A) \[3.2\,\,\pi \mu V\]
B) \[4.8\,\,\pi \mu V\]
C) \[0.8\,\,\pi \mu V\]
D) \[1.6\,\,\pi \mu V\]
Correct Answer: A
Solution :
| Key Idea According to Faraday's second law of electromagnetic induction the induced emf is given by rate of change of magnetic flux linked with the circuit. |
| Here, \[B=0.04\,T\] |
| and \[\frac{-\,dr}{dt}=2\,mm{{s}^{-1}}\] |
| Induced emf, \[e=\frac{-d\phi }{dt}=\frac{-BdA}{dt}=-B\frac{d(\pi {{r}^{2}})}{dt}\] |
| \[=-B\pi 2r\,\frac{dr}{dt}\] |
| Now, if \[r=2\,cm\] |
| \[e=-0.04\,\times \,\pi \,\times 2\times 2\times {{10}^{-2}}\times 2\times {{10}^{-3}}\] |
| \[=3.2\,\pi \mu V\] |
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