The magnetic field at the centre $O$ of the arc shown in the figure is

• A

  $2 I(\sqrt{2}+\pi) \times \frac{10^{-7}}{r}$
• Correct answer
  B

  $2 I\left(\sqrt{2}+\frac{\pi}{4}\right) \times \frac{10^{-7}}{r}$
• C

  $I(\sqrt{2}+\pi) \times \frac{10^{-7}}{r}$
• D

  $I\left(\sqrt{2}+\frac{\pi}{4}\right) \times \frac{10^{-7}}{r}$

Solution

Here, $a=\frac{r}{\sqrt{2}}$ Magnetic field at point $O$ due to $A B$ is $B_1=\frac{\mu_0}{4 \pi} \frac{I}{a}=\frac{\mu_0 I}{4 \pi(r / \sqrt{2})}$ Magnetic field at point $O$ due to $B C D$ is $B_2=\frac{\mu_0 I}{4 \pi r}\left(\frac{\pi}{2}\right)$ Magnetic field at point $O$ due to $D E$ is $B_3=\frac{\mu_0 I}{4 \pi a}=\frac{\mu_0 I}{4 \pi(r / \sqrt{2})}$ Resultant magnetic field at point $O$ is $\begin{aligned} B & =B_1+B_2+B_3 \\ & =\frac{\mu_0 I}{4 \pi(r / \sqrt{2})}+\frac{\mu_0 I}{4 \pi r}\left(\frac{\pi}{2}\right)+\frac{\mu_0 I}{4 \pi(r / \sqrt{2})} \\ & =\frac{\mu_0 I}{4 \pi r}\left(\sqrt{2}+\frac{\pi}{2}+\sqrt{2}\right)=\frac{\mu_0 I}{4 \pi r}\left(2 \sqrt{2}+\frac{\pi}{2}\right) \\ & =\frac{\mu_0 I I}{4 \pi r}\left(\sqrt{2}+\frac{\pi}{4}\right) \\ & =\frac{10^{-7} \times 2 I}{r}\left(\sqrt{2}+\frac{\pi}{4}\right) \quad\left[\because \frac{\mu_0}{4 \pi}=10^{-7}\right]\end{aligned}$

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