KCET · Physics · Magnetic Effects of Current
Current \(I\) is flowing in conductor shaped as shown in the figure. The radius of the curved part is \(r\) and the length of straight portion is very large. The value of the magnetic field at the centre \(O\) will be
- A \(\frac{\mu_{0} \mathrm{I}}{4 \pi \mathrm{r}}\left(\frac{3 \pi}{2}+1\right)\)
- B \(\frac{\mu_{0} I}{4 \pi r}\left(\frac{3 \pi}{2}-1\right)\)
- C \(\frac{\mu_{0} \mathrm{I}}{4 \pi \mathrm{r}}\left(\frac{\pi}{2}+1\right)\)
- D \(\frac{\mu_{0} \mathrm{I}}{4 \pi \mathrm{r}}\left(\frac{\pi}{2}-1\right)\)
Answer & Solution
Correct Answer
(A) \(\frac{\mu_{0} \mathrm{I}}{4 \pi \mathrm{r}}\left(\frac{3 \pi}{2}+1\right)\)
Step-by-step Solution
Detailed explanation
\(\mathrm{B}_{\mathrm{A}}=0\)
\(\begin{aligned}\mathrm{B}_{\mathrm{B}} &=\frac{\mu_{0}}{4 \pi} \frac{(2 \pi}{} \\&=\frac{\mu_{0}}{4 \pi} \frac{3 \pi \mathrm{I}}{2 \mathrm{r}} \\\mathrm{B}_{\mathrm{C}} &=\frac{\mu_{0} \mathrm{I}}{4 \pi \mathrm{r}} \otimes\end{aligned}\)
So, net magnetic field at the centre
\(=\mathrm{B}_{\mathrm{A}}+\mathrm{B}_{\mathrm{B}}+\mathrm{B}_{\mathrm{C}} \)
\(=0+\frac{\mu_{0}}{4 \pi} \frac{3 \pi \mathrm{I}}{2 \mathrm{r}}+\frac{\mu_{0} \mathrm{I}}{4 \pi \mathrm{r}}=\frac{\mu_{0}}{4 \pi} \frac{\mathrm{I}}{\mathrm{r}}\left(\frac{3 \pi}{2}+1\right)\)
\(\begin{aligned}\mathrm{B}_{\mathrm{B}} &=\frac{\mu_{0}}{4 \pi} \frac{(2 \pi}{} \\&=\frac{\mu_{0}}{4 \pi} \frac{3 \pi \mathrm{I}}{2 \mathrm{r}} \\\mathrm{B}_{\mathrm{C}} &=\frac{\mu_{0} \mathrm{I}}{4 \pi \mathrm{r}} \otimes\end{aligned}\)
So, net magnetic field at the centre
\(=\mathrm{B}_{\mathrm{A}}+\mathrm{B}_{\mathrm{B}}+\mathrm{B}_{\mathrm{C}} \)
\(=0+\frac{\mu_{0}}{4 \pi} \frac{3 \pi \mathrm{I}}{2 \mathrm{r}}+\frac{\mu_{0} \mathrm{I}}{4 \pi \mathrm{r}}=\frac{\mu_{0}}{4 \pi} \frac{\mathrm{I}}{\mathrm{r}}\left(\frac{3 \pi}{2}+1\right)\)
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