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The sliding rod on two rails is the standard way exam papers test electromagnetic induction together with mechanics. Finding the emf is the easy part. The harder questions ask how the rod moves over time: what speed it settles at, how far it coasts, how much charge flows and how much heat is produced. This post works through all four with one set of numbers, so you can check each answer against energy and momentum.
What does NCERT derive, and what does the syllabus list?
The NCERT Class 12 chapter Electromagnetic Induction (section 6.6) derives the motional emf of a rod of length l moving at speed v perpendicular to a uniform field B as ε = Blv, both from the rate of change of flux and from the Lorentz force on the free charges in the rod. Its Example 6.6 extends this to a rod rotating about one end, where adding Bv dr along the rod gives ε = ½BωR², which NCERT evaluates as 157 V for a 1 m rod at 50 rev/s in 1 T.
The Syllabus for JEE (Main) 2026 lists "Faraday's law, induced emf and current, Lenz's law" under Electromagnetic Induction, and the JEE (Advanced) 2026 Information Brochure lists "Faraday's law, Lenz's law" and self and mutual inductance. Check the syllabus for your exam year.
Why is a moving rod a tricky case?
Galili, Kaplan and Lehavi (2006) point out that teaching Faraday's law for moving conductors often relies on the ambiguous phrase "area change", and that the link between the changing field and the Lorentz force is usually taught separately. They advocate an expression for Faraday's law that shows both contributions explicitly. For a rod on rails, the field is constant, so the emf comes from the Lorentz force on charges moving with the rod. Keeping that picture in mind helps when a problem combines a moving rod with a changing field.
How does the rod move under a steady pull?
Rails a distance l apart are joined by a resistor R. A rod of mass m slides on them without friction, in a field B perpendicular to the plane of the rails. Ignore the resistance of the rod and rails.
Emf and current. ε = Blv, so I = Blv / R.
Magnetic force on the rod. F_B = BIl = B²l²v / R, opposing the motion (Lenz's law).
Equation of motion. With a steady applied force F: m dv/dt = F − B²l²v / R.
Terminal speed. The rod stops accelerating when F = B²l²v / R, so v_t = FR / (B²l²).
Approach. Solving the equation gives v(t) = v_t (1 − e^(−t/τ)), with time constant τ = mR / (B²l²).
Worked example 1. Let B = 0.50 T, l = 0.40 m, R = 0.20 Ω, m = 0.10 kg, F = 0.50 N.
B²l² = 0.25 × 0.16 = 0.040 T² m².
v_t = 0.50 × 0.20 / 0.040 = 2.5 m/s.
At terminal speed: ε = 0.50 × 0.40 × 2.5 = 0.50 V, I = 0.50 / 0.20 = 2.5 A, F_B = 0.50 × 2.5 × 0.40 = 0.50 N, equal to F as expected.
Power check: F v_t = 0.50 × 2.5 = 1.25 W, and I²R = 2.5² × 0.20 = 1.25 W. All the work you do becomes heat in the resistor.
τ = 0.10 × 0.20 / 0.040 = 0.50 s. After 3τ = 1.5 s, v = 2.5 × (1 − e⁻³) ≈ 2.38 m/s, about 95% of terminal speed.
What happens if the rod is pushed once and released?
Now set F = 0 and give the rod a starting speed v₀ = 4.0 m/s, with the same B, l, R and m.
Speed. m dv/dt = −B²l²v / R gives v = v₀ e^(−t/τ), with τ = 0.50 s as before.
Distance before stopping. Writing m v dv/dx = −B²l²v / R gives dv/dx = −B²l²/(mR), so the speed falls linearly with distance and the rod stops after x = m v₀ R / (B²l²) = 0.10 × 4.0 × 0.20 / 0.040 = 2.0 m.
Heat. All the kinetic energy becomes heat: ½ × 0.10 × 4.0² = 0.80 J.
Charge. q = ΔΦ / R = Blx / R = 0.50 × 0.40 × 2.0 / 0.20 = 2.0 C.
Momentum gives a second check on the charge. The magnetic impulse is ∫BIl dt = Blq, and it must equal m v₀. So q = m v₀ / (Bl) = 0.40 / 0.20 = 2.0 C. The two routes agree.
What about vertical rails?
If the rails are vertical and gravity pulls the rod down, replace F with mg. Terminal speed is v_t = mgR / (B²l²) = 0.10 × 9.8 × 0.20 / 0.040 = 4.9 m/s, with g = 9.8 m/s². The rest of the analysis is unchanged.
Which mistakes are most common?
Leaving out the magnetic force. It is what makes a terminal speed exist.
Using B²lv instead of B²l²v. The force has l twice: once from the emf, once from BIl.
Forgetting the rod's own resistance. If the rod has resistance r, use R + r in the current.
Expecting the speed to fall linearly with time. It falls exponentially with time but linearly with distance.
Getting the direction wrong. The magnetic force always opposes the rod's motion relative to the field.
For the direction of the induced current, our guide to Lenz's law and induction problems gives a four-step method. For circuits that also contain inductors, read self and mutual inductance.
A practice routine
For each rails problem, write the equation of motion before anything else.
Solve four cases: steady pull, released with v₀, vertical rails, and a rod with its own resistance.
Check every answer twice, with power for steady motion and with energy or momentum for coasting.
The idea of flux changing as the rod moves is easier to see than to read about. The Fermi Simulation Lab has interactive simulations for ideas easier to see than read, a good place to go once you have written the equation of motion yourself.
Take one rails problem from your book and add two parts: the terminal speed and the charge that flows before it stops.
