A conducting rod of length 0.32 m slides rightward on frictionless horizontal rails joined by a resistor on their left. Total circuit resistance is 0.6 Ω. A uniform field 0.44 T points into the page. A constant force 0.03 N pulls the rod right. Neglect self-inductance and rail/rod resistance beyond the stated total. Find the terminal speed and current, give current direction around the circuit, and show that mechanical input power equals resistor heating at terminal speed.

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Terminal speed 0.908 m s⁻¹; current 0.2131 A counterclockwise; input and heating power both 0.02724 W. As the rod moves right, loop area and into-page flux increase. Lenz's law requires an out-of-page induced field, so the induced conventional current is counterclockwise, upward through the moving rod. The motional emf is BLv, giving I = BLv/R. The field exerts a leftward force ILB = B²L²v/R on the rod, opposing its motion. At terminal speed this magnetic drag balances the applied force: F = B²L²v/R. Hence v_t = FR/(B²L²) = 0.908 m s⁻¹. The terminal current is F/(BL) = 0.2131 A. Mechanical power is Fv_t = 0.02724 W, while electrical heating is I²R = 0.02724 W. The equal powers check energy conservation: with no change in kinetic energy and negligible inductive storage, all mechanical input becomes thermal energy. At speeds above the terminal value the increased magnetic drag would slow the rod.

Terminal motion of a conducting rod · set 2

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