A) \[2\times {{10}^{-5}}\,m/s\]
B) \[4\times {{10}^{-5}}\,m/s\]
C) \[6\times {{10}^{-5}}\,m/s\]
D) \[6\times {{10}^{-5}}\,m/s\]
Correct Answer: B
Solution :
Given \[\theta ={{23}^{o}},\,B=2.6\,\,m\,T=2.6\times {{10}^{-6}}T\] and \[F=6.5\times {{10}^{-17}}N\] We know \[F=qVB\sin \theta \] \[6.5\times {{10}^{-17}}=1.6\times {{10}^{-19}}\times v\times 2.6\] \[\times {{10}^{-6}}\times \sin {{23}^{o}}\] \[v=\frac{6.5\times {{10}^{-17}}}{2.6\times {{10}^{-6}}\times 1.6\times {{10}^{-19}}\times 0.39}\] \[v=4\times {{10}^{5}}m/s\]
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