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In the fourth edition of Introduction to Electrodynamics, a scenario is described in example 5.3 where a rectangular loop of wire supporting a mass $m$ hangs vertically with one end in a uniform magnetic field $B$, which points into the page. A current runs through the loop of wire that initially keeps the loop suspended in air (i.e., the current generates a magnetic force that exactly cancels out the gravitational force on the loop).

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I understand the example up until the point where the current is increased from the equilibrium position. The loop begins to rise vertically since the magnetic force is greater than the gravitational force, and the charges in the wire acquire an upwards component to their velocity (making their total velocity $v$):

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This results in the magnetic force having a horizontal component that opposes the direction of the flow of current:

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A better picture of this from the textbook:

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However, don't the charges in the left and right vertical wires also acquire an upwards component to their velocity (such that the total velocity $v_L$ of the charges in the left wire becomes greater than the total velocity $v_R$ of the charges in the right wire)?

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For the charges that overlap with the magnetic field region, wouldn't the difference in velocity between the charges in the left and right wires create a magnetic force that displaces the entire square loop to the left (assuming it is a rigid object)?

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Does that mean that the component of the magnetic force which acts towards the left on the charges in the top wire ($quB$ in the fourth image) comes from the leftwards displacement of the square loop itself? How exactly does $quB$ contribute towards slowing down the current, if it just moves the square loop?

Any help is appreciated, thank you!

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    $\begingroup$ The forces on the right and left vertical wires cancel out and have no net effect if we assume that the loop is a rigid body. If it is not rigid then these forces would stretch the wire horizontally. I think that the force they're talking about comes from the upper horizontal wire that is moving upwards, so the charges in it have also a vertical component of velocity $\endgroup$
    – Matteo
    Commented Apr 13, 2023 at 8:31

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Always use the law ${\vec {F_{\text{L}}}}=I{\boldsymbol {\vec {\ell_i }}}\times {\vec {B}} $ where the $\vec{\ell_i}$ ($i=1..4$) are the four wire sections each in the same direction like the current. The force on the upper wire balances the gravity. The forces on the vertical wires are directed horizontal but opposite to each other, so they cancel and there is no net force in this direction. The argument why there should be a horizontal movement at all is unclear to me. The lower wire does not experience any force.

If you increase the current (slowly, so neglecting induction) the loop will gain height till the moment the lower wire enters the fild. At this moment a downward force will occur and the loop will drop down until the moment when the lower wire leaves the field again. I think that an (nonharmonic) oscillation is created.

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  • $\begingroup$ "The argument why there should be a horizontal movement at all is unclear to me." The OP is taking it from the net vertical motion after the current is increased. If the charges move up, they all experience a magnetic force to the left. $\endgroup$ Commented Apr 13, 2023 at 14:20

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