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(B) The magnetic force on a current-carrying wire is given by $\vec{F} = I(\vec{L} 	imes \vec{B})$, where $L$ is the length vector of the wire.Let th

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$9$

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(B) The magnetic force on a current-carrying wire is given by $\vec{F} = I(\vec{L} 	imes \vec{B})$, where $L$ is the length vector of the wire.Let the sides of the triangle be $EF = 5 \,cm$, $EG = 12 \,cm$, and $GF = 13 \,cm$.The magnetic field $\vec{B}$ is parallel to the side $GF$. Let $	heta$ be the angle between the side $EF$ and the side $GF$.From the geometry of the triangle, $\sin 	heta = \frac{EG}{GF} = \frac{12}{13}$.The force on the side $EF$ is $F_{EF} = I \cdot L_{EF} \cdot B \sin 	heta$.Given $I = 2 \,A$, $L_{EF} = 0.05 \,m$, $B = 0.75 \,T$, and $\sin 	heta = \frac{12}{13}$.$F_{EF} = 2 	imes 0.05 	imes 0.75 	imes \frac{12}{13} = 0.1 	imes 0.75 	imes \frac{12}{13} = 0.075 	imes \frac{12}{13} = \frac{0.9}{13} = \frac{9}{130} \,N$.Comparing this with $\frac{x}{130} \,N$, we get $x = 9$.

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