Solve coupled mechanical and inductive storage rather than assume instantaneous drag

Declare the coupled system and independent premises

The earlier driven-rod model neglected loop inductance. Here that approximation is deliberately removed. In a virtual uniform applied field, mass m=1 kg, constant loop inductance L=1 H and coupling B ell=1 T m are stipulated. Initial velocity is v0 and initial current is zero; no external mechanical drive acts. Resistance R is nonnegative. Geometry is idealized so the inductance remains constant as the rod moves; real rails and finite field boundaries would require a new model.

Derive the state or parameter relation

Choose current orientation so positive velocity produces positive emf. Then m vprime=−B ell I and L Iprime=B ell v−R I. With the unit parameters, eliminate I to obtain vdoubleprime+R vprime+v=0, with v(0)=v0 and vprime(0)=0. For alpha=R/2 less than one and omega=sqrt(1−alpha²), v=v0 exp(−alpha t)[cos(omega t)+alpha sin(omega t)/omega], while I=v0 exp(−alpha t)sin(omega t)/omega. At critical R=2, use the repeated-root limit v=v0(1+t)exp(−t) and I=v0 t exp(−t), not division by a zero frequency.

Challenge conservation and identification

Differentiate total kinetic plus inductive energy: d[(v²+I²)/2]/dt=−R I². Resistive conversion is initial energy minus this retained total. At R=0, velocity reverses while energy cycles between mechanical and inductive storage; setting I=v/R would be undefined and would erase the initial current state. Integrating vprime=−I yields net transported charge v0−v for these unit parameters. The macroscopic electromechanical account does not assert microscopic work by magnetic Lorentz force on a single charge. No real energized apparatus is requested.

Worked original case

For R=2 ohm and v0=2 m/s at t=2 s, velocity is 6 exp(−2)≈0.812011699 m/s and current is 4 exp(−2)≈0.541341133 A. Retained energy is 26 exp(−4)≈0.476206611 J, resistor conversion is 2−26 exp(−4)≈1.523793389 J, and net transported charge is 2−6 exp(−2)≈1.187988301 C.

Cumulative investigation

Before revealing values, compare R=0,1,2 with the same initial speed. Derive the two coupled equations, identify current continuity at release, locate a velocity reversal in the lossless case and reconcile the integrated energy loss with the resistor. State what source or geometric observation would invalidate constant L or uniform applied field.

Changed-case practice

Sources and exit evidence

Original synthesis checked 31 October 2026 against the primary component relation, inductive state and energy, RC state equations and capacitance geometry. Inter-stage derivations and evidence packets are original, not copied source experiments. The course workspace preserves first responses, source-bound reasoning, criterion critique, revision and a genuinely new frozen assignment. Actual laboratory technique, browser interaction, independent learner competence and credit require separate evidence; no physical circuit or field apparatus is requested.