan(\delta) = \frac{V}{H} \]
Where:
- \(V\) is the vertical component of the magnetic field.
- \(H\) is the horizontal component of the magnetic field.
- \(\delta\) is the angle of dip.

### Step 2: Use the given values
We are given:
- \(\delta = 37^\circ\)
- \(V = 6 \times 10^{-5} \, \text{T}\)

From the question, we know that:
\[ \tan(37^\circ) = \frac{3}{4} \]

### Step 3: Rearrange the formula to find \(H\)
From the formula:
\[ H = \frac{V}{\tan(\delta)} \]
Substituting the values we have:
\[ H = \frac{6 \times 10^{-5}}{\frac{3}{4}} = 6 \times 10^{-5} \times \frac{4}{3} \]

### Step 4: Calculate \(H\)
Now, perform the calculation:
\[ H = 6 \times 10^{-5} \times \frac{4}{3} = \frac{24 \times 10^{-5}}{3} = 8 \times 10^{-5} \, \text{T} \]

### Step 5: Find the total magnetic field \(B\)
The total magnetic field \(B\) can be calculated using the Pythagorean theorem:
\[ B = \sqrt{V^2 + H^2} \]
Substituting the values:
\[ B = \sqrt{(6 \times 10^{-5})^2 + (8 \times 10^{-5})^2} \]
Calculating each term:
\[ B = \sqrt{36 \times 10^{-10} + 64 \times 10^{-10}} = \sqrt{100 \times 10^{-10}} = 10 \times 10^{-5} = 1 \times 10^{-4} \, \text{T} \]

### Final Answer
The Earth's magnetic field at this place is:
\[ B = 1 \times 10^{-4} \, \text{T} \]

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