$A$ $100$ turn rectangular coil $ABCD$ (in $xy$-plane) is hung from one arm of a balance (Figure). $A$ mass of $500 \text{ g}$ is added to the other arm to balance the weight of the coil. $A$ current of $4.9 \text{ A}$ passes through the coil and a constant magnetic field of $0.2 \text{ T}$ acting inward (in $xz$-plane) is switched on such that only arm $CD$ of length $1 \text{ cm}$ lies in the field. How much additional mass '$m$' must be added to regain the balance?

(10 G) The magnetic force acting on the arm $CD$ creates a torque that disturbs the balance. To regain equilibrium,we must add an additional mass $m$ to the side of the coil.
The magnetic force $F_m$ on the arm $CD$ is given by $F_m = N I L B \sin(90^\circ) = N I L B$,where $N = 100$ is the number of turns,$I = 4.9 \text{ A}$ is the current,$L = 1 \text{ cm} = 0.01 \text{ m}$ is the length of the arm,and $B = 0.2 \text{ T}$ is the magnetic field.
$F_m = 100 \times 4.9 \times 0.01 \times 0.2 = 0.098 \text{ N}$.
This force acts downwards (by Fleming's Left-Hand Rule). To balance this,we add a mass $m$ such that the weight $mg$ equals the magnetic force $F_m$.
$mg = F_m$
$m \times 9.8 = 0.098$
$m = \frac{0.098}{9.8} = 0.01 \text{ kg} = 10 \text{ g}$.
Therefore,an additional mass of $10 \text{ g}$ must be added.