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Magnetic Field Surrounding a Straight Conductor
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The discussion focuses on understanding the derivation of the sine function in the context of the magnetic field surrounding a straight conductor. Participants clarify the relationship between the angles in the cross product ##\vec{ds} \times \hat r## and the angle ##\theta## depicted in a diagram. The connection is established through the definition of the cross product, leading to the equation involving sine functions. The clarification helps participants grasp the geometric relationships involved in the problem. Overall, the thread emphasizes the importance of understanding angles in vector calculations related to magnetic fields.
My attempt:
Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE.
PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##.
PE of part in the tube = ##\frac{m}{l}(l - h)gh##.
Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##.
Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...
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