Assume the dipole model of earth's magnetic field B which is given by `B_V`= vertical component of magnetic field `=(mu_0)/(4pi)(2Mcos theta)/(r^3)`, `B_H`=Horizontal component of magnetic field `=(mu_0)/(4pi)(sin theta M)/(r^3)`, `theta=90^@`-latitude as measured from magnetic equator.
Find loci of points for which dip angle is `+-45^@`.

Options

A. `tan^(-1)(3)`
B. `tan^(-1)(2)`
C. `tan^(-1)(0.5)`
D. `tan^(-1)(1)`

Explanation

To solve the problem, we need to find the loci of points where the dip angle is ±45°. The dip angle (also known as the angle of inclination) is defined as the angle made by the magnetic field with the horizontal plane.

### Step-by-Step Solution:

1. **Understanding the Dip Angle**:
The dip angle \( \theta \) is given by the relation:
\[ \tan(\theta) = \frac{B_V}{B_H} \]
where \( B_V \) is the vertical component and \( B_H \) is the horizontal component of the Earth's magnetic field.

2. **Substituting the Components**:
From the problem, we have:
\[ B_V = \frac{\mu_0}{4\pi} \cdot \frac{2M \cos \theta}{r^3} \]
\[ B_H = \frac{\mu_0}{4\pi} \cdot \frac{M \sin \theta}{r^3} \]

3. **Setting Up the Equation for ±45°**:
For a dip angle of \( 45° \):
\[ \tan(45°) = 1 \]
Thus, we can write:
\[ \frac{B_V}{B_H} = 1 \]

4. **Equating the Components**:
Substituting the expressions for \( B_V \) and \( B_H \):
\[ \frac{\frac{\mu_0}{4\pi} \cdot \frac{2M \cos \theta}{r^3}}{\frac{\mu_0}{4\pi} \cdot \frac{M \sin \theta}{r^3}} = 1 \]
The \( \frac{\mu_0}{4\pi} \) and \( \frac{1}{r^3} \) terms cancel out:
\[ \frac{2M \cos \theta}{M \sin \theta} = 1 \]

5. **Simplifying the Equation**:
Canceling \( M \) gives:
\[ \frac{2 \cos \theta}{\sin \theta} = 1 \]
Rearranging this, we have:
\[ 2 \cos \theta = \sin \theta \]

6. **Using Trigonometric Identity**:
Dividing both sides by \( \cos \theta \):
\[ 2 = \tan \theta \]

7. **Finding the Angle**:
Therefore, we can find \( \theta \) as:
\[ \theta = \tan^{-1}(2) \]

8. **Considering ±45°**:
Since we are interested in the loci for both \( +45° \) and \( -45° \), we can conclude that the points where the dip angle is \( ±45° \) correspond to the latitude where \( \theta = \tan^{-1}(2) \) and its symmetric counterpart.

### Final Answer:
The loci of points for which the dip angle is \( ±45° \) corresponds to the latitude \( \theta = \tan^{-1}(2) \).