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(D) The magnetic force on the conductor $EF$ is given by $\vec{F}_{mag} = i(\vec{l} 	imes \vec{B})$.Since the magnetic field $\vec{B}$ is directed ve

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Both $(A)$ and $(B)$

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(D) The magnetic force on the conductor $EF$ is given by $\vec{F}_{mag} = i(\vec{l} 	imes \vec{B})$.Since the magnetic field $\vec{B}$ is directed vertically upwards and the conductor $EF$ is on an incline,the angle between the length vector $\vec{l}$ (along $EF$) and the magnetic field $\vec{B}$ is $(90^\circ - 	heta)$.The magnitude of the magnetic force is $F_{mag} = i l B \sin(90^\circ - 	heta) = i l B \cos 	heta$.Using the right-hand rule,for the magnetic force to act up the inclined plane,the current $i$ must flow from $E$ to $F$.The component of the gravitational force acting down the inclined plane is $mg \sin 	heta$.For the conductor to be in equilibrium,the magnetic force must balance the component of the weight along the incline:$i l B \cos 	heta = mg \sin 	heta$Dividing both sides by $\cos 	heta$,we get:$i l B = mg 	an 	heta$.Thus,both conditions $(A)$ and $(B)$ are required for equilibrium.

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