Ok, e = △ b △ ts

Substituting the known quantity, we get e = 2× 0.5× 0.2 = 0.2v

I=ER=0.20.2= 1A

(2) Suppose PQ moves downwards at a uniform speed.

E'= BL(v 1+v2) For the movement of MN and PQ rods cutting magnetic induction lines in opposite directions at constant speed.

I′= E′R，

F'=BI'L，

Get f' = b2l2 (v 1+v2) r,

For the equilibrium state of PQ rod, there is F'=G,

Then b2l2 (v 1+v2) r = g,

Substituting the known quantity, we get12× 0.22× (5+v2) 0.1+0.1=1.2.

The result is v2 =1m/s.

Answer: (1) Keep the area of cyclic MNQP unchanged. When the magnetic induction intensity of the magnetic field decreases uniformly with a rate of change of 2T/s, the magnitude of the induced current in the loop is1a. ..

(2) Keep the magnetic induction intensity of the magnetic field constant at 1T, cut off the thin line OO', and make the metal bar MN move vertically upwards at a speed of 5m/s through external force, then the metal bar PQ will eventually move downwards at a speed of1m/s. 。