(20 14? As shown in the figure, the rectangular wireframe MNPQ with mass m has MN side length a and NP side length b; The resistance on MN side is R 1, and the resistance on PQ side is R2.

( 1)E =△φ△T = ab△B△T = ab△B△x△x△T = kab v0 B 0

I = er1+R2 = kabv0b0r1+R2.

Direction of induced current: PNMQ

(2) Ampere force on the wireframe is horizontal to the left, and Ampere force is:

F=aI△B=k2a2b2B20R 1+R2v

(3) According to Newton's second law:

f = f 1-F2 = m△v△t =-(kabb 0)2△x(r 1+R2)△t

Where m△v=-(kabB0)2△xR 1+R2.

mv0=(kabB0)2xmR 1+R2

The simultaneous solution is xm=mv0(R 1+R2)(kabB0)2.

(4) from m△v=-(kabB0)2△xR 1+R2:

m(v-v 0)=-(kabb 0)2x m2(r 1+R2)=- 12mv 0，v=v02

Q=△EK= 12mv20？ 12m(v2)2=38mv20

The heat generated by resistor R 1

QR 1 = r 1r 1+R2Q = 14mv 20

Answer: (1) The induced electromotive force in the wireframe is kabv0B0 in the position shown in the figure.

The magnitude of induced current is kabv0B0R 1+R2, and the direction is along NPQMN；;

(2) The direction of ampere force on the wireframe is horizontal to the left, and the expression of ampere force is K2A2B2B20R1+R2V;

(3) The maximum moving distance of wireframe MV0 (r1+R2) (kabb0) 2;

(4) When the wireframe moves to xm2, the joule heat generated by the resistor R 1 is 14mv20. ..