Derive braking motion and its heat ledger

Choose the moving recipient

A virtual rod of length l slides on ideal rails to close a resistive loop in a uniform perpendicular field. Its motional emf magnitude is B l v. With negligible circuit self-inductance, current magnitude is B l v/R. The magnetic force on the rod has magnitude I l B and points opposite its motion by Lenz’s law. The ideal rails and rod are a supplied model, not a construction activity.

Use Newton’s law after deriving the current

Writing gamma = B² l²/R gives m dv/dt = −gamma v. Separation and the initial speed v0 yield v(t) = v0 exp(−t/tau), tau = m/gamma. Integrating once more gives x(t) − x0 = tau v0[1 − exp(−t/tau)]. No finite time brings the exact exponential speed to zero. The ideal infinite-rail model has a finite limiting displacement; it does not model friction, stops or a field boundary.

Close the energy transfer without a magnetic-work shortcut

Resistive power is I²R = gamma v². Integrating gives heat = (m v0²/2)[1 − exp(−2t/tau)], exactly the lost kinetic energy. The charge-level magnetic force q v cross B does no work along a charge’s total instantaneous velocity. Mechanical-to-electrical transfer in a moving conductor includes the coupled forces and electric fields; saying magnetic point-force work equals Joule heat would omit that distinction.

Change the geometry and ask which segments feel force

For a separate rectangular loop entering a finite +z field region from the left, growing positive flux produces clockwise induced current. Only the immersed segments experience that external magnetic field; the leading vertical segment supplies the braking force, while horizontal contributions cancel in the ideal symmetric setup. When the whole loop translates within a uniform field, flux is constant and no current is induced in the zero-self-inductance resistive model. On leaving, current reverses but the force still opposes motion. An asymmetrically immersed loop may also experience torque, so translation alone is not universal.

Worked original case

With B = 2 T, l = 1 m, R = 2 ohm, m = 2 kg and v0 = 3 m/s, gamma = 2 kg/s and tau = 1 s. At t = 2 s, v ≈ 0.406006 m/s, force ≈ −0.812012 N and displacement ≈ 2.593994 m. Heat is 9(1 − exp(−4)) ≈ 8.835159 J, leaving the correct residual kinetic energy.

Investigation

Predict the effect of doubling B at fixed resistance before replay. Derive speed, displacement and heat independently, then retain an energy residual. Sketch entering, fully immersed and leaving loop cases, identifying current direction and exactly which segments feel the external field.

Changed-case practice

Sources and boundaries

Original teaching checked 31 October 2026 against current official Unit 13 and the relevant OpenStax discussion. No source exercises or diagrams are reproduced. All records and apparatus are virtual, supplied or generated; no real wiring, switching, magnetic braking or electrical measurement is requested. Public feedback and retained work do not establish independent mastery, supervised practical participation or AP authorization.