A square coil of side 10 cm with 20 turns carries 12 A. Its normal makes 30° with a uniform horizontal field of 0.80 T. The magnitude of the torque on the coil is:

1. Torque: $\tau = NIAB\sin\theta$, where θ is the angle between the normal and B. 2. Area: $A = (0.10)^2 = 1.0\times10^{-2}$ m². 3. Substitute $N=20$, $I=12$, $A=10^{-2}$, $B=0.80$, $\sin30^\circ=0.5$. 4. $\tau = 20\times12\times10^{-2}\times0.80\times0.5$. 5. $\tau = 20\times12\times0.01\times0.40 = 0.96$ N m. _Source: NCERT Class 12 Physics Ch 4 "Moving Charges and Magnetism", p.158_