A coil in the shape of an equilateral triangl | vedclass

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(C) The area of an equilateral triangle with side $a = 2 \,cm = 2 	imes 10^{-2} \,m$ is given by $A = \frac{\sqrt{3}}{4} a^2$.Substituting the value

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$2$

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(C) The area of an equilateral triangle with side $a = 2 \,cm = 2 	imes 10^{-2} \,m$ is given by $A = \frac{\sqrt{3}}{4} a^2$.Substituting the value of $a$, $A = \frac{\sqrt{3}}{4} 	imes (2 	imes 10^{-2})^2 = \frac{\sqrt{3}}{4} 	imes 4 	imes 10^{-4} = \sqrt{3} 	imes 10^{-4} \,m^2$.The torque $	au$ acting on a current-carrying coil in a magnetic field is given by $	au = I A B \sin 	heta$.Since the magnetic field is parallel to the plane of the coil, the angle between the area vector and the magnetic field is $	heta = 90^\circ$, so $\sin 90^\circ = 1$.Thus, $	au = I A B$.Given $	au = 2 \sqrt{3} 	imes 10^{-5} \,Nm$ and $B = 100 	imes 10^{-3} \,T = 10^{-1} \,T$.Substituting the values: $2 \sqrt{3} 	imes 10^{-5} = I 	imes (\sqrt{3} 	imes 10^{-4}) 	imes 10^{-1}$.$2 \sqrt{3} 	imes 10^{-5} = I 	imes \sqrt{3} 	imes 10^{-5}$.Solving for $I$, we get $I = 2 \,A$.

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