Physic Labs
Step 1 of 4: Mechanical–electrical power balance
Pđaˆˋu vaˋo=F⃗⋅v⃗,Pthoaˊt=I2R=εIP_{\text{đầu vào}} = \vec F \cdot \vec v,\qquad P_{\text{thoát}} = I^2R = \varepsilon I
Analysis

Consider a magnet pushed into a closed conducting loop (or a rod sliding on rails). To push the magnet at speed vv we need a force FF, mechanical power FvFv. In the loop of resistance RR an induced current II flows, dissipating power I2RI^2R. Energy conservation demands that every joule released as heat must be paid for by mechanical work put in.