A) \[\frac{{{\mu }_{0}}I}{2\pi a}\]
B) \[\frac{\sqrt{2}{{\mu }_{0}}I}{2\pi a}\]
C) \[\frac{{{\mu }_{0}}I}{2\sqrt{2}\pi a}\]
D) \[\frac{{{\mu }_{0}}I}{4\sqrt{2}\pi a}\]
Correct Answer: A
Solution :
\[{{B}_{A}}={{B}_{B}}=\frac{{{\mu }_{0}}I}{2\pi \times \sqrt{2}a}\] So, \[B=\sqrt{B_{A}^{2}+B_{B}^{2}}=\frac{{{\mu }_{0}}I}{2\pi a}\]You need to login to perform this action.
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