A) \[\frac{{{\mu }_{0}}I}{2\pi R}\left( \pi +1 \right)\]
B) \[\frac{{{\mu }_{0}}I}{2R}\]
C) \[\frac{{{\mu }_{0}}I}{2\pi R}\left( \pi +\frac{1}{\sqrt{2}} \right)\]
D) \[\frac{{{\mu }_{0}}I}{2\pi R}\left( \pi -\frac{1}{\sqrt{2}} \right)\]
Correct Answer: C
Solution :
\[{{B}_{0}}=\left( {{{\vec{B}}}_{0}} \right)_{1}^{\otimes }+\left( {{{\vec{B}}}_{0}} \right)_{2}^{\odot }+\left( {{{\vec{B}}}_{0}} \right)_{3}^{\odot }+\left( {{{\vec{B}}}_{0}} \right)_{4}^{\odot }\] \[=-\frac{{{\mu }_{0}}I}{4\pi R}\left( 1-\frac{1}{\sqrt{2}} \right)+\frac{{{\mu }_{0}}I}{2R}+\frac{{{\mu }_{0}}I}{4\pi R}\left( 1+\frac{1}{\sqrt{2}} \right)\] \[\vec{B}_{0}^{\odot }=\frac{{{\mu }_{0}}I}{2\pi R}{{\left( \pi +\frac{1}{\sqrt{2}} \right)}^{\odot }}\]You need to login to perform this action.
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