So the stress balance of wireframe abcd F=mgsinα+FA?
Ab side enters the magnetic field to cut the magnetic induction line, resulting in electromotive force e = bl1v.
Inductive current I = er = bl 1vr.
Amplitude force FA=BIl 1
F=mgsinα+B2l2 1vR
Substitute the data to get v=2m/s,
Therefore, when the wire frame enters the magnetic field, the speed of uniform motion V = 2m/s. 。
(2) When the wireframe abcd enters the magnetic field, it will do uniform acceleration linear motion;
In the process of entering the magnetic field, do uniform linear motion;
After entering the magnetic field, it moves to gh line, and still does uniform acceleration linear motion.
Before the wire frame enters the magnetic field, the wire frame is only subjected to the tensile force f of the thin wire, the supporting force of the inclined plane and the gravity of the wire frame.
From Newton's second law: F-mgsinα=ma
Acceleration of wireframe before entering magnetic field: a = f? mgsinαm=5m/s2
The moving time of the wire frame before entering the magnetic field is t1= va = 25s = 0.4s.
Time of uniform motion in the process of entering the magnetic field: T2 = L2v = 0.62s = 0.3s
After the wire frame completely enters the magnetic field, the stress on the wire frame is the same as that before entering the magnetic field.
So the acceleration at this stage is still: a=5m/s2.
s？ l2=vt3+ 12at23
Solution: T3 =1s.
Therefore, the time taken for the ab edge to reach the gh line from standstill is: t = t 1+t2+t3 = 1.7s?
(3)ε=△(BS)△t=0.5×0.62. 1？ 0.9 = 0.25 volts
According to problem (2), it takes 1.7s for ab edge to move from rest to gh line, so the time to generate heat due to magnetic field change is:
t4= 1.7s-0.9s=0.8s
q 1 =ε2t4R = 0.252×0.80 . 1J = 0.5J
Joule heat q = fal2+q1= (MGS in f-θ) L2+q1= 3.5j.
Therefore, the joule heat generated in the whole process is 3.5 J.