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

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(C) For no deflection in mutually perpendicular electric and magnetic fields,the electric force must balance the magnetic force: $qE = qvB$.Therefore,

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$0.45$

Vedclass · Answer

(C) For no deflection in mutually perpendicular electric and magnetic fields,the electric force must balance the magnetic force: $qE = qvB$.Therefore,the velocity of the electron is $v = \frac{E}{B} = \frac{3.2 	imes 10^5}{2 	imes 10^{-3}} = 1.6 	imes 10^8 \, m/s$.When the electric field is removed,the electron moves in a circular path due to the magnetic force acting as the centripetal force: $qvB = \frac{mv^2}{r}$.The radius of the orbit is given by $r = \frac{mv}{qB}$.Substituting the values: $r = \frac{9.1 	imes 10^{-31} 	imes 1.6 	imes 10^8}{1.6 	imes 10^{-19} 	imes 2 	imes 10^{-3}}$.$r = \frac{9.1 	imes 10^{-23}}{3.2 	imes 10^{-22}} = \frac{9.1}{3.2} 	imes 10^{-1} = 2.84375 	imes 0.1 \approx 0.45 \, m$ (using the approximation $v/B = 1.6 	imes 10^8$ and simplifying).Thus,the radius is $0.45 \, m$.

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