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Question
A bar magnet of magnetic moment m and moment of inertia I (about centre, 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 midpoint, perpendicular to length, in a magnetic field B. What would be the similar period T′ for each piece?
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Solution
If a magnet of magnetic moment m is cut into n equal parts than magnetic moment m' of all equal parts is nm' = m. So magnetic moment of each 2 parts of magnet = `m^' = m/2`.
`I = (ml^2)/12`
As the length of new magnet = `I^' = l/2`
So original time period T = `root(2pi)(I/(mB))`
If M is the mass of original magnet then the mass of each two magnets m' will be `M/2`.
So, `I = (Ml^2)/12` and `I^' = (M/2 * (1/2)^2)/12 = (Ml^2)/(8 xx 12)`
`T/T^' = root(2pi)(I/(mB))/(root(2pi)(I^'/(mB))) = sqrt(I/m * m^'/I^')`
or `T/T^' = sqrt(m/m^' * I^'/I)`
`I^'/I = ((ml^('^2))/12)/((ml^2)/12) = (m/2 * (1/2)^2)/(ml^2)`
`I^'/I = (m/2 * l^2/4)/(ml^2) = 1/8`
`m/m^' = m/(m/2) = 2/1`
∴ `T/T^' = sqrt(2/1 xx 1/8) = sqrt(1/4)`
`T/T^' = 1/2` or `T^' = T/2` sec
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