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?

Question

Philoid · Accepted Answer

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=W_coil

W_coil=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;

F_total = 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.