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
$T=2\pi \sqrt{\frac{I}{MH}}$
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
$M_1=M+2M=3M$
$T_1=2\pi \sqrt{\frac{1}{M_{1}H}}=2\pi \sqrt{\frac{I}{3MH}}\dots (i)$
When opposite poles of magnets are placed on same side, then net magnetic moment $M_2=2M-M=M$
$\therefore T_2=2\pi \sqrt{\frac{1}{M_{2}H}}=2\pi \sqrt{\frac{1}{MH}}\dots (ii)$
From eqs. (i) and (ii) we observe that
$T_1<T_2$