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A bar magnet of moment of inertia I is v...

A bar magnet of moment of inertia `I` is vibrated in a magnetic field of inducton is `0.4xx10^(-4)T`. The time period period of vibration is `12` sec. The magnetic moment of the magnet is `120Am^(2)`. The moment of inertia of the magnet is ("in"kgm^(2))` approximately

A

`172.8xx10^(-4)`

B

`2.1xx10^(-2)`

C

`1.57xx10^(2)`

D

`1728xx10^(-2)`

Text Solution

Verified by Experts

The correct Answer is:
A
`"Given,M 120Am"^(2),T=12s and B =0.4xx10^(-4)T`
We know that, `T= 2pi sqrt((I)/(MB)) ("squaring both sides")`
`So, I=(T^(2)MB)/(4pi^(2)) or I=((12)^(2)xx(120)xx(0.45xx10^(-4)))/(4xx3.14xx3.14)`
`" " I=175xx10^(-4)kgm^(2)`
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