PURE rolling of OA about A: the induced emf ACROSS OA will be: \[\left| \left. {\vec{e}} \right| \right.=\frac{B\omega {{(r)}^{2}}}{2}\] From LENZ law, 0 will be the negative end, while A will be the positive end. Hence \[{{V}_{0}}-{{V}_{A}}=-\frac{B\omega {{r}^{2}}}{2}\]

"> PURE rolling of OA about A: the induced emf ACROSS OA will be: \[\left| \left. {\vec{e}} \right| \right.=\frac{B\omega {{(r)}^{2}}}{2}\] From LENZ law, 0 will be the negative end, while A will be the positive end. Hence \[{{V}_{0}}-{{V}_{A}}=-\frac{B\omega {{r}^{2}}}{2}\]

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A conducting ring of radius r with a conducting spoke OA is in pure rolling on a horizontal surface in a region having a uniform magnetic field B as shown, v being the velocity of the centre of the ring. Then the potential difference \[{{V}_{0}}-{{V}_{A}}\]is:

Joint Entrance Exam - Main (JEE Main) Physics in Joint Entrance Exam - Main (JEE Main) 1 year ago

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[c] Considering PURE rolling of OA about A: the induced emf ACROSS OA will be: \[\left| \left. {\vec{e}} \right| \right.=\frac{B\omega {{(r)}^{2}}}{2}\] From LENZ law, 0 will be the negative end, while A will be the positive end. Hence \[{{V}_{0}}-{{V}_{A}}=-\frac{B\omega {{r}^{2}}}{2}\]

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Posted on 07 Aug 2024, this text provides information on Joint Entrance Exam - Main (JEE Main) related to Physics in Joint Entrance Exam - Main (JEE Main). Please note that while accuracy is prioritized, the data presented might not be entirely correct or up-to-date. This information is offered for general knowledge and informational purposes only, and should not be considered as a substitute for professional advice.

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