If the angles of dip at two places are \(30^{\circ}\) and \(45^{\circ}\) respectively, then the ratio of…
If the angles of dip at two places are \(30^{\circ}\) and \(45^{\circ}\) respectively, then the ratio of horizontal components of earth's magnetic field at the two places will be
\(\sqrt{3}: \sqrt{2}\)
\(1: \sqrt{2}\)
\(1: \sqrt{3}\)
\(1: 2\)
Solution
If \(B_e\) be the earth's net magnetic field, then the ratio of horizontal components of earth's magnetic field at the two places is given as
\(\begin{aligned}
\frac{H_1}{H_2} & =\frac{B_e \cos \delta_1}{B_e \cos \delta_2}=\frac{\cos \delta_1}{\cos \delta_2} \\
& =\frac{\cos 30^{\circ}}{\cos 45^{\circ}} \quad\left[\because \delta_1=30^{\circ}, \delta_2=45^{\circ}\right] \\
& =\frac{\sqrt{3} / 2}{1 / \sqrt{2}}=\frac{\sqrt{3}}{\sqrt{2}} \\
\therefore \quad H_1: H_2 & =\sqrt{3}: \sqrt{2}
\end{aligned}\)