A) \[\frac{3{{\mu }_{0}}}{2\pi }\]
B) \[\frac{{{\mu }_{0}}}{\pi }\]
C) \[\frac{\sqrt{3}{{\mu }_{0}}}{2\pi }\]
D) \[\frac{{{\mu }_{0}}}{2\pi }\]
Correct Answer: A
Solution :
When current flow in both wire in same direction then magnetic field at halfway due to P wire. \[{{\overrightarrow{B}}_{P}}=\frac{{{\mu }_{0}}{{I}_{1}}}{2\pi \frac{5}{2}}=\frac{{{\mu }_{0}}{{I}_{1}}}{\pi .5}=\frac{{{\mu }_{0}}}{\pi }\] (Where \[{{I}_{1}}=5\,A\]Amp) The direction of \[{{B}_{P}}\] is downward \[\odot \] Magnetic field at halfway due to Q wire \[{{\overrightarrow{B}}_{Q}}=\frac{{{\mu }_{0}}{{I}_{2}}}{2\pi \frac{5}{2}}=\frac{{{\mu }_{0}}}{2\pi }\] [upward \[\otimes \]] [Where \[{{I}_{2}}=2.5\] Amp.] Net magnetic field at halfway \[\overrightarrow{B}={{\overrightarrow{B}}_{P}}+{{\overrightarrow{B}}_{Q}}=\frac{{{\mu }_{0}}}{\pi }+\frac{{{\mu }_{0}}}{2\pi }=\frac{3{{\mu }_{0}}}{2\pi }\] (downward\[\odot \]) Hence net magnetic field at midpoint \[=\frac{3{{\mu }_{0}}}{2\pi }\]You need to login to perform this action.
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