For the two parallel guide rails, the resistance is not included, the direction of uniform electric field is perpendicular to the paper surface and inward, the magnetic induction B = 0.5T, the conductor bar AB and CD are 0.2m, and the resistance is 0.2m 1 Ω, and the gravity is 0.1N. Now pull the conductor bar AB upward with force to make it rise at a uniform speed (the conductor bar AB and CD are in good contact with the guide rail). At this time, CD is still. When AB rises, what is the tension on AB, (please explain)

For the two parallel guide rails, the resistance is not included, the direction of uniform electric field is perpendicular to the paper surface and inward, the magnetic induction B = 0.5T, the conductor bar AB and CD are 0.2m, and the resistance is 0.2m 1 Ω, and the gravity is 0.1N. Now pull the conductor bar AB upward with force to make it rise at a uniform speed (the conductor bar AB and CD are in good contact with the guide rail). At this time, CD is still. When AB rises, what is the tension on AB, (please explain)

It's very simple. Take two conductor bars as a system, and the total force is 0. Obviously, AB helps the bar to receive the component along the inclined plane of the gravity of the two bars (equivalent to a rope, invisible field force). In this way, the ampere force can be calculated according to the force balance, and everything can be solved easily. Finally, discuss with the same table, and think about it carefully, so that you can get the harvest

As shown in the figure, PQRS in the figure is a square wire frame, which enters into the uniform magnetic field with Mn as the boundary at a constant speed to the right, and the direction of the magnetic field is the same as that of the wire frame plane
The direction of the magnetic field is perpendicular to the plane of the wire frame, and the boundary Mn is 45 ° to the edge of the wire frame

When point P passes through boundary Mn, the induced current is the largest

As shown in the figure, in the uniform magnetic field with magnetic induction intensity of B and vertical downward direction, there is a square closed wire frame with length of L on one side and resistance of R (1) When the wire frame changes from position I (wire frame plane ⊥ magnetic induction line) to position II (wire frame plane ∥ magnetic induction line), if the angular velocity is ω, calculate the average induced electromotive force in the wire frame. (2) when the wire frame changes from position I to position III, calculate the electric quantity passing through the cross section of the wire

(1) According to the definition of induced electromotive force, the average induced electromotive force can be obtained: e = n △ t = bl2t = BL2 π 2 ω = 2bl2 ω π (2) according to Ohm's law of closed circuit, the current generated in the loop is: I = er; the amount of charge passing through the coil is: q = I △ t = △ r = 2bl2r

As shown in the figure, there is a metal ring with radius r = 10cm, resistance R = 1 Ω and mass m = 1kg on the smooth horizontal plane, which slides to a bounded magnetic field at the speed v = 10m / s. The direction of uniform magnetic field is perpendicular to the paper surface and inward, B = 0.5T. From the moment the ring enters the magnetic field to the moment half of it enters the magnetic field, the ring releases 32j of heat (1) In this case, the instantaneous power of the current in the ring; (2) the acceleration of the ring

(1) From the beginning of entering the magnetic field to half of entering the magnetic field, according to the law of conservation of energy: 12mv 2 = q + 12mv ′ 2, substituting into the data solution, V ′ = 6m / s, the induced electromotive force: e = BLV ′ = B · 2rv ′ = 0.5 × 2 × 0.1 × 6 = 0.6V, the instantaneous power of the ring P = E2R = 0.621 = 0.36w