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A 100 turn rectangular coil ABCD (in XY ...

A 100 turn rectangular coil ABCD (in XY plane) is hung from one arm of a balance (figure). A mass `500g` is added to the other arm to balance the weight of the coil. A current `4*9A` passes through the coil and a constant magnetic field of `0*2T` acting inward (in xz plane) is switched on such that only arm CD of length `1cm` lies in the field. How much additional mass 'm' must be added to regain the balance?

Text Solution

Verified by Experts

When the magnetic field is off and the weight added in one pan of a balance, balances the rectangular coil in the other pan of balance. Then
`Mgl=W_(coil)l` or `W_(coil)=Mg=500xx9*8N`
When current I is passed through the coil and the magnetic field is switched on, let m be the mass added in a pan to balance the beam. Then
`Mgl + mgl=W_(coil)l+IBLsin90^@l`
or `mgl=IBLl` or `m=(IBL)/(g) :. m=(4*9xx0*2xx(1xx10^-2))/(9*8)=10^-3kg=1g`
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