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Question: Answered & Verified by Expert
In the loop shown, the magnetic induction at the point $\mathrm{O}$ is
PhysicsMagnetic Effects of CurrentKCETKCET 2012
Options:
  • A zero
  • B $\frac{\mu_{0} I}{8}\left(\frac{R_{1}-R_{2}}{R_{1} R_{2}}\right)$
  • C $\frac{\mu_{0} \mathrm{I}}{8}\left(\frac{\mathrm{R}_{1}+\mathrm{R}_{2}}{\mathrm{R}_{\mathrm{1}} \mathrm{R}_{2}}\right)$
  • D $\frac{\mu_{0} \mathrm{I}}{8}\left(\frac{\mathrm{R}_{1} \mathrm{R}_{2}}{\mathrm{R}_{1}+\mathrm{R}_{2}}\right)$
Solution:
2002 Upvotes Verified Answer
The correct answer is: $\frac{\mu_{0} \mathrm{I}}{8}\left(\frac{\mathrm{R}_{1}+\mathrm{R}_{2}}{\mathrm{R}_{\mathrm{1}} \mathrm{R}_{2}}\right)$
Net magnetic field at $O$
$$
\begin{aligned}
B &=B_{1}+B_{2}=\frac{-\pi / 2}{2 \pi}-\frac{\mu_{0} I}{2 R_{1}}+\frac{\pi / 2}{2 \pi} \cdot \frac{\mu_{0} I}{2 R_{2}} \\
&=\frac{\mu_{0} I}{8}\left(\frac{R_{1}+R_{2}}{R_{1} R_{2}}\right)
\end{aligned}
$$

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