Consider the inferences given below in respec | vedclass

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(B) The magnetic force on a current-carrying wire is given by $\vec{F} = I(\vec{L} 	imes \vec{B})$.$1$. For segment $AC$: The length vector $\vec{L}_

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$B$ only

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(B) The magnetic force on a current-carrying wire is given by $\vec{F} = I(\vec{L} 	imes \vec{B})$.$1$. For segment $AC$: The length vector $\vec{L}_{AC}$ is vertical (upward) and $\vec{B}$ is horizontal (right). The angle between them is $90^{\circ}$. The force is $F_{AC} = I(L \cos 30^{\circ})B \sin 90^{\circ} = I(L \frac{\sqrt{3}}{2})B = \frac{\sqrt{3}}{2} ILB$. By the right-hand rule, the force is directed into the page $(\otimes)$. Thus, statement $A$ is true.$2$. For segment $AB$: The length vector $\vec{L}_{AB}$ is at an angle of $30^{\circ}$ with the vertical. The angle between $\vec{L}_{AB}$ and $\vec{B}$ is $90^{\circ} - 30^{\circ} = 60^{\circ}$. The force is $F_{AB} = I(L)B \sin 60^{\circ} = I L B \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2} ILB$. By the right-hand rule, the force is directed out of the page $(\odot)$. Thus, statement $B$ is false.$3$. For segment $CB$: The length vector $\vec{L}_{CB}$ is parallel to $\vec{B}$ (angle $0^{\circ}$ or $180^{\circ}$), so the force $F_{CB} = 0$. $4$. Total force on a closed loop in a uniform magnetic field is always zero. Thus, statement $C$ is true.Since only statement $B$ is not true, the correct option is $B$ only.

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