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 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

  $\frac{\mu_0 n I}{2 R}$
• C

  $\frac{\mu_0 I}{2 \pi R}(\pi+1)$
• Correct answer
  D

  $\frac{\mu_0 I}{2 \pi R}(\pi-1)$

Solution

Magnetic field due to long wire at $O$ point $B_1=\frac{\mu_0}{2 \pi}\left(\frac{I}{R}\right)$ (upward direction) Magnetic field due to loop at $O$ point $\begin{aligned} & B_2=\frac{\mu_0}{4 \pi} \cdot \frac{I \cdot 2 \pi R}{R^2} \\ & B_2=\frac{\mu_0}{2} \cdot \frac{I}{R} \quad \text { (in upward direction) } \end{aligned}$ Hence, resultant magnetic field at centre $O$ $\begin{aligned} B & =B_1+B_2 \\ B & =\frac{\mu_0 I}{2 \pi \cdot R}(\pi+1) \mathrm{T} \end{aligned}$

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