How does the magnetic field around an electrified straight wire generate The correct principle,

How does the magnetic field around an electrified straight wire generate The correct principle,

The moving charge generates a magnetic field, and there is a moving charge in the electrified guide line, so it generates a magnetic field around it

Mn and PQ are smooth parallel guides with a distance of L = 0.2m. The angle between the plane of the guide rail and the horizontal plane is θ = 30 ° and the guide rail is in a uniform magnetic field with magnetic induction intensity of B = 1t and direction perpendicular to the plane of the guide rail. A constant resistance with resistance of R = 2 Ω is connected at both ends of M and P of the two guide rails, and the rest resistance is ignored. A conductor bar with mass of M = 0.2kg is placed perpendicular to the guide rail and has good contact with the guide rail A force F is applied to the conductor bar parallel to the guide rail to make the conductor bar slide down the guide rail to the end from the static position ab. during the sliding process, the conductor bar is always perpendicular to the guide rail, the acceleration is a = 4m / S2, and it slides to the CD position after time t = 1s. During the process from AB to CD, the resistance heating is q = 0.1j, and G is 10m / S2 (1) (2) the work done by the force F in the process of the conductor rod sliding from AB to CD

(1) The velocity of conductor bar at CD is: v = at = 4 & nbsp; M / s, the electromotive force generated by cutting the magnetic induction line is e = BLV = 0.8V, the loop induced current is I = Er = 0.4A, the ampere force on the conductor bar at CD is f a = bil = 0.08n, according to the left-handed rule, the ampere force direction is parallel to the inclined plane upward, according to Newton's second law: mgsin θ + f-f a = ma, the solution is f = -0.12n, then the applied force on the conductor bar is 0.12n, and the direction is parallel to the guide plane (2) the distance between AB and CD: x = 12at2 = 2m, according to the law of conservation of energy: mgxsin θ + WF = q + 12mv2, the solution is WF = - 0.3j

As shown in the figure, the distance between two sufficiently long parallel and smooth metal guides Mn and PQ is l, the angle between the guide plane and the horizontal plane α = 30 ° and the resistance of the guide is ignored. The uniform magnetic field with magnetic induction intensity B is vertical to the guide plane, the metal bar AB with length L is vertical to Mn and PQ and placed on the guide rail, and it always has good contact with the guide rail. The mass of the metal bar is m and the resistance is r Connect the right end circuit, the resistance of the bulb RL = 4R, the setting resistance R1 = 2R, the resistance of the resistance box is adjusted to R2 = 12R, and the acceleration of gravity is g. now release the metal rod from static state, try to find: (1) what is the maximum speed of the metal rod sliding? (2) When the sliding distance of the metal bar is S0, the speed just reaches the maximum, and the Joule heat q? Produced by the whole circuit during the sliding process of the metal bar from static to 2s0 is calculated?

(1) When the metal rod slides at a constant speed, the maximum speed is set as VM. When the maximum speed is reached, according to the equilibrium condition, there is & nbsp; Mgsin θ = f a and f a = ILB, I = Er total, e = blvmr total = R1 + r2rlr2 + RL + r = 2R + 12R · 4r12r + 4R + r = 6R The maximum velocity of the simultaneous solution is: VM = 3mgrb2l2 (2) known from the conservation of energy, Mg · 2s0sin30 ° = q + 12mv2m solution, q = mgs0-12mv2m = mgs0-9m3g2r22b4l4 A: (1) the maximum velocity of the metal rod sliding is 3mgrb2l2. (2) the Joule heat Q generated by the whole circuit is mgs0-9m3g2r22b4l4 in the process of the metal rod sliding from static to 2s0

The magnetic induction intensity B = 1t, the width of parallel guide rail L = 1m, Mn and PQ metal rods are in the order of 1m / s
This is the exercise after class in the textbook. The magnetic induction intensity B = 1t and the width of the parallel guide rail L = 1m
Mn and PQ metal rods move to the right near the guide rail at the speed of 1m / s, r = 1 Ω, other resistances are ignored
(1) A moving wire generates an induced electromotive force, which is equivalent to a power supply
(2) What is the direction of the current passing through R? What is the magnitude?
(3) What is the direction of current passing through Mn and PQ metal rods?

1. The equivalent diagram is that two power supplies are connected in parallel and then connected in series with resistance
2. From top to bottom, current = E / r = 1a
3. From n to m, from Q to P,