$A$ square conducting frame of side $a$ and a long straight wire carrying current $I$ are placed as shown in the figure. If the frame is moved to the right with a constant velocity $V$,then the induced $emf$ in the frame is proportional to:

  • A
    $\frac{1}{x^2}$
  • B
    $\frac{1}{(2x - a)^2}$
  • C
    $\frac{1}{(2x + a)}$
  • D
    $\frac{1}{(2x - a)(2x + a)}$

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