A) 45 m
B) 4.5 m
C) 0.45 m
D) 0.045 m
Correct Answer: C
Solution :
For no deflection in mutually perpendicular electric and magnetic field \[v=\frac{E}{B}=\frac{3.2\times {{10}^{5}}}{2\times {{10}^{-3}}}=1.6\times {{10}^{8}}\,m/s\]. If electric field is removed then due to only magnetic field radius of the path described by electron \[r=\frac{mv}{qB}=\frac{9.1\times {{10}^{-31}}\times 1.6\times {{10}^{8}}}{1.6\times {{10}^{-19}}\times 2\times {{10}^{-3}}}=0.45\,m\]You need to login to perform this action.
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