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A bar magnetic with poles 25cm apart and...

A bar magnetic with poles `25cm` apart and pole strength `14*4A.m` rests with its centre on a frictionless pivot. It is held in equilibrium at `60^@` to a uniform magnetic field of induction `0*25T` by applying a force F at right angles to its axis, `10cm` from the pivot. Calculate the value of F. What will happen if the force is removed?

Text Solution

Verified by Experts

Here, `2l=25cm=25xx10^-2m`
`m=14.4Am, theta=60^@`
`B=0*25T, r=10cm=0*1m`, `F=?`
In equilibrium, torque on the magnet due to magnetic field B is balanced by torque due to external force F applied at `r=0*1m` from the centre O.
i.e., `vecMxxvecB=vecrxxvecF`
`MBsin theta=rF sin 90^@`
`F=(MBsintheta)/(r)=((mxx2l)Bsintheta)/(r)`
`=(14.4xx25xx10^-2xx0*25sin60^@)/(0*1)`
`=(14.4xx6.25)/(0*1)sqrt3/2xx10^-2=7*794N`

If force `vecF` is removed, the torque `vecMxxvecB` will become unbalnced. The magnet will execute oscillatory motion about the direction of `vecB`, on its pivot O. Further, as `sintheta!=theta`, therefore oscillations of the magnet will not be simple harmonic.
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