$A$ 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:

  • A
    $\frac{vBR}{2}\left(1 + \frac{1}{\sqrt{2}}\right)$
  • B
    $\frac{vBR}{\sqrt{2}}$
  • C
    $\frac{vBR}{\sqrt{2}}\left(1 - \frac{1}{\sqrt{2}}\right)$
  • D
    $\frac{vBR}{\sqrt{2}}\left(1 + \frac{1}{\sqrt{2}}\right)$

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