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(C) The magnetic force on a current-carrying wire is given by $\vec{F} = I(\vec{L} 	imes \vec{B})$.Given:Length of each side $L = 100 \,cm = 1 \,m$Cu

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$4 \,N$,normal to the side towards the centre of the triangle

Vedclass · Answer

(C) The magnetic force on a current-carrying wire is given by $\vec{F} = I(\vec{L} 	imes \vec{B})$.Given:Length of each side $L = 100 \,cm = 1 \,m$Current $I = 2 \,A$Magnetic field $B = 2.0 \,T$The magnitude of the force on each side is $F = I L B \sin(	heta)$.Since the wire is in the plane of the paper and the magnetic field is perpendicular to the plane,the angle $	heta$ between the length vector and the magnetic field is $90^\circ$.Therefore,$F = 2 \,A 	imes 1 \,m 	imes 2.0 \,T 	imes \sin(90^\circ) = 4 \,N$.Using Fleming's Left-Hand Rule,the direction of the force on each side is perpendicular to the side and directed towards the centre of the triangle.

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