You reproduce Thompsons e-m experiment by applying an accelerating vol.docx

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You reproduce Thompsons e/m experiment by applying an accelerating voltage of 295 V to the apparatus. (a) At what fraction of the speed of light do the electrons move? (b) If the deflecting electric field has a magnitude of 8.0Â·10 6 N/C, what is the strength of the magnetic field? (c) If the beam radius r is measured to be 5 cm, what must be the strength of the magnetic field produced by the Helmholtz coils? Hint: you know the value of e/m . Solution given V = 295 V we know, q = 1.6*10^-19 C m = 9.1*10^-31 kg a) Workdone on the electron = gain in kinetic energy q*V = (1/2)*m*v^2 ==> v = sqrt(2*q*V/m) = sqrt(2*1.6*10^-19*295/(9.1*10^-31)) = 1.018*10^7 m/s = 1.018*10^7/(3*10^8) = 0.0339*c <<<<<<<<------------------Answer (here c is light speed) b) for no deflection Fb = Fe q*v*B = q*E B = E/v = 8*10^6/(1.018*10^7) = 0.786 T <<<<<<<<------------------Answer c) we know, r = m*v/(B*q) B = m*v/(r*q) = 9.1*10^-31*1.018*10^7/(0.05*1.6*10^-19) = 1.16*10^-3 T <<<<<<<<------------------Answer .

$You reproduce Thompsons e/m experiment by applying an accelerating voltage of 295 V to the apparatus. (a) At what fraction of the speed of light do the electrons move? (b) If the deflecting electric field has a magnitude of 8.0Â·10 6 N/C, what is the strength of the magnetic field? (c) If the beam radius r is measured to be 5 cm, what must be the strength of the magnetic field produced by the Helmholtz coils? Hint: you know the value of e/m . Solution given V = 295 V we know, q = 1.6*10^-19 C m = 9.1*10^-31 kg a) Workdone on the electron = gain in kinetic energy q*V = (1/2)*m*v^2 ==> v = sqrt(2*q*V/m) = sqrt(2*1.6*10^-19*295/(9.1*10^-31)) = 1.018*10^7 m/s = 1.018*10^7/(3*10^8) = 0.0339*c <<<<<<<<------------------Answer (here c is light speed) b) for no deflection Fb = Fe q*v*B = q*E$

B = E/v
= 8*10^6/(1.018*10^7)
= 0.786 T <<<<<<<<------------------Answer
c) we know, r = m*v/(B*q)
B = m*v/(r*q)
= 9.1*10^-31*1.018*10^7/(0.05*1.6*10^-19)
= 1.16*10^-3 T <<<<<<<<------------------Answer

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1. You reproduce Thompsons e/m experiment by applying an accelerating voltage of 295 V to the apparatus. (a) At what fraction of the speed of light do the electrons move? (b) If the deflecting electric field has a magnitude of 8.0Â·10 6 N/C, what is the strength of the magnetic field? (c) If the beam radius r is measured to be 5 cm, what must be the strength of the magnetic field produced by the Helmholtz coils? Hint: you know the value of e/m . Solution given V = 295 V we know, q = 1.6*10^-19 C m = 9.1*10^-31 kg a) Workdone on the electron = gain in kinetic energy q*V = (1/2)*m*v^2 ==> v = sqrt(2*q*V/m) = sqrt(2*1.6*10^-19*295/(9.1*10^-31)) = 1.018*10^7 m/s = 1.018*10^7/(3*10^8) = 0.0339*c <<<<<<<<------------------Answer (here c is light speed) b) for no deflection Fb = Fe q*v*B = q*E

2. B = E/v = 8*10^6/(1.018*10^7) = 0.786 T <<<<<<<<------------------Answer c) we know, r = m*v/(B*q) B = m*v/(r*q) = 9.1*10^-31*1.018*10^7/(0.05*1.6*10^-19) = 1.16*10^-3 T <<<<<<<<------------------Answer