NEET-XII-Physics
38: Electromagnetic Induction
- #21-aFind the average emf induced in the coil. (b) If the coil is a closed one (with the two ends joined together) and has a resistance of 4.00 Ω, calculate the net charge crossing a cross-section of the wire of the coil.Ans : The emf induced in the coil is given by
`` e=-\frac{N∆\varphi }{∆t}=\frac{N(\stackrel{\to }{{B}_{f}}.{\stackrel{\to }{A}}_{f}-\stackrel{\to }{{B}_{i}}.{\stackrel{\to }{A}}_{i})}{T}``
`` =\frac{NB.A(\,\mathrm{\,cos\,}{0}^{\,\mathrm{\,o\,}}-\,\mathrm{\,cos\,}60°)}{T}``
`` =\frac{50\times 2\times {10}^{-1}\times \,\mathrm{\,\pi \,}(0.02{)}^{2}}{2\times 0.1}``
`` =5\times 4\times {10}^{-5}\times \,\mathrm{\,\pi \,}``
`` =2\,\mathrm{\,\pi \,}\times {10}^{-2}\,\mathrm{\,V\,}=6.28\times {10}^{-3}\,\mathrm{\,V\,}`` (b) The current in the coil is given by
`` i=\frac{e}{R}=\frac{6.28\times {10}^{-3}}{4}``
`` =1.57\times {10}^{-3}\,\mathrm{\,A\,}``
The net charge passing through the cross section of the wire is given by
`` Q=it=1.57\times {10}^{-3}\times {10}^{-1}``
`` =1.57\times {10}^{-4}\,\mathrm{\,C\,}``
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- #21-bIf the coil is a closed one (with the two ends joined together) and has a resistance of 4.00 Ω, calculate the net charge crossing a cross-section of the wire of the coil.Ans : The current in the coil is given by
`` i=\frac{e}{R}=\frac{6.28\times {10}^{-3}}{4}``
`` =1.57\times {10}^{-3}\,\mathrm{\,A\,}``
The net charge passing through the cross section of the wire is given by
`` Q=it=1.57\times {10}^{-3}\times {10}^{-1}``
`` =1.57\times {10}^{-4}\,\mathrm{\,C\,}``
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