A) \[2.5\times {{10}^{-6}}\]coulomb
B) \[2\times {{10}^{-6}}\]coulomb
C) \[{{10}^{-6}}\]coulomb
D) \[8\times {{10}^{-6}}\]coulomb
Correct Answer: A
Solution :
\[\varphi =BA\] Þ change in flux \[d\varphi =B.dA\]= \[0.05\ (101-100)\ {{10}^{-4}}\] \[={{5.10}^{-6}}\]Wb. Now, charge \[dQ=\frac{d\varphi }{R}=\frac{5\times {{10}^{-6}}}{2}=2.5\times {{10}^{-6}}\]C.You need to login to perform this action.
You will be redirected in
3 sec