Final answer:
The magnitude of the magnetic field B that balances the gravitational force on the rod moving down the inclined plane with a constant velocity and with a current flowing through it is given by B = mg/Ilsinθ.
Step-by-step explanation:
The student is asking about the magnitude of the magnetic field B that acts on a conducting rod moving with a constant velocity down a smooth inclined plane while a current is flowing through it. The force due to the magnetic field that acts on the current-carrying conductor can be described by the equation F = IlBsinθ, where I is the current, l is the length of the conductor, B is the magnetic field, and θ is the angle between the direction of the current and the magnetic field. Since the rod is moving with constant velocity, the magnetic force should balance out the component of the gravitational force along the plane.
The component of the gravitational force along the incline is mg sin θ, and thus the magnetic force acting on the rod will be F = mg sin θ. As per the equation for magnetic force, the magnetic force can also be expressed as F = IlB since the current is perpendicular to the magnetic field (θ = 90 degrees, sin θ = 1). Setting these two expressions for force equal to each other and solving for B yields B = mg/Ilsin θ. Given that sin θ is in the θ, they cancel out, and the correct answer is B = mg/Ilsin θ or option D.