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?

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?

I = E / (R + R) = 0.3125a, f = bil = 0.3125n, the direction is inclined downward, and the angle with the guide plane is 30 degrees
FN=mg+Fcos30=0.77N

There is a U-shaped guide rail nmpq on the horizontal plane, the width between them is L = 1m, and the electromotive force between M and P is e = 10V (excluding internal resistance). Now a metal bar AB with mass of M = 1kg and resistance of R = 2 Ω is placed perpendicular to the guide rail, and a large range of uniform magnetic field is added. The magnetic induction intensity is b = 0.5T, and the angle between the direction and the horizontal plane is θ = 37 ° and points to the upper right Q: (1) when AB rod is stationary, what are the supporting force and friction force on AB rod? (2) If the size and direction of B can be changed, what is the minimum size of B to make the support force of AB bar zero? What is the direction of B at this time?

(1) According to the left-hand rule, the direction of Ampere force on the rod AB is perpendicular to the rod oblique upward, and the force cross-section is: FX = FMO - fsin θ = 0 & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & n

There is a U-shaped guide rail nmpq on the horizontal plane, the width between them is L = 1m, and the electromotive force between M and P is e = 10V (excluding internal resistance). Now a metal bar AB with mass of M = 1kg and resistance of R = 2 Ω is placed perpendicular to the guide rail, and a large range of uniform magnetic field is added. The magnetic induction intensity is b = 0.5T, and the angle between the direction and the horizontal plane is θ = 37 ° and points to the upper right Q: (1) when AB rod is stationary, what are the supporting force and friction force on AB rod? (2) If the size and direction of B can be changed, what is the minimum size of B to make the support force of AB bar zero? What is the direction of B at this time?

(1) According to the left-hand rule, the direction of Ampere force on the bar AB is perpendicular to the bar oblique upward, and the force cross-section is as follows: FX = FMO - fsin θ = 0 & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; ① FY = FN + fcos θ - Mg = 0 & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp;            ②F=BIL=BERL                           ③ The formula of solution (1) and (2) is: FN = mg blecos θ r = 8N; F friction = blersin θ = 1.5n. Therefore, when AB rod is stationary, the support force and friction force on AB rod are 8N and 1.5n respectively. (2) if the support force on AB rod is zero, the static friction force must be zero= Bil = mg, that is: bminerl = mg. The minimum magnetic induction is obtained: bmin = mgrel = 2T. According to the left-handed rule, the direction of B should be horizontal to the right. Therefore, to make the support force of AB bar zero, the size of B should be at least 2T, and the direction should be horizontal to the right

As shown in figure 9-1.14, the U-shaped guide rail without resistance is placed horizontally, the width of the guide rail is L = 0.5m, the left end is connected to the power supply, the electromotive force is e = 6V, the internal resistance is r = 0.9 Ω and the variable resistance is R. a conductor bar Mn with a resistance of 0.1 Ω is placed on the guide rail perpendicular to the guide rail, and a mass of M = 20g is hoisted by a horizontal light rope through a fixed pulley. The whole device is in a vertical uniform magnetic field, and the resistance value of the variable resistance is changed, In the range of 1 Ω≤ R ≤ 5 Ω, Mn can be at rest, and the magnetic induction of uniform magnetic field can be calculated. (g = 10m / S2)
(in the figure, the magnetic field is upward and the current is from inside to outside, i.e. from m to n)

When r = 1, I = 3A, then f direction is right, F + Mg = bil
When r = 5, I = 1a, then f direction is left, F + bil = mg
So B = 2mg / L (I1 + I2) = 0.2T