COIL is \[M=NIA=16\times 0.75\times \pi \times {{(0.1)}^{2}}=0.377\,A{{m}^{2}}\] If K be the moment of INERTIA of the coil about its axis of rotation, then its period of oscillation in a magnetic FIELD B is given by \[T=2\pi \sqrt{\frac{K}{MB}}\] or its frequency v is\[=\frac{1}{T}=\frac{1}{2\pi }\sqrt{\frac{MB}{K}}\] This gives \[K=\frac{MB}{4{{\pi }^{2}}{{v}^{2}}}\] Given that \[B=5.0\times {{10}^{-2}}T\], \[M=0.377\,A-{{m}^{2}}\]and \[v=2{{s}^{-1}}\] \[\THEREFORE \,\,\,\,\,\,\,K=\frac{0.377\times 5.0\times {{10}^{-2}}}{4\times {{(3.14)}^{2}}\times {{(2)}^{2}}}=1.2\times {{10}^{-4}}\,kg{{m}^{2}}\]

"> COIL is \[M=NIA=16\times 0.75\times \pi \times {{(0.1)}^{2}}=0.377\,A{{m}^{2}}\] If K be the moment of INERTIA of the coil about its axis of rotation, then its period of oscillation in a magnetic FIELD B is given by \[T=2\pi \sqrt{\frac{K}{MB}}\] or its frequency v is\[=\frac{1}{T}=\frac{1}{2\pi }\sqrt{\frac{MB}{K}}\] This gives \[K=\frac{MB}{4{{\pi }^{2}}{{v}^{2}}}\] Given that \[B=5.0\times {{10}^{-2}}T\], \[M=0.377\,A-{{m}^{2}}\]and \[v=2{{s}^{-1}}\] \[\THEREFORE \,\,\,\,\,\,\,K=\frac{0.377\times 5.0\times {{10}^{-2}}}{4\times {{(3.14)}^{2}}\times {{(2)}^{2}}}=1.2\times {{10}^{-4}}\,kg{{m}^{2}}\]

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A circular coil of 16 turns and radius 10cm carries a current of 0.75 A and rest with its plane normal to an external magnetic field of \[5.0\times {{10}^{-2}}T\]. The coil is free to rotate about its stable equilibrium position with a frequency of \[2.0\,{{s}^{-1}}\] Compute the moment of inertia of the coil about its axis of rotation.                 

Joint Entrance Exam - Main (JEE Main) Physics in Joint Entrance Exam - Main (JEE Main) . 9 months ago

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[b] The magnetic moment of the COIL is \[M=NIA=16\times 0.75\times \pi \times {{(0.1)}^{2}}=0.377\,A{{m}^{2}}\] If K be the moment of INERTIA of the coil about its axis of rotation, then its period of oscillation in a magnetic FIELD B is given by \[T=2\pi \sqrt{\frac{K}{MB}}\] or its frequency v is\[=\frac{1}{T}=\frac{1}{2\pi }\sqrt{\frac{MB}{K}}\] This gives \[K=\frac{MB}{4{{\pi }^{2}}{{v}^{2}}}\] Given that \[B=5.0\times {{10}^{-2}}T\], \[M=0.377\,A-{{m}^{2}}\]and \[v=2{{s}^{-1}}\] \[\THEREFORE \,\,\,\,\,\,\,K=\frac{0.377\times 5.0\times {{10}^{-2}}}{4\times {{(3.14)}^{2}}\times {{(2)}^{2}}}=1.2\times {{10}^{-4}}\,kg{{m}^{2}}\]

Posted on 26 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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