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  • A bar magnet of magnetic moment m and moment of inertia I (about center, perpendicular to length) is cut into two equal pieces, perpendicular to length.

A bar magnet of magnetic moment m and moment of inertia I (about center, perpendicular to length) is cut into two equal pieces, perpendicular to length. Let T be the period of oscillations of the original magnet about an axis through the mid-point, perpendicular to length, in a magnetic field B. What would be the similar period T’ for each piece?

Answers (1)

Time period in this type of S.H.M is

 T=2\pi \sqrt{\frac{I}{MB} }

where, T=time period

 I=moment of Inertia 

 m=mass of magnet

  B=magnetic field  

I=\frac{ml^2}{12}

 When the magnet is cut into two equal pieces, perpendicular to length the M.O.I of each piece

 of magnet about an axis perpendicular to the length passing through its centre is

 I'=\frac{m/2}{12}\left (\frac{l}{2} \right )^2\times =\frac{ml^2}{12}\times \frac{1}{}8=\frac{I}{}8

  Magnetic dipole momen

t M'=\frac{M}{}2

 Its time period of oscillation is

T'=2\pi \sqrt{\frac{I'}{M'B} }=2\pi \sqrt{\frac{\frac{I}{8}}{(\frac{M}{2})B}}=\frac{2\pi }{2}\sqrt{\frac{I}{MB}}

T'=\frac{T}{2}

Posted by

infoexpert24

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