A) \[\frac{{{\mu }_{0}}I}{4\pi r}\left( \frac{3\pi }{2}+1 \right)\]
B) \[\frac{{{\mu }_{0}}I}{4\pi r}\left( \frac{3\pi }{2}-1 \right)\]
C) \[\frac{{{\mu }_{0}}I}{4\pi r}\left( \frac{\pi }{2}+1 \right)\]
D) \[\frac{{{\mu }_{0}}I}{4\pi r}\left( \frac{\pi }{2}-1 \right)\]
Correct Answer: A
Solution :
\[{{B}_{A}}=0\] \[{{B}_{B}}=\frac{{{\mu }_{0}}}{4\pi }\frac{(2\pi -\pi /2)I}{\tau }\otimes \] \[=\frac{{{\mu }_{0}}}{4\pi }\frac{3\pi I}{2\,r}\] \[{{B}_{C}}=\frac{{{\mu }_{0}}I}{4\pi r}\otimes \] So, net magnetic field at the centre \[={{B}_{A}}+{{B}_{B}}+{{B}_{C}}\] \[=0+\frac{{{\mu }_{0}}}{4\pi }\frac{3\pi I}{2\,r}+\frac{{{\mu }_{0}}I}{4\pi r}=\frac{{{\mu }_{0}}}{4\pi }\frac{I}{r}\left( \frac{3\pi }{2}+1 \right)\]You need to login to perform this action.
You will be redirected in
3 sec