Question

The vertical and horizontal components of Earth's magnetic field at a place are \( 2 \times 10^{-5} \, \text{T} \) and \( 2\sqrt{3} \times 10^{-5} \, \text{T} \) respectively. The angle of dip and resultant Earth's magnetic field is

• A. \( \tan^{-1} (\sqrt{3}) \), \( 2 \times 10^{-10} \, \text{T} \)
• B. \( \tan^{-1} (\sqrt{3}) \), \( 2 \times 10^{-5} \, \text{T} \)
• C. \( \tan^{-1} \left( \frac{1}{\sqrt{3}} \right) \), \( 4 \times 10^{-10} \, \text{T} \)
• D. \( \tan^{-1} \left( \frac{1}{\sqrt{3}} \right) \), \( 4 \times 10^{-5} \, \text{T} \)

Hint:

The angle of dip is found using the ratio of the vertical to horizontal components of the magnetic field, and the resultant magnetic field is the vector sum of the components.

Answer 1

The Correct Option is D

Step 1: Using the components of the magnetic field.
The resultant magnetic field \( B \) is calculated using the vector sum of the vertical and horizontal components: \[ B = \sqrt{B_v^2 + B_h^2} \] where \( B_v = 2 \times 10^{-5} \, \text{T} \) and \( B_h = 2\sqrt{3} \times 10^{-5} \, \text{T} \). Substituting these values: \[ B = \sqrt{(2 \times 10^{-5})^2 + (2\sqrt{3} \times 10^{-5})^2} = 4 \times 10^{-5} \, \text{T} \] Step 2: Finding the angle of dip.
The angle of dip \( \delta \) is given by: \[ \tan \delta = \frac{B_v}{B_h} \] Substituting the values: \[ \tan \delta = \frac{2 \times 10^{-5}}{2\sqrt{3} \times 10^{-5}} = \frac{1}{\sqrt{3}} \] Thus, \[ \delta = \tan^{-1} \left( \frac{1}{\sqrt{3}} \right) \] Step 3: Conclusion.
Thus, the angle of dip is \( \tan^{-1} \left( \frac{1}{\sqrt{3}} \right) \) and the resultant magnetic field is \( 4 \times 10^{-5} \, \text{T} \), corresponding to option (D).