$A$ conducting circular loop of copper is placed as shown in the figure. The cross-sectional area of part $abc$ is $A$ and that of part $adc$ is $A/3$. The magnetic field at point $O$ is:

  • A
    $\frac{\mu_0 I}{8 R} \otimes$
  • B
    $\frac{3 \mu_0 I}{11 R} \otimes$
  • C
    $\frac{\mu_0 I}{16 R} \odot$
  • D
    Zero

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