A) \[12.8\times {{10}^{-13}}N,\,1.1\times {{10}^{-4}}m\]
B) \[1.28\times {{10}^{-14}}N,\,1.1\times {{10}^{-3}}m\]
C) \[1.28\times {{10}^{-13}}N,\,1.1\times {{10}^{-3}}m\]
D) \[1.28\times {{10}^{-13}}N,\,1.1\times {{10}^{-4}}m\]
Correct Answer: D
Solution :
F = evB\[=1.6\times {{10}^{-19}}\times 4\times {{10}^{6}}\times 2\times {{10}^{-1}}\]\[=1.28\times {{10}^{-13}}N\] Also \[\frac{m{{v}^{2}}}{r}=evB\Rightarrow r=\frac{mv}{eB}\] \[\Rightarrow \ r=\frac{9\times {{10}^{-31}}\times 4\times {{10}^{6}}}{1.6\times {{10}^{-19}}\times 2\times {{10}^{-1}}}=1.1\times {{10}^{-4}}m\]
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