#### 100 turn rectangular coil ABCD (in X-Y plane) is hung from one arm of a balance (shown in figure). A mass 500 g is added to the other arm to balance the weight of the coil. A current 4.9 A passes through the coil and a constant magnetic field of 0.2 T acting inward (in x-z plane) is switched on such that only arm CD of length 1 cm lies in the field. How much additional mass m must be added to regain the balance?

The concept of the magnetic force on a conductor placed in the region of an external uniform magnetic field is used. The external magnetic field causing the magnetic force on CD must balance its weight.

And the net torque present on the spring balance must be zero to attain equilibrium.

At t=0 ,the external magnetic field is off. Let us consider the sepration of each hung from the mid-point be l.

$Mgl = W_{coil}l$

$0.5gl = W_{coil}l$

$W_{coil}=0.5 \times 9.8$

By taking the moment of the about the mid-point , we get the weigh of the coil. And let 'm' be the mass which is added to regain the balance . When the magnetic field is switched on.

$Mgl +mgl=W_{coil}l+(IlB\sin 90^{\circ})l$

$Mgl =(IlB)l$

$m=\frac{BIL}{g}=\frac{0.2 \times 4.9 \times 1 \times 10^{-2}}{9.8}=10^{-3}kg =1g$

Therefore 1g of additional mass must be added to regain the balance.