A coil in the shape of an equilateral triangle of side 10cm lies in...

$τ = \overset{\to}{M} \times \overset{\to}{B}$

$= IAB$

$= \left(0 . 2\right) \times \left(\dfrac{\sqrt{3}{a}^{2}}{4}\right) \times \left(2 \times {10}^{- 2}\right) = \dfrac{0 . 2 \times \sqrt{3} \times {10}^{- 2} \times 2 \times {10}^{- 2}}{4}$

$= \sqrt{3} \times {10}^{- 5}Nm$