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(C) The magnetic flux through the loop initially is $\phi_i = B \cdot A \cdot \cos(0^o) = BA$.After rotating the loop by $180^o$,the normal vector to

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$= 2AB/R$

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(C) The magnetic flux through the loop initially is $\phi_i = B \cdot A \cdot \cos(0^o) = BA$.After rotating the loop by $180^o$,the normal vector to the loop points in the opposite direction relative to the magnetic field,so the final flux is $\phi_f = B \cdot A \cdot \cos(180^o) = -BA$.The change in magnetic flux is $\Delta \phi = \phi_f - \phi_i = -BA - BA = -2BA$.According to Faraday's law,the induced emf is $\varepsilon = -\frac{d\phi}{dt}$.Since $\varepsilon = I \cdot R = R \frac{dq}{dt}$,we have $R \frac{dq}{dt} = -\frac{d\phi}{dt}$.Integrating both sides,the total charge $q$ is given by $q = \int dq = -\frac{1}{R} \int d\phi = -\frac{\Delta \phi}{R}$.Substituting the change in flux,$q = -\frac{-2BA}{R} = \frac{2BA}{R}$.

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