Question

The horizontal component of the earth’s magnetic field is \( 3.6 \times 10^{-5} \) tesla where the dip angle is 60°. The magnitude of the earth’s magnetic field is:

• A. \( 2.8 \times 10^{-4} \) tesla
• B. \( 2.1 \times 10^{-5} \) tesla
• C. \( 7.2 \times 10^{-5} \) tesla
• D. \( 3.6 \times 10^{-5} \) tesla

Hint:

The magnitude of the earth's magnetic field can be calculated using the horizontal component and dip angle.

Answer 1

The Correct Option is C

Step 1: The total magnetic field \( B \) can be related to the horizontal component \( B_H \) and the dip angle \( \theta \) using the formula: \[ B_H = B \cos \theta. \] Step 2: Given that \( B_H = 3.6 \times 10^{-5} \, \text{T} \) and \( \theta = 60^\circ \), we can solve for \( B \): \[ B = \dfrac{B_H}{\cos \theta} = \dfrac{3.6 \times 10^{-5}}{\cos 60^\circ} = \dfrac{3.6 \times 10^{-5}}{0.5} = 7.2 \times 10^{-5} \, \text{T}. \]
Final Answer: \[ \boxed{7.2 \times 10^{-5} \, \text{T}} \]