An electron beam passes through a magnetic fi | vedclass

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Question

(C) Given:$B = 2 	imes 10^{-3} \, Wb/m^2$$E = 1.0 	imes 10^4 \, V/m$Since the path of the electron remains undeviated,the electric force must balanc

Vedclass · Accepted Answer

$5 	imes 10^6 \, m/s, 1.43 \, cm$

Vedclass · Answer

(C) Given:$B = 2 	imes 10^{-3} \, Wb/m^2$$E = 1.0 	imes 10^4 \, V/m$Since the path of the electron remains undeviated,the electric force must balance the magnetic force:$qE = qvB \Rightarrow v = \frac{E}{B}$$v = \frac{1.0 	imes 10^4}{2 	imes 10^{-3}} = 0.5 	imes 10^7 = 5 	imes 10^6 \, m/s$If the electric field is removed,the electron moves in a circular path due to the magnetic force acting as the centripetal force:$\frac{mv^2}{r} = qvB \Rightarrow r = \frac{mv}{qB}$Using $m = 9.1 	imes 10^{-31} \, kg$ and $q = 1.6 	imes 10^{-19} \, C$:$r = \frac{9.1 	imes 10^{-31} 	imes 5 	imes 10^6}{1.6 	imes 10^{-19} 	imes 2 	imes 10^{-3}}$$r = \frac{45.5 	imes 10^{-25}}{3.2 	imes 10^{-22}} = 14.218 	imes 10^{-3} \, m \approx 1.43 \, cm$Thus,the speed is $5 	imes 10^6 \, m/s$ and the radius is $1.43 \, cm$.

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