Question

A magnet having a magnetic dipole moment ' M is placed in two magnetic fields ' \( B_1 \) ' and ' \( B_2 \) ' respectively. If it is displaced slightly from the equilibrium position, it oscillates $60$ times in $20$ second in field ' \( B_1 \) ' and $60$ times in $30$ second in field ' \( B_2 \) '. The ratio of field ' \( B_1 \) ' to that of ' \( B_2 \) ' is

• A. $3 : 2$
• B. $9 : 4$
• C. $2 : 3$
• D. $4 : 9$

Hint:

- $T \propto \frac{1}{\sqrt{B}}$ - Use oscillations/time to find period

Answer 1

The Correct Option is B

Concept: Time period of oscillation of magnet: \[ T = 2\pi \sqrt{\frac{I}{MB}} \Rightarrow T \propto \frac{1}{\sqrt{B}} \]

Step 1: Find time periods.
\[ T_1 = \frac{20}{60} = \frac{1}{3}\ \text{s}, \quad T_2 = \frac{30}{60} = \frac{1}{2}\ \text{s} \]

Step 2: Use relation.
\[ \frac{T_1}{T_2} = \sqrt{\frac{B_2}{B_1}} \] \[ \frac{1/3}{1/2} = \frac{2}{3} = \sqrt{\frac{B_2}{B_1}} \]

Step 3: Square both sides.
\[ \frac{4}{9} = \frac{B_2}{B_1} \Rightarrow \frac{B_1}{B_2} = \frac{9}{4} \]