An infinitely long straight conductor is bent into the shape as shown below. It carries a current of I ampere and the radius of the circular loop is $\mathrm{R}$ metre. Then, the magnitude of magnetic induction at the centre of the circular loop is

• A

  $\frac{\mu_{0} I}{2 \pi R}$
• B

  $\frac{\mu_{0} \mathrm{nI}}{2 \mathrm{R}}$
• Correct answer
  C

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

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

Solution

Magnetic field due to long wire at $O$ point
$$B_{1}=\frac{\mu_{0}}{2 \pi}\left(\frac{l}{R}\right)$$(upward) Magnetic field due to loop at $O$ point $B_{2}=\frac{\mu_{0}}{4 \pi} \cdot \frac{I \cdot 2 \pi R}{R^{2}}$ $\Rightarrow B_{2}=\frac{\mu_{0}}{2} \cdot \frac{I}{R} \quad$ (in upward direction) Resultant magnetic field at centre $O$ $B=B_{1}+B_{2}$ $\Rightarrow B=\frac{\mu_{0} I}{2 \pi \cdot R}(\pi+1) T$ Resultant magnetic field at centre $O$

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