Question

A compass needle oscillates 20 times per minute at a location with dip \( 45^\circ \) and magnetic field \( B_1 \). It oscillates 30 times per minute at another location where dip is \( 30^\circ \) and field is \( B_2 \). Find the ratio \( \frac{B_1}{B_2} \).

• A. \( \frac{\sqrt{3}}{4\sqrt{2}} \)
• B. \( \frac{4\sqrt{2}}{9\sqrt{3}} \)
• C. \( \frac{\sqrt{3}}{2\sqrt{3}} \)
• D. \( \frac{2\sqrt{2}}{3\sqrt{3}} \)

Hint:

Use \( B_H = B \cos \theta \) and \( T \propto \frac{1}{\sqrt{B_H}} \) for magnetic dip and oscillation relation.

Answer 1

The Correct Option is D

Time period \( T \propto \frac{1}{\sqrt{B_H}} \), where \( B_H = B \cos \theta \) is horizontal component.
\[ \frac{T_1}{T_2} = \frac{1}{\sqrt{B_1 \cos 45^\circ}} \cdot \sqrt{B_2 \cos 30^\circ} = \sqrt{\frac{B_2 \cdot \frac{\sqrt{3}}{2}}{B_1 \cdot \frac{1}{\sqrt{2}}}} = \sqrt{\frac{B_2 \cdot \sqrt{3}}{2B_1 / \sqrt{2}}} \] Given: \( T_1 = \frac{60}{20} = 3\, \text{s}, T_2 = \frac{60}{30} = 2\, \text{s} \Rightarrow \frac{3}{2} = \sqrt{\frac{B_2 \sqrt{3}}{B_1 \sqrt{2}}} \Rightarrow \frac{B_1}{B_2} = \frac{2\sqrt{2}}{3\sqrt{3}} \)