A coil of 150 turns, each of area 0.003 m², carries 1.2 A in a field of 0.2 T.

Step 1. $NBIA = 150 \times 0.2\ \text{T} \times 1.2\ \text{A} \times 0.003\ \text{m}^2 = 0.108\ \text{N m}$.

Step 2. At $0^\circ$. $\tau = 0.108\ \text{N m}$.

Step 3. At $60^\circ$. $\tau = 0.108\ \text{N m} \times 0.5 = 0.054\ \text{N m}$.

Step 4. At $90^\circ$. $\tau = 0\ \text{N m}$: the forces pull straight outward.