A) \[4.8\pi \mu \,V\]
B) \[0.8\pi \mu \,V\]
C) \[1.6\pi \mu \,V\]
D) \[3.2\pi \mu \,V\]
Correct Answer: D
Solution :
[d] Induced emf in the loop is given by \[e=-B.\frac{dA}{dt}\] where A is the area of the loop. \[e=-B.\frac{d}{dt}(\pi {{r}^{2}})=-B\,\pi 2r\frac{dr}{dt}\] \[r=2cm=2\times {{10}^{-2}}m\] \[dr=2mm=2\times {{10}^{-3}}m\], \[dt=1s\] \[e=-\,0.04\times 3.14\times 2\times 2\times {{10}^{-2}}\times \frac{2\times {{10}^{-3}}}{1}V\] \[=0.32\pi \times {{10}^{-5}}V=3.2\pi \times {{10}^{-6}}V=3.2\pi \mu V\]
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