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In the given figure, the magnetic field at $O$.
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Verified Answer
The correct answer is:
$\frac{3}{8} \frac{\mu_{0} I}{r}+\frac{\mu_{0} I}{4 \pi r}$
According to the figure, magnetic field at point is given as
$B_{\text {net }}=$ magnetic field due to straight wire $A B+$
magnetic field due to straight wire $C D+$ magnetic field due circular wire $B C$
Since, point $O$ is along the length of the wire $B A$. So, $B_{A B}=0$
$\begin{aligned}
B_{\text {net }} &=0+\frac{\mu_{0}}{4 \pi} \cdot \frac{I}{r}+\frac{270}{360}\left(\frac{\mu_{0} I}{2 r}\right) \\
&=\frac{\mu_{0} I}{4 \pi r}+\frac{3}{4} \cdot \frac{\mu_{0} I}{2 r} \\
&=\frac{\mu_{0} I}{4 \pi r}+\frac{3 \mu_{0} I}{8 r}(\text { downward })
\end{aligned}$
$B_{\text {net }}=$ magnetic field due to straight wire $A B+$
magnetic field due to straight wire $C D+$ magnetic field due circular wire $B C$
Since, point $O$ is along the length of the wire $B A$. So, $B_{A B}=0$
$\begin{aligned}
B_{\text {net }} &=0+\frac{\mu_{0}}{4 \pi} \cdot \frac{I}{r}+\frac{270}{360}\left(\frac{\mu_{0} I}{2 r}\right) \\
&=\frac{\mu_{0} I}{4 \pi r}+\frac{3}{4} \cdot \frac{\mu_{0} I}{2 r} \\
&=\frac{\mu_{0} I}{4 \pi r}+\frac{3 \mu_{0} I}{8 r}(\text { downward })
\end{aligned}$
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