# Rail Induction Models - sliding rods, capacitors, inclines, two rods

> Conducting rods on rails in a magnetic field: coasting, constant force or power, capacitor, incline and two-rod models with v–t, i–t, energy and analysis.

- URL: https://qilingtools.com/en/physics/rail-induction
- Category: Physics Tools (https://qilingtools.com/en/physics)
- Tags: [Physics](https://qilingtools.com/en/tags/physics), [Education](https://qilingtools.com/en/tags/education)
- Last updated: 2026-10-29
- Price: Free, no sign-up
- Privacy: Runs locally in the browser; files are never uploaded

The five rod-on-rails models that dominate electromagnetic-induction exam problems: a rod with a resistor given an initial speed (with or without friction), a rod pulled by a constant force or constant power to its terminal speed, a rod charging a capacitor at constant acceleration, a rod released on inclined rails (field normal to the incline or vertical), and two rods on equal or different-width rails. Each is simulated by numerical integration; the animation shows the current direction from the right-hand rule and arrows for the magnetic and applied forces, alongside v–t and i–t graphs and live energy bars. A worked analysis substitutes your numbers, analytic results (time constant, terminal speed, charge, heat) are checked against the simulation and energy conservation, and classic presets and random practice are included.

## How to use

1. Pick a model: rod + resistor, constant force/power, rod + capacitor, inclined rails or two rods
2. Choose a classic case or set B, L, R, m, v₀, μ, F, C and the angle
3. Press Play to watch the rod, current and forces; scrub the time slider
4. Read the v–t and i–t graphs, energy bars and the step-by-step analysis
5. Try random problems on the Practice tab, then watch the animation for the answer

## Features

- Five rail models with 18 classic presets; RK4 simulation checked against analytic formulas
- Animated current direction, magnetic force, applied force and velocity, with switchable field direction
- v–t and i–t graphs with a time cursor and live bars for kinetic energy, Joule heat, friction heat and capacitor energy
- Impulse method for charge and distance, terminal speeds, the capacitor's constant acceleration, two-rod final speeds
- Step-by-step analysis with your numbers and an energy-balance check
- Random practice: sliding distance, charge, maximum speed, acceleration, two-rod final speed and heat

## FAQ

### How far does a rod slide after being given a push?

Only the magnetic force acts, so the deceleration fades. The impulse theorem gives −BL·q = 0 − mv₀, so q = mv₀/(BL); since q = ΔΦ/R = BLx/R, the distance is x = mv₀R/(B²L²), and all the initial kinetic energy becomes Joule heat. Constant-acceleration formulas do not work here because the acceleration keeps changing.

### What is the terminal speed under a constant force?

As the rod speeds up the current and the opposing force grow until F = μmg + B²L²v/R, so v = (F − μmg)R/(B²L²). Then the input power equals electrical plus friction power: Fv = I²R + μmgv. For constant power, solve P/v = μmg + B²L²v/R.

### Why does a rod connected to a capacitor accelerate uniformly?

With no resistance the capacitor voltage always equals BLv, so the charging current is i = CBL·Δv/Δt = CBLa. Newton's law F − BiL = ma then gives a = F/(m + CB²L²), a constant, and a constant current. The work done by F goes partly into kinetic energy and partly into the capacitor's field.

### When is momentum conserved for two rods?

On rails of the same width both rods carry the same current, so the magnetic forces are equal and opposite and momentum is conserved: the common final speed is m₁v₀/(m₁ + m₂) and the lost kinetic energy is Joule heat. On rails of different widths the forces BIL₁ and BIL₂ differ, so apply the impulse theorem to each rod and use the final condition L₁v₁ = L₂v₂ (no current).

### What changes on an incline if the field is vertical instead of normal to the rails?

With the field normal to the incline the force is up the slope and v_T = mgR(sinθ − μcosθ)/(B²L²). With a vertical field only the speed component v cosθ cuts the field, the force is horizontal, only F cosθ acts along the slope and F sinθ presses into it, increasing friction, giving v_T = mgR(sinθ − μcosθ)/(B²L²cosθ(cosθ + μsinθ)).

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