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Let E be the electric field of magnitude 6.0 xx 10^(6) N C^(-1) and B be the magnetic field magnitude 0.83 T. Suppose an electron is accelerated with a potential of 200 V, will it show zero deflection ? If not, at what potential will it show zero deflection . |
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Answer» SOLUTION :Electric field , E = `6.0 xx 10^(6) N C^(1)` and magnetic field , B = 0.83 T Then, v = `(E)/(B) = (6.0 xx 10^(6))/(0.83) = 7.23 xx 10^(6) ms^(-1)` when an electron goes with this velocity, it shows null deflection.SINCE the accelerating potential is 200 V , the electron acquires kinetic energy because of this accelerating potential Hence, `(1)/(2) mv^(2) = eV rArr v = sqrt((eV)/(2m))` Since the mass of the electron, m = 9.1`xx 10^(-31)` kg and charge of an electron, |q| = e = 1.6 `xx 10^(-19) C` The velocity due to accelerating potential 200 V `v_(200) = sqrt( (2(1.6 xx 10^(-19))(200))/((9.1 xx 10^(-31) )))= 8.39 xx 10^(6) m s^(-1)` Since the speed `v_(200) gt `v, the electron is DEFLECTED towards direction of Lorentz force. So, in order to have null deflection , the potential , we have to supply is V = `(1)/(2) (mv^(2))/(e ) = ((9.1 xx 10^(-31)) xx (7.23 xx 10^(6))^(2) )/(2 xx (1.6 xx 10^(-19)) ` V = 148.65 V |
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