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(A) The magnetic force on a current-carrying wire is given by $\vec{F} = i(\vec{L} 	imes \vec{B})$,where $\vec{L}$ is the displacement vector from th

Vedclass · Accepted Answer

$24$

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(A) The magnetic force on a current-carrying wire is given by $\vec{F} = i(\vec{L} 	imes \vec{B})$,where $\vec{L}$ is the displacement vector from the start to the end of the wire segment.For the upper triangle $ACD$,the path is $C 	o A 	o D$. The displacement vector $\vec{L}_{CAD}$ is the vector from $C$ to $D$,which is the same as the segment $CD$.Similarly,for the lower triangle $CDE$,the path is $C 	o E 	o D$. The displacement vector $\vec{L}_{CED}$ is also the vector from $C$ to $D$.There is also a wire segment directly from $C$ to $D$ with current $i$ flowing from $C$ to $D$.Thus,the total force is the sum of forces on three parallel segments,each of length $l = 1 \, m$ carrying current $i = 2 \, A$ in the same direction from $C$ to $D$.$F_{net} = F_{CAD} + F_{CED} + F_{CD} = 3 	imes (i \cdot l \cdot B)$$F_{net} = 3 	imes (2 \, A 	imes 1 \, m 	imes 4 \, T) = 3 	imes 8 \, N = 24 \, N$.

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