MHT CET202526 Apr 2025Evening ShiftPhysicsElectromagnetic InductionActual
A long rectangular conducting loop of width ' ', mass ' m ' and resistance ' R ' is placed partly in a perpendicular magnetic field ' B '. It is pushed downwards with velocity ' V ' so that it may continue to fall freely. The velocity ' V ' is
Options
- Amg R ^2 ~B
- BB ^2 ^2 R mg
- Cmg R B ^2 ^2
- Dmg B ^2 R ^2
Correct answer
C. mg R B ^2 ^2
Step-by-step solution
The velocity V at which the loop falls freely is determined by the terminal velocity condition where the magnetic force balances the gravitational force. The induced electromotive force due to the motion in the magnetic field B is e = B V . Using Ohm’s law, the induced current becomes I = e R = B V R . The resulting upward magnetic force is F_m = B I = B^2 ^2 V R . Setting this equal to the gravitational force mg gives B^2 ^2 V R = mg . Solving for V yields V = mgR B^2 ^2 . Among the options, this corresponds to ch