A rectangular conducting loop, PQRS is hanging from a thread as shown. Thread is now cut. Compute velocity with which the loop falls as function of time t. Show that if side PS is large enough the loop ultimately falls with constant speed. (m = mass of loop, r = Resistance of loop, PQ = l, PS = b)

Explanation

Strategy : At any time t, let current in loop (direction of which is given by Lnez's law) and velocity of the loop are `i` and v respectively. Forces acting on the loop are mg (downwards) and magnetic force Bil (upward).
(i) By Newton's second law,
F = ma
`rArr mg-Bil=(mdv)/(dt)`
`rArr mg-B((Bvl)/(r))l=(mdv)/(dt)`

`rArr (dt)/(m)=(dv)/(mg-(B^(2)vl^(2))/(r))`
`rArr int_(0)^(t)(dt)/(m)=int_(0)^(v)(dv)/(mg-(B^(2)vl^(2))/(r))`
`rArr(t)/(m)[(ln[mg-(B^(2)vl^(2))/(r)])/(-(B^(2)l^(2))/(r))}_(0)^(v)`
`rArr (-B^(2)l^(2))/(mr)t=ln[(mg-(B^(2)vl^(2))/(r))/(mg)]`
`rArr v=(rmg)/(B^(20l^(2))[1-e^((-B^(2)l^(2)t)/(mr))]`
(ii) Velocity variation can be shown graphically, as in the graph here,

Therefore after long-time it can be seen the loop will fall with constant speed (called terminal speed) equal to `(rmg)/(B^(2)l^(2))`.