A) \[\frac{{{10}^{-19}}}{{{\mu }_{0}}}\]
B) \[{{10}^{-19}}\,{{\mu }_{0}}\]
C) \[2\,\,\times \,\,{{10}^{-10}}\,{{\mu }_{0}}\]
D) \[\frac{2\,\,\times \,\,{{10}^{-10}}}{{{\mu }_{0}}}\,\]
Correct Answer: B
Solution :
\[i=\frac{q}{T}=\frac{2\times 1.6\times {{10}^{-19}}}{2}=1.6\times {{10}^{-19}}\,A\] \[\therefore \,\,\,\,\,\,B=\frac{{{\mu }_{0}}i}{2r}=\frac{{{\mu }_{0}}\times 1.6\times {{10}^{-19}}}{2\times 0.8}={{\mu }_{o}}\times {{10}^{-19}}\]
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