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

Answer:

The time period of bar magnet                

T=2πIMH−−−−−√                

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                

M1=M+2M=3M                

Misplaced &                

=2πI3MH−−−−−−√                                          

....

(i) When opposite poles of magnets are placed on same side, then net magnetic moment                

M2=2M−M=M                

(ii)  ∴T2=2πIM2H−−−−−√=2πIMH−−−−−√.... (ii)                

(iii)  From Eqs. (i) and (ii), we observe that    

T1<T2