Two bar magnets having geometry with magnetic moments $M$ and $2 M$ are first placed in such a way that…

Two bar magnets having geometry with magnetic moments $M$ and $2 M$ are first placed in such a way that their similar poles are same side then its time period of oscillation is $T_1$. Now the polarity of one of the magnet is reversed then the time period of oscillation is $T_2$, so:
  1. $T_1 < T_2$
  2. $T_1=T_2$
  3. $T_1 > T_2$
  4. $T_2=\infty$

Solution

We can write
$T=2 \pi \sqrt{\frac{1}{M B}} \Rightarrow T \propto \frac{1}{\sqrt{M}}$
Case I : $M_1=2 M+M$
Case II : $M_2=2 M-M$
$\begin{aligned}
& \therefore \frac{T_1}{T_2}=\sqrt{\frac{M}{3 M}}=\frac{1}{\sqrt{3}} \\
& \Rightarrow T_2=\sqrt{3} T_1
\end{aligned}$

Asked in: NEET 2002

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