Magnetic field problem of steady current A square wire frame, remove one side, the remaining three sides are a, the mass of each side is m, the wire frame is in the uniform magnetic field B. take the place without side as the axis, the wire frame can rotate around this axis, and the wire frame is connected with the current I. if the wire frame is in the horizontal position, it is just balanced, then the magnetic induction intensity B should be? (the direction of B is horizontal to the left, which is perpendicular to the axis, but parallel to both sides of the wire frame, can you imagine?) The answer is: 2mg / AI But I made 3mg / AI, who can help me explain why? Is the answer wrong or am I wrong

Magnetic field problem of steady current A square wire frame, remove one side, the remaining three sides are a, the mass of each side is m, the wire frame is in the uniform magnetic field B. take the place without side as the axis, the wire frame can rotate around this axis, and the wire frame is connected with the current I. if the wire frame is in the horizontal position, it is just balanced, then the magnetic induction intensity B should be? (the direction of B is horizontal to the left, which is perpendicular to the axis, but parallel to both sides of the wire frame, can you imagine?) The answer is: 2mg / AI But I made 3mg / AI, who can help me explain why? Is the answer wrong or am I wrong

The side perpendicular to the magnetic field is subjected to two forces: the vertical upward ampere force F = IAB, the generated anti clockwise moment M1 = iaba; the vertical downward gravity mg produces clockwise moment M2 = mga, the other two sides equal to the magnetic field are not subjected to ampere force, but the vertical downward gravity 2mg produces clockwise moment m3 = 2

Magnetic field of steady current
A hexagon current carrying wire is connected with current I. The current direction is clockwise and the side length of the wire is a. the magnitude and direction of the magnetic induction intensity at the center of the current carrying wire are calculated?

In this simple way, six sides are symmetrical, and the magnetic field of each side is U0 * I / (4 * pi * b) * (cos60-cos120), where b = root sign (3) * A / 2, the magnetic field direction is inward, multiply by 6, and the result is root sign (3) * U0 * I / (a * PI)

Not only formulas, but also all knowledge points
Such as principle, experiment, all kinds of knowledge involved

Er. Even if I tell you this estimate, it's useless. The important thing is to understand SA, understand SA~
Besides, if I tell you, it will take a long time ~ (if it's mm, haha, maybe I'll think about it ~ cough, be serious ~)
Of course, if the reward score is very high, it's estimated that someone will tell you. Although I don't think it's necessary, I still have to read by myself. If you want my help,