mgsinα=μmgcosα？ ①
The dynamic friction coefficient is obtained: μ=tanα? ②
Therefore, the dynamic friction coefficient between the guide bar and the guide rail is μ = tan α.
(2) The conductor bar decelerates under the action of Ampere force and finally stops on the guide rail. The work done by friction is equal to the work done by gravity, so the Joule heat in the whole circuit is obtained from the law of conservation of energy.
Q= 12mv20？ ③
According to the circuit knowledge, the series currents of resistors R and R are always equal, so the heat on resistor R
QR=RR+rQ=mRv202(R+r) ④
Therefore, the Joule heat generated on the resistor R during the whole movement is QR = MRV 202 (R+R).
(3) Let the speed of the conductor bar at a certain moment during deceleration be υ, take a short time △t, and a small displacement △x will occur. In the time △t, the change of magnetic flux is △φ, including:
△φ= BL△x⑤
Current in the circuit: I = er+r = △φ (r+r) △t6.
Ampere force on conductor bar: F=BIL ⑦
△t is very small, the ampere force is constant, and it is positive along the inclined plane. According to the momentum theorem,
-F△t=m△v ⑧
At the same time ⑤ ⑧ Solve:? B2L2△x(R+r)=m△v？ ⑨
The sum of the two sides of Formula 9 is ∑(B2L2△x? (R+r))？ =∑(m△v) ⑩
The total sliding distance of the conductor bar is obtained as follows: x = ∑△ x = m (r+r) B2L2 ∑△ v = MV0 (r+r) B2L2?
Therefore, the maximum distance that the guide bar moves on the guide rail is: x = mv0 (r+r) b2l2.