Q. The time period of a thin bar magnet in earth's magnetic field is T . If the magnet is cut into two equal parts perpendicular to its lenght the time period of the each part in the same will be ..??????????????plz ans it soon..:)

The time period of an oscillating magnet is given as

T = 2π (I / MB)^{1/2}

here

I is the moment of inertia = (1/2)ml^{2}

M is the magnetic moment

B is the magnetic field strength

now, as the magnet is cut in to two equal parts, the new length becomes l' = l/2 and thus, the magnetic moment will also be halved. So, the new time period will be

T' = 2π (I' / M'B)^{1/2}

or

T' = 2π ((I/4)/[(M/2)*B])

now, T'/T = 2π (I' / M'B)^{1/2 }/ 2π (I / MB)^{1/2}

or

T'/T = 1/2

or

the new time period will be

T' = 2T

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