A) 45 m
B) 4.5 m
C) 0.45 m
D) 0.045 m
Correct Answer: C
Solution :
Since, electron has no deflection in electric and magnetic, field so magnetic force on electron=electric force on electron i.e., \[Bev=eE\] or \[v=\frac{E}{B}\] Given, \[E=3.2\times {{10}^{5}}\,V/m,\] \[B=2.0\times {{10}^{-3}}\,Wb/{{m}^{2}}\] \[\therefore \] \[v=\frac{3.2\times {{10}^{5}}}{2.0\times {{10}^{-3}}}=1.6\times {{10}^{8}}\,m/s\] When electric field is switched off, then electron will move on circular path of radius \[r=\frac{mv}{eB}\] \[=\frac{9.1\times {{10}^{-31}}\times 1.6\times {{10}^{8}}}{1.6\times {{10}^{-19}}\times 2\times {{10}^{-3}}}\] \[=0.45\,m\]You need to login to perform this action.
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