Physics · Electromagnetic Induction · NEET
EMF is work done per unit charge. The magnetic force on a charge in the rod is qvB (all three at right angles). This force pushes the charge along the whole length l, so work done W = force × distance = qvB × l = qvBl. EMF = W/q = Bvl. Each of B, v, l appears only once, so there is no square. If you ever get B v l squared, you have multiplied one quantity twice — check your setup.
Use the length of the rod itself (the side sitting across the field, at right angles to the velocity). It is the part that actually cuts field lines. The distance travelled (x) is NOT the l in the formula. In the flux method x changes with time and its rate dx/dt is the speed v; the fixed rod length l stays as l. NEET loves loops with two different sides (like 8 cm and 3 cm) to test exactly this.
No — it is Faraday's law seen from the moving-rod side. Faraday's law says ε = -dΦ/dt. For a rod on rails, flux Φ = B·l·x, and only x changes, so dΦ/dt = B·l·(dx/dt) = Blv. So ε = Blv is Faraday's law applied to a changing area. NCERT also derives the same Bvl using the Lorentz force qvB on the charges, which is why both roads reach the same result.
A lone rod moving through B already develops an EMF across its ends — charges pile up until the electric force balances the magnetic push. But current flows only if the rod is part of a closed circuit (rails plus resistor). So: EMF exists on an isolated rod; current needs a closed path. The NEET 2026 loop question asks for EMF across a small cut, which is the open-circuit EMF = Bvl.
They are two views of one physics. The Lorentz force view watches charges inside the rod feel qvB and do work qvBl. The flux view watches the circuit's area shrink or grow and uses ε = -dΦ/dt. Because the rod's motion is what changes the area, dΦ/dt works out to Blv. Same v, same B, same l — so both must agree. NCERT presents both to build confidence, not to give two answers.
A rectangular wire loop of sides 8 cm and 3 cm with a small cut is moving out of a region of uniform magnetic field of magnitude 0.3 T directed normal to the plane of the loop. The EMF developed across the cut, if the velocity of the loop is 2 cm/s in a direction normal to the shorter side of the loop, will be:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
ε = Bvl, where B is the magnetic field, v is the speed of the rod, and l is the rod's length. It assumes B, v, and l are mutually perpendicular. If v makes an angle θ with B, only the perpendicular part counts, so ε = Bvl sinθ.
A charge q in the moving rod feels magnetic force qvB along the rod. Moving the charge across the full length l does work W = qvBl. Since EMF is work per unit charge, ε = W/q = Bvl. This is NCERT's second derivation and matches the flux method.
Use Lenz's law or the right-hand rule. The current flows so its magnetic effect opposes the change in flux. In the rod, the magnetic force qv×B pushes positive charge to one end, making that end the positive terminal, so current in the external circuit flows from the low to high potential end inside the rod.
No. The field B is steady and uniform. The flux changes only because the circuit's area changes as the rod moves. This is the key difference from static EMF, where the conductor is still and B itself changes with time.
Because the EMF is produced by the motion of the conductor through a steady magnetic field, not by a changing field. The moving charges feel a magnetic force, which acts like a battery inside the rod.
