An electron (mass = 9.1 times 10^{-31} kg; c | vedclass

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(C) For no deflection in mutually perpendicular electric and magnetic fields,the velocity $v$ of the electron is given by $v = \frac{E}{B}$.Substituti

Vedclass · Accepted Answer

$0.45$

Vedclass · Answer

(C) For no deflection in mutually perpendicular electric and magnetic fields,the velocity $v$ of the electron is given by $v = \frac{E}{B}$.Substituting the given values: $v = \frac{3.2 	imes 10^5}{2 	imes 10^{-3}} = 1.6 	imes 10^8 \ m/s$.When the electric field is removed,the electron moves in a circular path due to the magnetic force acting as the centripetal force.The radius $r$ of the path is given by $r = \frac{mv}{qB}$.Substituting the values: $r = \frac{9.1 	imes 10^{-31} 	imes 1.6 	imes 10^8}{1.6 	imes 10^{-19} 	imes 2 	imes 10^{-3}} = \frac{9.1 	imes 10^{-23}}{3.2 	imes 10^{-22}} = 0.45 \ m$.

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