A 100 turn rectangular coil ABCD (in XY plane) is hung from one arm of a balance (Fig. 4.4). A mass 500g 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 xz 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?
Initially a mass of 500g has been added to balance the weight of the coil. Hence the weight of the coil is going to be,
Mg=Wcoil
Wcoil=500 × 10-3× 9.8 N
=0.5 × 9.8 N
Now a current of4.9 Amp has been started. So the force due to the magnetic interaction is going to be,
F=i(lB) along(lxB)
=4.9 × (0.01 × 0.2) N
=9.8 × 0.001N
But there are 100 turns. So the total force will be;
Ftotal = 100 × 9.8 × 0.001N
=9.8 × 0.1 N
Let us assume the mass to be added is m.
Hence, mg=9.8 × 0.1
m=0.1kg
So, the mass to be added is 0.1kg.