A) \[q{{B}_{0}}\].
B) \[\frac{{{q}_{0}}{{B}_{0}}{{a}^{2}}}{2r}\]
C) \[\frac{q,{{B}_{0}}r}{2}\]
D) zero
Correct Answer: B
Solution :
\[\oint{\vec{E}.d\vec{l}\,=\frac{-d\phi }{dt}=\frac{-dBA}{dt}={{B}_{0}}\pi {{a}^{2}}}\] \[\Rightarrow \,\,E2\pi r\,={{B}_{0}}\pi {{a}^{2}}\,\Rightarrow \,E=\frac{{{B}_{0}}{{a}^{2}}}{2r}\] \[F=qE=\frac{q\,{{B}_{0}}\,{{a}^{2}}}{2r}\]
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