A horizontal rod of mass 10 , g and length 10 | vedclass

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(C) The forces acting on the rod are the gravitational force $(mg)$ acting downwards and the magnetic force $(F_m = ilB)$ acting horizontally. For the

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$1$

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(C) The forces acting on the rod are the gravitational force $(mg)$ acting downwards and the magnetic force $(F_m = ilB)$ acting horizontally. For the rod to remain stationary on the smooth inclined plane,the component of the gravitational force down the plane must be balanced by the component of the magnetic force up the plane.The component of gravitational force down the plane is $F_g = mg \sin 60^\circ$.The component of magnetic force up the plane is $F_m \cos 60^\circ = (ilB) \cos 60^\circ$.Equating these two forces for equilibrium:$mg \sin 60^\circ = ilB \cos 60^\circ$$B = \frac{mg \sin 60^\circ}{il \cos 60^\circ} = \frac{mg}{il} 	an 60^\circ$Given: $m = 10 \, g = 0.01 \, kg$,$l = 10 \, cm = 0.1 \, m$,$i = 1.73 \, A$,$g = 10 \, m/s^2$,$	an 60^\circ = \sqrt{3} \approx 1.73$.$B = \frac{0.01 	imes 10}{0.1 	imes 1.73} 	imes 1.73 = \frac{0.1}{0.173} 	imes 1.73 = 1 \, T$.

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