How to distinguish the direction of magnetic field around the electrified straight wire
Judging by the right-hand spiral rule, the thumb direction refers to the current direction, the four fingers point to the magnetic field direction, and there is no fixed N / S pole
What is usually used to determine the direction of the magnetic field around an electrified straight wire?
The result of the formula is the cross product operation of two vectors (IDL * r). In mathematics, the cross product of two vectors (L, R) is a = L * r, the size of a is the area of the parallelogram line composed of LR, the direction is from the minimum angle of fingers to R, and the direction of thumb is the tangent direction of the point B
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- 1. How does the magnetic field around an electrified straight wire generate The correct principle,
- 2. 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?
- 3. Put a parallel metal guide rail in the uniform magnetic field, and the guide rail is connected with the large coil m, as shown in the figure. Put a wire AB on the guide rail, and the magnetic induction line is perpendicular to the plane of the guide rail. In order to make the small closed coil n surrounded by M produce clockwise induced current, the motion of the wire may be () A. Uniform right movement B. accelerated right movement C. uniform left movement D. accelerated left movement
- 4. 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)
- 5. As shown in Figure 11, half of a rectangular coil ABCD with an area of S is in a uniform magnetic field with a magnetic induction intensity of B. at this time, the magnetic flux passing through the coil is WB
- 6. \X0c a single turn rectangular coil ABCD is in a uniform magnetic field. At this time, the magnetic flux passing through the coil is 0.06wb \X0c a single turn rectangular coil ABCD is in a uniform magnetic field. At this time, the magnetic flux passing through the coil is 0.06wb. Now pull the coil out of the magnetic field from the boundary at a constant speed within 0.2S, and the magnetic flux becomes 0. Calculate the induced electromotive force in the coil. Given that the resistance of the coil is 0.2, how much current is in the coil in this process?
- 7. The square wire frame ABCD, with 10 turns and 20 cm side length, rotates uniformly around the rotation axis perpendicular to direction B in a uniform magnetic field with 0.2T magnetic induction, and the rotation speed is 120 R / min. when the wire frame turns 90 degrees from the position parallel to the magnetic field, the change of magnetic flux in the coil is -- WB, and the average change rate of magnetic flux in the coil is -- WB / s, The average induced electromotive force is -- V
- 8. A conducting rod AB is horizontally placed on two metal guide rails and is in a vertical upward uniform magnetic field. As shown in the figure, the mass of the conducting rod is 1.2kg and the length is 1m. When 3A current is applied to the conducting rod, it can slide on the guide rail at a uniform speed. If the current intensity increases to 5A, the conducting rod can obtain an acceleration of 2m / S2, and the magnetic induction intensity of the device is calculated
- 9. When the distance L of the metal guide rail is fixed horizontally, the resistance is neglected. The left end of the guide rail is connected with a resistance R. the uniform magnetic field B is perpendicular to the plane of the guide rail downward. The conductor bar with mass m, resistance R and length L is vertically placed on the guide rail to keep good contact with the guide rail. The dynamic friction factor between the conductor bar AB and the guide rail is μ (1) The direction and magnitude of the induced current in the conductor bar AB; (2) the electric power of the resistance R; (3) the magnitude of the external force F
- 10. When the smooth parallel metal guide rails are placed horizontally with a spacing of 0.5m, the electromotive force of the power supply is e = 1.5V, the internal resistance is r = 2.0 ohm, and the resistance of the metal bar is r = 2.8 ohm, which is perpendicular to the parallel guide rails, and the other resistances are not included. When the metal bar is in a uniform magnetic field with magnetic induction intensity B = 2.0T and the direction is 60 ° to the horizontal direction of water, the circuit will be connected, What is the magnitude and direction of the ampere force on the metal bar? If the mass of the metal bar is m = 5 * 10 negative quadratic kg, what is its pressure on the orbit?
- 11. How to judge the force direction of the electrified conductor in the magnetic field
- 12. An electrified wire is subjected to a force in a magnetic field, and the direction of the force follows______ Direction and direction______ It all has to do with the direction
- 13. In the uniform magnetic field with magnetic induction intensity of 4.0 * 10 ^ 2T, there is a 4cm long electrified straight wire AB perpendicular to the direction of magnetic field If the ampere force of AB is 1.0 * 10 ^ - 2n and the direction is perpendicular to the paper surface and outward, the magnitude and direction of the current in the wire are the same
- 14. A small section of electrified wire is 1 in length and 5A in current intensity. If it is placed at a certain point in the magnetic field and the magnetic field force is 0.1N, then the magnetic induction intensity at that point must be? To explain in detail, the answer is that B is greater than or equal to 2T. I think it should be less than or equal to 2T. Why?
- 15. In the uniform magnetic field with magnetic induction of 0.5T, there is an electrified wire AB which is perpendicular to the magnetic direction and 0.3m long. The current in the wire is 2a and the direction is a In the uniform magnetic field with magnetic induction of 0.5T, there is an electrified wire AB which is perpendicular to the magnetic direction and 0.3m long. The current in the wire is 2a and the direction is a to B. the magnitude and direction of Ampere force on the wire AB are calculated
- 16. The direction of Ampere force F is not only perpendicular to the magnetic field, but also perpendicular to the conducting wire, that is, f is perpendicular to the plane where Bi is, but the direction of B and I is not necessarily perpendicular
- 17. In the equatorial plane along the east-west direction to place a straight wire, through the west to east current, then this wire by the geomagnetic field action direction is what direction
- 18. A 2m long straight wire is horizontally placed along the east-west direction with the current of 1a. Try to calculate the ampere force on the wire in the geomagnetic field
- 19. The output power of a generator is 10kW, the output voltage is 500V, the total resistance of the conductor is 5 Ω, the power loss is 5% of the total power, and the voltage required by the user is 220V. The turn ratio of step-up transformer and step-down transformer is calculated
- 20. When the voltage at both ends of a resistance wire is u, the power consumed is p. if the voltage at both ends of the resistance wire is increased by one fifth, what is the original voltage?