A coil in the shape of an equilateral triangl | vedclass

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(A) The area of an equilateral triangle with side length $a = 10 \, cm = 0.1 \, m$ is given by $A = \frac{\sqrt{3}}{4} a^2 = \frac{\sqrt{3}}{4} (0.1)^

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$3$

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(A) The area of an equilateral triangle with side length $a = 10 \, cm = 0.1 \, m$ is given by $A = \frac{\sqrt{3}}{4} a^2 = \frac{\sqrt{3}}{4} (0.1)^2 = \frac{\sqrt{3}}{4} 	imes 0.01 \, m^2$.The magnetic moment of the coil is $M = I A = 0.2 	imes \frac{\sqrt{3}}{4} 	imes 0.01 = 0.05 	imes \sqrt{3} 	imes 0.01 = 5 \sqrt{3} 	imes 10^{-4} \, Am^2$.The torque acting on the coil is $	au = M B \sin 	heta$,where $	heta$ is the angle between the normal to the coil and the magnetic field. When the plane of the coil is parallel to the magnetic field,the normal to the coil is perpendicular to the magnetic field,so $	heta = 90^{\circ}$.Thus,$	au = M B \sin 90^{\circ} = M B = (5 \sqrt{3} 	imes 10^{-4}) 	imes (20 	imes 10^{-3} \, T) = 100 \sqrt{3} 	imes 10^{-7} = \sqrt{3} 	imes 10^{-5} \, Nm$.Comparing this with $\sqrt{x} 	imes 10^{-5} \, Nm$,we get $\sqrt{x} = \sqrt{3}$,which implies $x = 3$.

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