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A bar magnet of magnetic moment M and moment of inertia I is freely suspended such that the magnetic axial line 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$)

\[\begin {array} {1 1} (a)\;\frac{MB_H \theta}{I} & \quad (b)\;\frac{IB_H \theta}{M} \\ (c)\;\frac{M \theta}{IB_H} & \quad (d)\;\frac{I \theta}{MB_H} \end {array}\]

1 Answer

$ (a)\;\frac{MB_H \theta}{I}$
answered Nov 7, 2013 by pady_1

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