A) 1.6cm
B) 16cm
C) 12cm
D) 1.2cm
Correct Answer: A
Solution :
Induced emf is given by \[|e|=\frac{d\phi }{dt}=B\frac{dA}{dt}\] \[(\because \phi =BA)\] \[=\pi B\frac{d}{dt}({{r}^{2}})=\pi B.2r\frac{dr}{dt}\] Given, \[\frac{dr}{dt}={{10}^{-2}}\]units, \[B={{10}^{-3}}\]units, \[e=1\mu V\] \[\therefore \] \[1\times {{10}^{-6}}=3.14\times {{10}^{-3}}\times 2\times r\times {{10}^{-2}}\] or \[r=\frac{{{10}^{-6}}}{3.14\times {{10}^{-5}}\times 2}\] =0.016 m=1.6 cm
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