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Consider a current carrying wire shown in the figure. If the radius of the curved part of the wire is $R$ and the linear parts are assumed to be very long, then the magnetic induction of the field at the point $O$ is
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The correct answer is:
$\frac{\mu_0}{4 \pi} \frac{i}{R}(2+\pi)$
Magnetic field at point $O=$ Magnetic field due to wire l + Magnetic field due to semi-circle + Magnetic field due to wire 2
$$
\therefore B_{\text {at } O}=\frac{\mu_0 i}{4 \pi R}+\frac{\mu_0 i}{4 R}+\frac{\mu_0 i}{4 \pi R}=\frac{\mu_0 i}{4 \pi R}(2+\pi)
$$
$$
\therefore B_{\text {at } O}=\frac{\mu_0 i}{4 \pi R}+\frac{\mu_0 i}{4 R}+\frac{\mu_0 i}{4 \pi R}=\frac{\mu_0 i}{4 \pi R}(2+\pi)
$$
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