Work done by power supply to maintain a constant current

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The discussion revolves around the relationship between work done by a power supply and changes in potential energy in a system where a coil flips in a magnetic field. It questions whether the equation should include a negative sign, suggesting that the work done by the power supply should counteract the increase in potential energy. The original answer guide omits the negative sign, leading to confusion about the correct formulation. Participants emphasize the importance of maintaining constant current and the implications of potential energy changes. Clarification is sought on whether the negative sign is necessary to accurately represent the work-energy relationship in this context.
mymodded
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Homework Statement
A coil with N turns and area A, carrying a constant current, i, flips in an external magnetic field, ##\overrightarrow{B_{ext}}##, so that its dipole moment switches from opposition to the field to alignment with the field. During this process, induction produces a potential difference that tends to reduce the current in the coil. Calculate the work done by the coil's power supply to maintain the constant current.
Relevant Equations
$$U = NiAB\sin(\Theta)$$
Since we want the current to stay constant, the change in potential energy that's caused by the coil flipping in the magnetic field should be "undone" by the work done by the power supply, so shouldn't ##W = -\Delta U## ? the answer guide did it without the negative sign, so I'm wondering if the negative sign should be included or not and because we don't really want to add more potential energy.
 
Thread 'Correct statement about size of wire to produce larger extension'
The answer is (B) but I don't really understand why. Based on formula of Young Modulus: $$x=\frac{FL}{AE}$$ The second wire made of the same material so it means they have same Young Modulus. Larger extension means larger value of ##x## so to get larger value of ##x## we can increase ##F## and ##L## and decrease ##A## I am not sure whether there is change in ##F## for first and second wire so I will just assume ##F## does not change. It leaves (B) and (C) as possible options so why is (C)...

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