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(A) The magnetic force $F_m$ on the conductor $EF$ of length $l$ carrying current $i$ is given by $F_m = i(\vec{l} 	imes \vec{B})$. Since the magneti

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$Bil = mg\, 	an\, 	heta$

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(A) The magnetic force $F_m$ on the conductor $EF$ of length $l$ carrying current $i$ is given by $F_m = i(\vec{l} 	imes \vec{B})$. Since the magnetic field $B$ is directed vertically upwards and the conductor is on an inclined plane,the force $F_m$ acts horizontally. The magnitude of this force is $F_m = Bil$.For the conductor to be in equilibrium,the net force along the incline must be zero.The forces acting on the conductor are:$1$. Gravitational force $mg$ acting vertically downwards.$2$. Magnetic force $F_m = Bil$ acting horizontally.$3$. Normal reaction $N$ from the rails perpendicular to the incline.Resolving the forces along the incline:The component of gravitational force along the incline is $mg\, \sin\, 	heta$ (downwards).The component of magnetic force along the incline is $F_m\, \cos\, 	heta = Bil\, \cos\, 	heta$ (upwards).For equilibrium,$Bil\, \cos\, 	heta = mg\, \sin\, 	heta$.Therefore,$Bil = mg\, \frac{\sin\, 	heta}{\cos\, 	heta} = mg\, 	an\, 	heta$.

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