KCET · Physics · Magnetic Effects of Current
In the given figure, the magnetic field at \(O\).

- A \(\frac{3}{4} \frac{\mu_{0} I}{r}+\frac{\mu_{0} I}{4 \pi r}\)
- B \(\frac{3}{10} \frac{\mu_{0} I}{r}-\frac{\mu_{0} I}{4 \pi r}\)
- C \(\frac{3}{8} \frac{\mu_{0} I}{r}+\frac{\mu_{0} I}{4 \pi r}\)
- D \(\frac{3}{8} \frac{\mu_{0} I}{r}-\frac{\mu_{0} I}{4 \pi r}\)
Answer & Solution
Correct Answer
(C) \(\frac{3}{8} \frac{\mu_{0} I}{r}+\frac{\mu_{0} I}{4 \pi r}\)
Step-by-step Solution
Detailed explanation
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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