A) 2 mm
B) 1 mm
C) \[\frac{\sqrt{3}}{2}mm\]
D) \[0\,.\,5\,mm\]
Correct Answer: D
Solution :
This radius of the helical path of the electron in the uniform magnetic field is \[r=\frac{m{{v}_{\bot }}}{eB}=\frac{mv\sin \theta }{eB}\] \[=\frac{\left( 2.4\times {{10}^{-23}}\,kg\,m/s \right)\times \sin \,30{}^\circ }{\left( 1.6\times {{10}^{-19}}C \right)\times 0\,.\,15T}\] \[=5\times {{10}^{-4}}m=0\,.\,5\times {{10}^{-3}}m=0.5\,mm\]
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