A) \[2\times {{10}^{-19}}{{\mu }_{0}}\]
B) \[{{10}^{-19}}/{{\mu }_{0}}\]
C) \[{{10}^{-19}}{{\mu }_{0}}\]
D) \[2\times {{10}^{-20}}/{{\mu }_{0}}\]
Correct Answer: C
Solution :
[c] \[B=\frac{{{\mu }_{0}}}{4\pi }\frac{2\pi i}{r}\]where \[i=\frac{2e}{t}=\frac{2\times 1.6\times {{10}^{-19}}}{2}=1.6\times {{10}^{-19}}\,A\] \[\therefore \,\,B=\frac{{{\mu }_{0}}i}{2r}=\frac{{{\mu }_{0}}\times 1.6\times {{10}^{-19}}}{2\times 0.8}={{\mu }_{0}}\times {{10}^{-19}}T\]
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