Question

Two bar magnets having same geometry with magnetic moments $M$ and $2M$, are firstly 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 time period of oscillation is $T_2$, then :-

• A. $T_1 < T_2$
• B. $T_1 = T_2$
• C. $T_1 > T_2$
• D. $T_2 = \infty$

Answer 1

Answer: (A)

The time period of bar magnet
$T=2 \pi \sqrt{\frac{I}{M H}}$
where $M =$ magnetic moment of magnet
$I=$ moment of inertia
and $H =$ horizontal component of magnetic field
when same poles of magnets are placed on same side, then net magnetic moment
$M_{1}=M+2 M=3 M$
$T_{1}=2 \pi \sqrt{\frac{1}{M_{1} H}}=2 \pi \sqrt{\frac{I}{3 M H}}\ldots(i)$
When opposite poles of magnets are placed on same side, then net magnetic moment $M_{2}=2 M-M=M$
$\therefore T_{2}=2 \pi \sqrt{\frac{1}{M_{2} H}}=2 \pi \sqrt{\frac{1}{M H}} \ldots(ii)$
From eqs. (i) and (ii) we observe that
$T_{1}< T_{2}$