A) 2 mm
B) 1mm
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 \[r=\frac{mv\bot }{eB}=\frac{mv\,\sin \,\theta }{eB}=\frac{(2.4\times {{10}^{-23}}kg\,m/s)\times sin\,30{}^\circ }{(16\times {{10}^{-19}}C)\times 0.15\,T}\]\[=5\times {{10}^{-4}}m=0.5\times {{10}^{-3}}m=0.5\,mm\]You need to login to perform this action.
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