A conducting rod of length 0.500 m slides right at constant 4.00 m s⁻¹ on horizontal rails in a uniform 0.800 T magnetic field into the page. The total circuit resistance is 2.00 Ω. Determine the motional emf, current magnitude and direction, magnetic force on the rod, external force required, and mechanical power. Verify power conservation and explain what changes if the rod speed doubles.

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emf = 1.60 V; current = 0.800 A counterclockwise; magnetic force = 0.320 N left; external force = 0.320 N right; power = 1.28 W. Doubling speed doubles current and force and quadruples power. The swept-loop flux increases into the page. Lenz's law requires an induced field out of the page, so the current is counterclockwise. Motional emf is ε = BLv = (0.800)(0.500)(4.00) = 1.60 V. Ohm's law gives I = ε/R = 0.800 A. The rod's current direction and field give a magnetic force opposing motion. Its magnitude is F = BIL = (0.800)(0.800)(0.500) = 0.320 N left. Constant speed requires an external 0.320 N force right. Mechanical input power is Fv = (0.320)(4.00) = 1.28 W, equal to I²R = (0.800)²(2.00) = 1.28 W. Since ε and I scale with v, magnetic drag scales with v and required mechanical power Fv scales with v². Doubling speed therefore gives twice the current and force but four times the heating power.

Mechanical power and motional emf

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