A) \[2\,m\,m\]
B) \[1\,m\,m\]
C) \[\frac{\sqrt{3}}{2}m\,m\]
D) \[0.5\,m\,m\]
Correct Answer: D
Solution :
Radius of the helical path of the electron in the field \[r=\frac{mv\sin \theta }{qB}\] \[r=\frac{2.4\times {{10}^{-23}}\sin {{30}^{o}}}{1.6\times {{10}^{-19}}\times 0.15}\] \[=\frac{2.4\times {{10}^{-23}}\times \frac{1}{2}}{1.6\times {{10}^{-19}}\times 0.15}\] \[=5\times {{10}^{-4}}m=0.5\,\,mm\]
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