At this time, the ampere force on the conductor is F =BIL, I=ER+r, and E=BLv.
Get f amp = b2l2vr+r。
According to the equilibrium condition, f = f a+μ g.
V=5m/s is obtained by substituting into the solution.
(2) Let the distance that the conductor moves along the frame from rest be x 。
By e = △φ△t
q=I? △t,get
q =△φR+R = BLxR+R
X=q(R+r)BL=2×(2+2)0.8×0.5=20m。
According to the law of conservation of energy, draw a conclusion.
Fx=μmgx+Q+ 12mv2
Substitute the solution to get q =1.5j.
So the electric heat generated by the metal bar is QR =12q = 0.75j.
Answer: (1) The speed of the conductor bar is 5m/s when it moves at a uniform speed;
(2) The electric heat generated by the conductor bar is 0.75 J from the beginning to the end of uniform motion.
Teacher's final teaching summary 600 words 1
A semester passes in a hurry, and the county bureau will issue a unified test