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The case that has the highest torque on the loop
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[QUOTE="VKint, post: 6005394, member: 176160"] Unfortunately, no. :) In your analysis of A, B, and C, you're only accounting for one side of each rectangle--the one labeled with an "I" in the diagram. In fact, in the formula ##| \mathbf{\tau} | = IAB \sin(\theta)##, the angle ##\theta## is meant to be taken between the current loop's normal vector and the magnetic field, *not* between the current and ##\mathbf{B}##. This is actually a consequence of the second formula you gave, ##\mathbf{\tau} = \mathbf{\mu} \times \mathbf{B}##, since the magnetic moment of a current loop is ##IA \mathbf{n}## (where ##\mathbf{n}## is the positively-oriented unit norm to the loop). So actually, ##\theta = 90^{\circ}## in each of choices A, B, and C. Given this, what should the answer be? EDIT: Your answer was correct, but your reasoning was not. Apologies for the lack of clarity. [/QUOTE]
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The case that has the highest torque on the loop
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