1) The electron gun in a TV picture tube accelerates electrons between two paral
ID: 1285177 • Letter: 1
Question
1) The electron gun in a TV picture tube accelerates electrons between two parallel plates 1.20 cm apart with a 25.0 kV potential difference between them. The electrons enter through a small hole in the negative plate, accelerate, and then exit through a small hole in the positive plate. Assume that the holes are small enough not to affect the electric field or the potential. (a) What is the electric field between the plates? (b) What is the exit speed of an electron if its entry speed is close to zero?
2)The Hall voltage across a conductor in a 55.0 mT magnetic field is 1.90 mV. When used with the same current in a different magnetic field, the voltage across the conductor is 2.80 mV. What is the strength of the second magnetic field?
Explanation / Answer
1) a) E = V/d
= 25000/0.012
= 2.083*10^6 N/c <<<<<<<<<<<<------------ANswer
b) a = F/m
= q*E/m
= 1.6*10^-19*2.083*10^6/9.1*10^-31
= 3.663*10^17 m/s^2
v^2-u^2 = 2*a*d
here u = 0
so, v = sqrt(2*a*d)
= sqrt(2*3.663*10^17*0.012)
= 2.965*10^8 m/s <<<<<<<<<<<<------------ANswer
2) we know vH is proportional to B
so, vH2/vH1 = B2/B1
B2 = B1*(vH2/vH1)
= 55*10^-3*(2.8/1.9)
= 81 mT <<<<<<<<<<<<------------ANswer
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