1.

A vibration magnetometer consists of two identical bar magnets placed one over the other such thatthey are perpendicular and bised each other. The time period of oscillation in a horizontal magneticfield 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 :-

Answer»

`2^(1//4)`
`2^(1//2)`
2
`2^(5//4)`

Solution :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When one of the magnets is withdrawn, the time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


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