At a given palce on the earth's surface, the horizontal component of earth's magnetic field is `3 xx 10^(-5)T` and resultant magnetic field is `6 xx 10^(-5)T`. The angle of dip at this place is

Options

A. `30^(@)`
B. `40^(@)`
C. `50^(@)`
D. `60^(@)`

Answer

To find the angle of dip (δ) at a given place on the Earth's surface, we can use the relationship between the horizontal component of the Earth's magnetic field (Bh), the vertical component (Bv), and the resultant magnetic field (B). The formula we will use is:

\[ \tan(\delta) = \frac{Bv}{Bh} \]

Where:
- Bh = Horizontal component of the magnetic field
- Bv = Vertical component of the magnetic field
- B = Resultant magnetic field

From the problem, we have:
- Bh = \(3 \times 10^{-5} \, T\)
- B = \(6 \times 10^{-5} \, T\)

### Step 1: Calculate the vertical component (Bv)
Using the Pythagorean theorem, we know that:

\[ B^2 = Bh^2 + Bv^2 \]

Rearranging gives us:

\[ Bv^2 = B^2 - Bh^2 \]

Substituting the values:

\[ Bv^2 = (6 \times 10^{-5})^2 - (3 \times 10^{-5})^2 \]

Calculating each term:

\[ (6 \times 10^{-5})^2 = 36 \times 10^{-10} \]
\[ (3 \times 10^{-5})^2 = 9 \times 10^{-10} \]

Now, substituting back:

\[ Bv^2 = 36 \times 10^{-10} - 9 \times 10^{-10} = 27 \times 10^{-10} \]

Taking the square root to find Bv:

\[ Bv = \sqrt{27 \times 10^{-10}} = 3\sqrt{3} \times 10^{-5} \, T \]

### Step 2: Calculate the angle of dip (δ)
Now we can find the angle of dip using the tangent function:

\[ \tan(\delta) = \frac{Bv}{Bh} \]

Substituting the values:

\[ \tan(\delta) = \frac{3\sqrt{3} \times 10^{-5}}{3 \times 10^{-5}} = \sqrt{3} \]

### Step 3: Find δ
Now, we know that:

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

Thus, we conclude:

\[ \delta = 60^\circ \]

### Final Answer
The angle of dip at this place is \(60^\circ\).

---