Concepts Of Physics MCQ Edition [Volume 2]PhysicsElectromagnetic Induction
At t = 0 , a conducting wire ab of length l , resistance r , and mass m begins sliding down a smooth, vertical, thick pair of connected rails. A uniform magnetic field B is present in the space, directed perpendicular to the plane of the rails. After steady state is reached, what is the relationship between the rate of heat developed in the wire and the rate at which the gravitational potential energy decreases?
Options
- AThe rate of heat developed is twice the rate of potential energy decrease.
- BThey are exactly equal.
- CThe rate of heat developed is zero.
- DThe rate of heat developed is half the rate of potential energy decrease.
Correct answer
D. The rate of heat developed is half the rate of potential energy decrease.
Step-by-step solution
When the wire slides down with a velocity v , the induced motional emf is E = Bvl . The induced current in the circuit is I = E r = Bvl r . The upward magnetic force acting on the wire is F_m = IlB = B^2l^2v r . At steady state, the wire moves with a constant terminal velocity v_t , so the net force is zero. Equating the downward gravitational force to the upward magnetic force: mg = B^2l^2v_t r The rate of decrease of gravitational potential energy is: P_ grav = mgv_t The rate of heat developed in the wire is: P_