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(A) The forces acting on the rod are its weight $(mg)$ acting vertically downwards, the normal force $(N)$ perpendicular to the inclined plane, and th

Vedclass · Accepted Answer

$30$

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(A) The forces acting on the rod are its weight $(mg)$ acting vertically downwards, the normal force $(N)$ perpendicular to the inclined plane, and the magnetic force $(F_m = ILB)$ acting horizontally (since the magnetic field $B$ is vertical and the current $I$ is horizontal).To keep the rod stationary, the component of the gravitational force along the inclined plane must be balanced by the component of the magnetic force along the inclined plane.The component of the gravitational force along the incline is $mg \sin 45^{\circ}$.The magnetic force $F_m = ILB$ acts horizontally. Its component along the inclined plane is $F_m \cos 45^{\circ} = ILB \cos 45^{\circ}$.Equating these two components for equilibrium:$mg \sin 45^{\circ} = ILB \cos 45^{\circ}$Since $\sin 45^{\circ} = \cos 45^{\circ}$, we can simplify the equation:$mg = ILB$Given the linear density $\lambda = \frac{m}{L} = 0.45\,kg/m$, $g = 10\,m/s^2$, and $B = 0.15\,T$:$I = \frac{mg}{LB} = \left(\frac{m}{L}ight) \frac{g}{B}$Substituting the values:$I = \frac{0.45 	imes 10}{0.15} = \frac{4.5}{0.15} = 30\,A$Therefore, the minimum current required is $30\,A$.

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