A) \[\text{1}0\times \text{1}{{0}^{\text{6}}}\text{ m}/\text{s},\text{ 2}.\text{43 cm}\]
B) \[\text{2}.\text{5}\times \text{1}{{0}^{\text{6}}}\text{ m}/\text{s},\text{ }0.\text{43 cm}\]
C) \[\text{5}\times \text{l}{{0}^{\text{6}}}\text{m}/\text{s},\text{ 1}.\text{42cm}\]
D) None of the above
Correct Answer: C
Solution :
Since, path of electron remains undeviated, hence force due to magnetic field = force due to electric field \[\therefore \] \[F=qE=qvB\] \[\Rightarrow \] \[v=\frac{E}{B}\] Given, \[B=2\times {{10}^{-3}}Wb/{{m}^{2}}\], \[E=1\times {{10}^{4}}V/m\] \[\therefore \] \[v=\frac{1\times {{10}^{4}}}{2\times {{10}^{-3}}}=5\times {{10}^{6}}m/s\]. When electric field is removed, electron transverses a circular path of radius r given by \[\Rightarrow \] \[r=\frac{mv}{eB}\] Given, \[m=9.1\times {{10}^{-31}}kg\], \[v=5\times {{10}^{6}}m/s\], \[e=1.6\times {{10}^{-19}}C\] \[B=2\times {{10}^{-3}}Wb/{{m}^{2}}\] \[\therefore \] \[r=\frac{9.1\times {{10}^{-31}}\times 5\times {{10}^{6}}}{1.6\times {{10}^{-19}}\times 2\times {{10}^{-3}}}\] = 1.42 cm
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