A) \[2\pi ({{R}_{0}}+t)B\]
B) \[\pi \left( {{R}_{0}}+t \right)B\] clockwise
C) \[2\pi \left( {{R}_{0}}+t \right)B\] anticlockwise
D) zero
Correct Answer: C
Solution :
\[e=B\,\,\frac{dA}{dt}\] \[=B\,\,\frac{d}{dt}\,\,=\,\,(\pi {{r}^{2}})\] \[=\,\,B\,\pi \,\frac{d}{dt}\,\,\,{{({{R}_{0}}+t)}^{2}}\] \[=\,\,B\,\pi \,\frac{d}{dt}\,\,\,({{R}_{0}}^{2}+{{t}^{2}}+2\,\,{{R}_{0}}t)\] \[=\,\,B\pi \left( 2t+2{{R}_{0}} \right)\] \[=\text{ }2\pi B\left( t+{{R}_{0}} \right)\] As flux in -ve z direction is increasing hence induced current should be anticlockwise.
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