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?

Explanation

For equilibrium/balance, net torque should also be equal to zero
When the field is off `sumt=0` considering the separation of each hung from mid-point be l
`Mgl=W_("coil")l`
`500gl=W_("coil")l`
`W_("coil")=500xx9.8N`
Taking moment of force about mid-point, we have the weight of coil
When the magnetic field is switched on
`Mgl+mgl=W_("coil")l+IBL sin90^(@)I`
`mgl=BILl`
`m=(BIL)/(g)=(0.2xx4.9xx1xx10^(-2))/(9.8)=10^(-3)kg=1g`
Thus, 1 g of additional mass must be added tor eagin the balance.