As shown in the figure,a uniform straight wir | vedclass

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(D) The length of the wire is $L = 30 \sqrt{3} 	ext{ cm} = 0.3 \sqrt{3} 	ext{ m}$.Since it is an equilateral triangle,the length of each side is $l

Vedclass · Accepted Answer

$0.6 	ext{ N}$

Vedclass · Answer

(D) The length of the wire is $L = 30 \sqrt{3} 	ext{ cm} = 0.3 \sqrt{3} 	ext{ m}$.Since it is an equilateral triangle,the length of each side is $l = L / 3 = 0.1 \sqrt{3} 	ext{ m}$.The magnetic field $\vec{B}$ is parallel to side $BC$. The angle between the side $AC$ and the magnetic field $\vec{B}$ is $60^\circ$ (as the triangle is equilateral).The force on a current-carrying conductor is given by $\vec{F} = I(\vec{l} 	imes \vec{B})$,so the magnitude is $F = I l B \sin 	heta$.Here,$I = 2 	ext{ A}$,$l = 0.1 \sqrt{3} 	ext{ m}$,$B = 2 	ext{ T}$,and $	heta = 60^\circ$.$F = 2 	imes (0.1 \sqrt{3}) 	imes 2 	imes \sin(60^\circ) = 0.4 \sqrt{3} 	imes (\sqrt{3} / 2) = 0.2 	imes 3 = 0.6 	ext{ N}$.

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