Two bar magnets oscillate in a horizontal plane in earth's magnetic field with time periods of $3 \mathrm{~s}$ and $4 \mathrm{~s}$ respectively. If their moments of inertia are in the ratio of $3: 2$, then the ratio of their magnetic moments will be:
Solution
<p>$T = 2\pi \sqrt {{I \over {M{B_H}}}}$</p>
<p>$$ \Rightarrow {{{T_1}} \over {{T_2}}} = \sqrt {{{{I_1}} \over {{I_2}}}} \sqrt {{{{M_2}} \over {{M_1}}}} $$</p>
<p>$\Rightarrow {3 \over 4} = \sqrt {{3 \over 2}} \sqrt {{{{M_2}} \over {{M_1}}}}$</p>
<p>$$ \Rightarrow {{{M_1}} \over {{M_2}}} = {3 \over 2} \times {{16} \over 9} = {8 \over 3}$$</p>
About this question
Subject: Physics · Chapter: Magnetism · Topic: Bar Magnet and Magnetic Field
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