A long straight conductor is bent into shape as shown. If it carries $1 \mathrm{~A}$ and its radius is $R$, then magnetic ficld $B$ at the centre of circular coil

• A

  $\infty$
• B

  Zero
• C

  $\frac{\mu_{0} i(\pi+1)}{2 \pi R}$
• Correct answer
  D

  $\frac{\mu_{0} i(\pi-1)}{2 \pi R}$

Solution

Magnetic field at centre $O=$ magnetic field due to reaction $A B+$ magnetic field due to section $P Q R$ $$ B_{0}=B_{A B}+B_{P Q R} $$
but $B_{A B}=\frac{\mu_{0} i}{2 \pi R}$ (upside the plane of paper) and $B_{P Q R}=\frac{\mu_{0} i}{2 R}$ (inside the plane of paper) Hence $B_{0}=\frac{\mu_{0} i}{2 R}-\frac{\mu_{0} i}{2 \pi R}$ $$ B_{0}=\frac{\mu_{0} i(\pi-1)}{2 \pi R} $$

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