Recent content by Dr. 104

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    Find Chemical Potential of Ideal Gas: Gibbs-Duhen Relation

    Homework Statement Find the chemical potential for an ideal gas as a function of temperature and pressure. Use the "Gibbs-Duhen relation". Homework Equations \mu=\frac{\partial U}{\partial N} dU=TdS-pdV+\sum\limits_{i}\mu_{i}dN_{i} U=Q+W Gibbs-Duhen relation...
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    Torque experienced by a current loop in uniform magnetic field

    Yes that is absolutely correct. and that is why the book chose that direction. However, I also was pointing out that if you chose to use q<0 which would be the actual electrons then the velocity vector will be the other direction. (because the electrons will be moving in the opposite direction...
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    Torque experienced by a current loop in uniform magnetic field

    Here's the thing. This is somewhat of a bad question because its wording includes electromagnetism concepts, but really it is just a geometry question. So to be honest, if I just looked at this as a geometry problem I would have given the same answer you had in mind originally, that it is...
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    Torque experienced by a current loop in uniform magnetic field

    Should that matter? Nothing I said really requires current to be a vector anyway. If it makes you feel better you can think of the quantity \vec{J} which is the current density, and that is a vector. Anyway, my whole argument was based on the direction the "positive" charges of the conventional...
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    Torque experienced by a current loop in uniform magnetic field

    Pretty much. If you think about it, the only relevant thing here is the moving charges. This is because of the lorentz force law: \vec{F} = q\left( \vec{E}+\vec{v}\times\vec{B}\right) So what matters here is going to be the \vec{v}\times\vec{B} and \vec{v} is determined by the direction of the...
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    Torque experienced by a current loop in uniform magnetic field

    Right, that's correct. Sorry it took me so long to see this.
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    Torque experienced by a current loop in uniform magnetic field

    This is where you went wrong. Read what you just wrote again. If you were correct then that would be your final answer for both current elements, there is no need to add \Theta again, because you are looking for the angles that each of the two current elements makes with B, which is what you...
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    Newton's second law and ice skating

    Yes that looks correct to me.
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