Physics · Electromagnetic Induction · NEET
Moving the rod makes an emf ε = Blv, which drives a current I = Blv/R through the rod. But the rod now carries a current I while sitting inside the field B, so it feels a magnetic force F = BIl. By Lenz's law this force points opposite to the motion (backward), trying to stop the rod. So the very current the rod creates turns around and pushes back on it.
Start with F = BIl. Replace I with the induced current I = Blv/R. So F = B × (Blv/R) × l = B²l²v/R. Notice the force grows with speed v: the faster you push, the harder it pushes back. This is why the rod reaches a steady (terminal-like) behaviour if only gravity or a fixed pull acts.
Yes, exactly. To keep the rod at constant speed you apply force F = B²l²v/R, so your power is P = Fv = B²l²v²/R. The heat made in the resistor is I²R = (Blv/R)² × R = B²l²v²/R. They are identical. Every joule of your muscle/engine work becomes a joule of heat in the resistor — energy is conserved (Lenz's law in action).
No. The magnetic force on the moving charges is always perpendicular to their velocity, so magnetic forces never do work. The energy comes only from the external agent (your hand) that pushes the rod. The magnetic field is just the middle-man that converts your mechanical work into electrical energy and then heat.
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, with velocity 2 cm/s in a direction normal to the shorter side. The emf developed across the cut is:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
F = BIl = B²l²v/R, where B is the field, l the rod length, v its speed and R the circuit resistance. It always points opposite to the motion (Lenz's law).
At constant speed, yes exactly: P_applied = Fv = B²l²v²/R and heat = I²R = B²l²v²/R. They match because kinetic energy is unchanged and there is no energy store.
The heat I²R appears in whatever carries the resistance — usually the external resistor R. All of the mechanical work you do ends up as this Joule heat.
Higher speed gives more emf (ε = Blv), so more current (I = Blv/R), so more force (F = BIl). Force rises linearly with v, which is why a rod pulled by gravity approaches a steady speed.
Current I = Blv/R halves, the force F = B²l²v/R halves, and the heat power B²l²v²/R halves. Higher resistance means weaker braking and less heat for the same speed.
