As shown in the figure, the smooth U-shaped metal guide rail MNN ′ m ′ is horizontally fixed in the vertical uniform magnetic field, the magnetic induction intensity is B, the width of the guide rail is l, and its length is long enough. A resistance of R is connected between M ′ and m, and the rest of the resistance is ignored. A metal bar AB with mass of M and resistance of R can just be placed on the guide rail and has good contact with the guide rail The results are as follows: (1) at the beginning of motion, what is the instantaneous current I in the rod and the instantaneous voltage u at both ends of the rod? (2) When the velocity of the rod decreases from v0 to V010, what is the Joule heat Q produced in the rod?

As shown in the figure, the smooth U-shaped metal guide rail MNN ′ m ′ is horizontally fixed in the vertical uniform magnetic field, the magnetic induction intensity is B, the width of the guide rail is l, and its length is long enough. A resistance of R is connected between M ′ and m, and the rest of the resistance is ignored. A metal bar AB with mass of M and resistance of R can just be placed on the guide rail and has good contact with the guide rail The results are as follows: (1) at the beginning of motion, what is the instantaneous current I in the rod and the instantaneous voltage u at both ends of the rod? (2) When the velocity of the rod decreases from v0 to V010, what is the Joule heat Q produced in the rod?

(1) At the beginning of the movement, the induced electromotive force in the rod is: e = blv0, the instantaneous current in the rod: I = E2R = blv02r, the instantaneous voltage at both ends of the rod: u = IR = 12blv0 (2) known by the law of energy conservation, the Joule heat generated by the closed circuit in this process: Q total = 12mv02-12m (110v0) 2 = 99200mv02

The U-shaped metal guide rail, which is fixed horizontally and long enough, is in the vertical upward uniform magnetic field, and the metal bar AB is placed on the guide rail,
The U-shaped metal guide rail, which is fixed horizontally and long enough, is in the vertical upward uniform magnetic field. A metal bar AB is placed on the guide rail. At the beginning, the bar moves to the right at the initial horizontal velocity V0, and finally it is stationary on the guide rail,
The work done by a ampere force on AB rod, etc
B work done by current, etc
C produces the same total heat
The momentum changes of DAB rods are equal

A. S is different
B is the same as a
C. Conceptual confusion
D is right

As shown in the figure, two smooth parallel guide rails are placed in a uniform magnetic field, the magnetic field is perpendicular to the plane of the guide rail, the metal bar AB can move freely along the guide rail, the left end of the guide rail is connected with a certain resistance R, and the resistance of the metal bar and the guide rail is ignored. Under the action of external force F parallel to the guide rail, the metal bar starts to move along the guide rail from static state. If the pulling force remains constant, after the time TL, the speed is V, plus If the power of the pulling force is kept constant, after time T2, the velocity becomes V, the acceleration is A2, and finally the uniform motion is 2V, then ()
A. tl=t2B. tl＜t2C. a2=3alD. a2=4al

In both cases, the final motion is uniform, so there are: F = bil & nbsp; = 2b2l2vr & nbsp; & nbsp; ① when the pull force is constant: F − b2l2vr = MA1 & nbsp; & nbsp; ② from the solution of ① and ②: A1 = b2l2vrm. If the power of the pull force is constant and the speed is V, the pull force is F1, then there are: P = f1v = f2.2v, so: F1 = 4b2l2v & nbsp; R, F1 − b2l2v & nbsp; R = ma2, the solution is: A2 = 3b2l2vrm, so there is A2 = 3A1, so C is correct and D is wrong; when the power of the pulling force is constant, the pulling force gradually decreases with the increase of the speed, and finally the pulling force is the minimum in the uniform motion, and the minimum value is equal to the pulling force in the first case, so when the speed reaches 2V, T1 > T2, so AB is wrong

As shown in figure a, the plane of sufficiently long smooth parallel metal guide rails Mn and PQ forms an angle of 30 ° with the horizontal plane. The distance between the two guide rails is L = 0.50m, and one end is connected with a resistance R = 1.0 Ω. The metal bar AB with mass m = 0.10kg is placed on the guide rail and perpendicular to the rail, with resistance R = 0.25 Ω. The whole device is located in a uniform magnetic field with magnetic induction intensity B = 1.0T, and the magnetic field direction is perpendicular to the plane of the guide rail and downward with T = 0 At any moment, apply an external force F parallel to the guide rail upward to make it move upward along the inclined plane from static state. The relationship between the current in the circuit and time t during the movement is shown in Fig. B. the resistance of other parts in the circuit is ignored, G is taken as 10m / S2, and the solution is as follows:
(1) At the end of 4.0 s, the instantaneous velocity of the metal bar AB and the instantaneous power of the final force F are obtained

(1) The induced electromotive force (EMF) produced by cutting magnetic induction line with conductor rod is e = BLV, which can be obtained from Ohm's law of closed circuit. The circuit current is I = E & nbsp; R + r = BLVR & nbsp; + R, which can be obtained from figure B. when t = 4S, I = 0.8A, that is BLVR & nbsp; + r = 0.8A, the solution is v = 2m / S; (2) because B, l, R and R are fixed values