Question

Two bar magnets having same geometry with magnetic moments M and 2M, are firstly placed in such a way that their similer poles are same side then its time period of osccilation is T1 . Now the polarity of one of the magnet is reversed then time period of osccilation is T2 , then : -

Answer 1

time period of oscillation of bar magnet in earth's magnetic field is given by, $T=2\pi\sqrt{\frac{I}{MH}}$

where, I is moment of inertia, M is net magnetic moment and H is horizontal component of earth's magnetic field.

case 1 : when bar magnets are placed in such a way that their similar poles are on same side .

then, net magnetic moment, $M_1=M+2M=3M$

so, time period, $T_1=2\pi\sqrt{\frac{I}{3MH}}$

case 2 : when bar magnets are placed in such a way that their similar poles are on opposite sides .

then, net magnetic moment, $M_2=2M-M=M$

so, time period, $T_2=2\pi\sqrt{\frac{I}{MH}}$

it is clear that , $T_1 < T_2$

Answer 2

T1<T2, Time period of oscillation of bar magnet in earth's magnetic field  is – 2π√I/MH

Here  I = moment of inertia

M =magnetic moment  

H = horizontal component of earth's magnetic field.

Situation 1 : when bar magnets have  similar poles are on same side .

Net magnetic momentum= M1= M+2M= 3M  

Time period= 2π√I/3MH

Situtaion 2 : When bar magnets have similar poles are on opposite sides .

Net magnetic moment= 2M-M= M

Time period=  2π√I/MH

It clearly shows that T2>T1