Question

The horizontal component of earth's magnetic field at any place is \(\sqrt{3}\) times its vertical component. What will be the value of angle of dip at that place?

Hint:

Remember the definition \(\tan(\delta) = B_V / B_H\). A common mistake is to flip the fraction. A simple way to check is to think about the poles and the equator. At the magnetic equator, \(B_V = 0\), so \(\delta = 0^\circ\). At the magnetic poles, \(B_H = 0\), so \(\tan(\delta)\) is infinite and \(\delta = 90^\circ\). This confirms \(B_V\) must be in the numerator.

Answer 1

Step 1: Understanding the Concept:
The Earth's total magnetic field (\(B_E\)) at any point can be resolved into two components: a horizontal component (\(B_H\)) and a vertical component (\(B_V\)). The angle that the total magnetic field vector makes with the horizontal direction is called the angle of dip or inclination (\(\delta\)).
Step 2: Key Formula or Approach:
The relationship between the components and the angle of dip (\(\delta\)) is given by trigonometry:
\[ \tan(\delta) = \frac{B_V}{B_H} \] Step 3: Detailed Explanation:
We are given the following relationship in the problem:
The horizontal component (\(B_H\)) is \(\sqrt{3}\) times the vertical component (\(B_V\)).
Mathematically, this is written as:
\[ B_H = \sqrt{3} \, B_V \] Now, we substitute this into the formula for the angle of dip:
\[ \tan(\delta) = \frac{B_V}{B_H} = \frac{B_V}{\sqrt{3} \, B_V} \] The \(B_V\) term cancels out from the numerator and the denominator:
\[ \tan(\delta) = \frac{1}{\sqrt{3}} \] To find the angle \(\delta\), we take the inverse tangent (arctan) of both sides:
\[ \delta = \arctan\left(\frac{1}{\sqrt{3}}\right) \] From our knowledge of standard trigonometric values, we know that \(\tan(30^\circ) = \frac{1}{\sqrt{3}}\).
Therefore, the angle of dip is:
\[ \delta = 30^\circ \] Step 4: Final Answer:
The value of the angle of dip at that place is 30\(^{\circ}\).