A) 2 mm
B) 1 mm
C) \[\frac{\sqrt{3}}{2}mm\]
D) 0.5 mm
Correct Answer: D
Solution :
(d) 0.5 mm. The radius of the helical path of the electron in the uniform magnetic field is \[r=\frac{m{{v}_{\bot }}}{eB}=\frac{mv\,\sin \theta }{eB}\] \[=\frac{(2.4\times {{10}^{-23}}kg\,m{{s}^{-1}})\times sin\,30{}^\circ }{(16\times {{10}^{-19}}C)\times (0.15\,T)}\] \[=5\times {{10}^{-4}}m=0.5\times {{10}^{-3}}m=0.5\,mm.\]You need to login to perform this action.
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