The magnitude of the magnetic field $(B)$ due to loop $A B C D$ at the origin $(O)$ is

• A

  zero
• Correct answer
  B

  $\frac{\mu_0(b-a)}{24 a b}$
• C

  $\frac{\mu_0 \mathrm{I}}{4 \pi}\left[\frac{b-a}{a b}\right]$
• D

  $\frac{\mu_o \mid}{4 \pi}\left[2(b-a)+\frac{\pi}{3}(a+b)\right]$

Solution

Net magnetic field due to loop $A B C D$ at $O$ is $$ \begin{aligned} & B=B_{A B}+B_{B C}+B_{C D}+B_{D A} \\ & =0+\frac{\mu_o I}{4 \pi a} \times \frac{\pi}{6}+0-\frac{\mu_o I}{4 \pi b} \times \frac{\pi}{6}=\frac{\mu_o I}{24 a}-\frac{\mu_o I}{24 b}=\frac{\mu_o I}{24 a b}(b-a) \end{aligned} $$

Hence, the correct answer is option: 'B'.