A bar magnet of magnetic moment M and moment of inertia I is freely suspended such that the magnetic axial line is in the direction of magnetic meridian. If the magnet is displaced by a very small angle (theta), the angular acceleration is (magnetic induction of earth's horizontal field= `B_(H)`)

Options

A. `(MB_(H)theta)/(I)`
B. `(IB_(H)theta)/(M)`
C. `(Mtheta)/(IB_(H)`
D. `(Itheta)/MB_(H)`

Answer

A

Explanation

The angular acceleration of a displaced bar magnet,
`tau=MB_(H)"sin"theta"`
Negative sign indicates torque is acting in opposite direction
`tau=Ialpha`
When theta is small `"sin"theta=theta`
`Ialpha=-MB_(H)theta`
`therefore alpha=(MB_(H)theta)/(I)`