Magnetic Braking Rod on Rails: Terminal-Speed Misconception
A 0.50 m long conducting rod slides on two long, parallel, frictionless rails. The rails are connected at one end by a 0.90 Ω resistor; the rod itself has resistance 0.10 Ω, and the rails have negligible resistance. A uniform magnetic field B = 0.80 T points perpendicular to the plane of the rails. At t = 0 the rod is moving away from the resistor at speed v0 = 3.0 m/s, and no external force is applied. The rod has mass 0.20 kg. A student predicts that the rod will slow down and then settle at a constant terminal speed because the magnetic braking force will balance the driving force. Another student says the rod simply stops, and asks where the initial kinetic energy goes. The lab group also wants to know the initial current, the force on the rod, and how the dissipated energy is divided between the fixed resistor and the rod. Ignore self-inductance and any contact resistance. The rails are long enough that the rod does not leave them during the motion.
1. Calculate the initial motional emf, current, magnetic force on the rod, and initial acceleration. State the direction of the magnetic force relative to the velocity.
Read the worked answer
The motional emf is ε = B L v = (0.80)(0.50)(3.0) = 1.2 V. The total circuit resistance is R_tot = 0.90 + 0.10 = 1.0 Ω, so I = ε/R_tot = 1.2 A. The magnetic force magnitude is F = B I L = (0.80)(1.2)(0.50) = 0.48 N.
Newton's second law gives a = F/m = 0.48/0.20 = 2.4 m/s². By Lenz's law, the induced current produces a magnetic field opposing the increase in flux, so the magnetic force on the rod is opposite to its velocity. The rod slows down.
2. Determine the total electrical energy dissipated after the rod comes to rest, and the fraction deposited in the fixed resistor versus the rod. Explain why the split does not depend on the instantaneous speed.
Read the worked answer
The only initial energy is kinetic: K = 1/2 m v0² = 0.5(0.20)(3.0²) = 0.90 J. With no external work and negligible radiation or inductance, all of it is dissipated as joule heating.
The rod and fixed resistor are in series, so the same current I flows through both at every instant. Instantaneous power is I²R, and the ratio P_resistor/P_rod = 0.90/0.10 = 9. Integrating over time preserves the ratio because I(t) is common. Thus the fixed resistor gets 0.90×0.90 J = 0.81 J and the rod gets 0.10×0.90 J = 0.090 J.
3. Evaluate the terminal-speed prediction. Write the differential equation for v(t) and describe the long-time behavior, including the condition under which a nonzero terminal speed would occur.
Read the worked answer
With no external applied force, F = -B²L²v/R_tot, so m dv/dt = -(B²L²/R_tot)v. This gives exponential decay, v(t)=v0 exp(-t/τ), with τ = m R_tot/(B²L²) = (0.20)(1.0)/[(0.80²)(0.50²)] = 0.20/0.16 = 1.25 s. The speed approaches zero asymptotically; there is no nonzero terminal speed.
The student's prediction would require a persistent external force F_ext. Then steady state occurs when F_ext = B²L²v_T/R_tot, so v_T = F_ext R_tot/(B²L²). Here F_ext = 0, hence v_T = 0.