A vibration magnetometer consists of two identical bar magnets placed one over the other such that they are perpendicular and bised each other. The time period of oscillation in a horizontal magnetic field is `2^(5//4)` sec. one of the magnets is removed and if the other magnet oscillates in the same field, then time period in seconds is :-

Options

A. `2^(1//4)`
B. `2^(1//2)`
C. 2
D. `2^(5//4)`

Answer

C

Explanation

Magnetic moment is a vector quantity . If the magnetic moments of two magnets are M each then, the net magnetic moment when the magnets are placed perpendicular to each other, is
`M_"eff"=sqrt(M^2+M^2)=Msqrt2`
and the moment of inertia is 2I.
Thus, `T=2pisqrt(I/(Msqrt(2H)))`
When one of the magnets is withdrawn, the time
`T'=2pisqrt(I(MH))`
`therefore (T')/T=sqrt(1/2^(1//2))` or `T'=T/2^(1//4) = 2^(5//4-1//4)` = 2 sec