Question
A conducting rod of length m moves with velocity m/s perpendicular to a uniform magnetic field T. The rod slides along two frictionless rails closed by a resistor . Find (a) the induced EMF, (b) the current in the loop, (c) the force needed to keep the rod moving with constant velocity and (d) the power dissipated in . Confirm energy conservation.
Solution — Step by Step
The current-carrying rod in a magnetic field experiences . By Lenz’s law, this opposes , so the external agent must apply equal force in the direction of motion.
Power supplied by external force: W. They match — all mechanical work converts to heat in .
EMF V, A, N, W.
Why This Works
The motional-EMF setup is the cleanest demonstration of energy conservation in electromagnetism. The external agent does work; that work appears as electrical energy in the circuit, which then dissipates as heat in . No magnetic force on a moving charge ever does work — the role of is just to redirect the energy.
The trap most students fall into: they compute but forget that this is the force the agent must overcome. The agent’s force equals it in magnitude (constant velocity means zero net force) but points the other way.
Alternative Method
Skip the force calculation entirely. Use . Then N. This is the JEE Advanced shortcut.
Common Mistake
Students compute EMF correctly but then write the current as — fine — and immediately use the same expression for force: . They forget to verify with units or check the Lenz direction. If the rod were moving with the field instead of cutting it, EMF would be zero. Always confirm before using .