A) \[0.8\times {{10}^{-17}}N\]
B) \[0.8\times {{10}^{-16}}N\]
C) \[1.6\times {{10}^{-17}}N\]
D) \[1.6\times {{10}^{-16}}N\]
Correct Answer: C
Solution :
\[B=\frac{{{\mu }_{0}}}{4\pi }\cdot \frac{2i}{r}\] \[={{10}^{-7}}\times \frac{2\times 10}{0.1}\] \[=2\times {{10}^{-5}}wb/{{m}^{2}}\] \[F=gvB\sin \theta \] \[=(1.6\times {{10}^{-19}})(5\times {{10}^{6}})(2\times {{10}^{-5}})\sin {{90}^{o}}\] \[=1.6\times {{10}^{-17}}\]newton.You need to login to perform this action.
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