Answer to Question #159797 in Electric Circuits for zain ul abdeen

Question

 An accelerating voltage of 2 500 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. (a) What are the magnitude and direction of the deflection on the screen caused by the Earth’s gravitational field? (b) What are the magnitude and direction of the deflection on the screen caused by the vertical component of the Earth’s magnetic field, taken as 20.0 mT down? Does an electron in this vertical magnetic field move as a projectile, with constant vector acceleration perpendicular to a constant northward component of velocity? Is it a good approximation to assume it has this projectile motion? Explain. 

Answer 1

$V=2500 V, x=35cm$

a)

$\Delta K+\Delta U=0$

$\frac{1}{2}mv_f^2+qV=0\implies v_f=\sqrt{-\frac{2qV}{m}}$

$v_f=\sqrt{-\frac{-(1.6\cdot10^{-19})\cdot2500}{9.1\cdot10^{-31}}}=2.96\cdot10^7m/s$

$t=x/v_f=\frac{35}{2.96\cdot10^7}=1.18\cdot10^{-8}s$

$\Delta y=\frac{1}{2}a_yt^2$

$\Delta y=\frac{1}{2}\cdot9.8\cdot1.18\cdot10^{-8}=6.85\cdot10^{-16}m$

$tan \theta=\Delta/x\implies \theta=tan^{-1}(\Delta/x)$

$\theta=tan^{-1}(\frac{6.85 \cdot 10^{-15}}{35})=1.96\cdot10^{-15}rad$ vertically downwards

b) $B=20\mu T$

$F=qvB\implies ma=qvB\implies a=qvB/m$

$a=\frac{1.6\cdot10^{-19}\cdot2.96\cdot10^7\cdot20}{9.1\cdot10^{-31}}=1.041\cdot10^{14}m/s^2$

$\Delta z=\frac{1}{2}at^2$

$\Delta z=\frac{1}{2}\cdot1.041\cdot10^{14}\cdot1.18\cdot10^{-8}=0.7cm$

$\theta=tan^{-1}(\Delta z/x)=\theta=tan^{-1}(0.7/35)=0.02rad$ right direction

An electron in this vertical magnetic field move as a projectile, because thereis only acceleration that is acting vertically.

It is not a good approximation to assume that it is projecile motion, because in this case the magnetic field and gravitational field act on the same axis.