An accclerating voltage of 2.50 times 10^3 V is applied to an electron gun, prod

Question

An accclerating voltage of 2.50 times 10^3 V is applied to an electron gun, producing a beam of electrons originally traveling horizontally north in vacuum toward the center of a viewing screen 35.0 cm away. What are (a) the magnitude and (b) the direction of the deflection on the screen caused by the Earth's gravitational field? What are (c) the magnitude and (d) the direction of the deflection on the screen caused by the vertical component of the Earth's magnetic field, taken as 20.0JUJ down? (e) Docs an electron in this vertical magnetic field move as a projectile, with constant vector acceleration perpendicular to a constant northward component of velocity? (0 Is it a good approximation to assume it has this projectile motion? Explain.

Explanation / Answer

V = 2500 V
E = eV = 1.6*10^-19*2500 J = 0.5mv^2 = 0.5*9.1*10^-31*v^2
v = 9.376*10^6 m/s

1. a = g
let time taken to reach 35 cm be t
v = 0.35/t
t = 3.732*10^-8 s
s = 0.5*9.8*t^2 = 68.278*10^-16 m
Direction, downwards
3. F = q(v x B)
a = F/m = 1.6*10^-19 ( 9.376*10^6 * 20*10^-6) / 9.1*10^-31
s = 0.5*a*t^2 = 2.296 cm
direction, eastwards [ from left hand rule]
5. As the acc is always perpendicular to velocity, its not safe to assume it as projectile motion as the direction of acc is not constant and keeps on changing with velocity. but for small values it can be approximated to be as a projectile (which would be an approximation)

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