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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

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The correct answer is:
$\frac{\mu_{0} I}{2 \pi R}(\pi+1)$
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 $O$

$$B_{1}=\frac{\mu_{0}}{2 \pi}\left(\frac{l}{R}\right)$$(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 $O$
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