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(C) The induced $emf$ across any conductor moving in a magnetic field is given by $\varepsilon = BvL_{eff}$,where $L_{eff}$ is the effective length of

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$\frac{vBR}{\sqrt{2}}\left(1 - \frac{1}{\sqrt{2}}\right)$

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(C) The induced $emf$ across any conductor moving in a magnetic field is given by $\varepsilon = BvL_{eff}$,where $L_{eff}$ is the effective length of the conductor perpendicular to the velocity vector $\vec v$.For the arc $PQ$,the effective length is the projection of the chord $PQ$ perpendicular to the velocity $\vec v$.The coordinates of $P$ are $(0, R)$ and the coordinates of $Q$ are $(R \sin 45^\circ, R \cos 45^\circ) = (R/\sqrt{2}, R/\sqrt{2})$.The projection of the chord $PQ$ perpendicular to the velocity (which is along the $x$-axis) is the difference in the $y$-coordinates of $P$ and $Q$.$L_{eff} = y_P - y_Q = R - \frac{R}{\sqrt{2}} = R\left(1 - \frac{1}{\sqrt{2}}\right)$.Therefore,the induced $emf$ across arc $PQ$ is $\varepsilon = BvR\left(1 - \frac{1}{\sqrt{2}}\right) = \frac{BvR}{\sqrt{2}}(\sqrt{2} - 1)$.

$A$ conducting ring of radius $R$ is placed in a uniform inward magnetic field $\vec B$ as shown. If the ring is moving with velocity $\vec v$ in its plane,the induced $emf$ across the arc $PQ$ will be:

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