At a certain place the angle of dip is `30^(@)` and the horizontal component of earth's magnetic field is `0.50` oersted. The earth's total magnetic field is

Options

A. `sqrt(3)`
B. `1`
C. `1/sqrt(3)`
D. `1/2`

Explanation

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

### Step 1: Understand the relationship between the components of the Earth's magnetic field.
The total magnetic field \( B \) can be related to the horizontal component \( B_H \) and the angle of dip \( \phi \) using the formula:
\[ B_H = B \cdot \cos(\phi) \]
where:
- \( B_H \) is the horizontal component of the magnetic field.
- \( B \) is the total magnetic field.
- \( \phi \) is the angle of dip.

### Step 2: Rearrange the formula to solve for the total magnetic field \( B \).
From the relationship, we can rearrange the formula to find \( B \):
\[ B = \frac{B_H}{\cos(\phi)} \]

### Step 3: Substitute the known values into the equation.
We know:
- \( B_H = 0.50 \) oersted
- \( \phi = 30^\circ \)

Now, we need to calculate \( \cos(30^\circ) \):
\[ \cos(30^\circ) = \frac{\sqrt{3}}{2} \]

Substituting these values into the equation gives:
\[ B = \frac{0.50}{\frac{\sqrt{3}}{2}} = 0.50 \cdot \frac{2}{\sqrt{3}} = \frac{1}{\sqrt{3}} \text{ oersted} \]

### Step 4: Final answer.
Thus, the total magnetic field \( B \) at that place is:
\[ B = \frac{1}{\sqrt{3}} \text{ oersted} \]