An accelerating voltage of 2.42 times 10^3 V is applied to an electron gun, prod

Question

An accelerating voltage of 2.42 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 37.0 cm away. (a) What is the magnitude of the defection on the screen caused by the Earth's gravitational field? (b) What is the direction of the deflection on the screen caused by the Earth's gravitational field? up down east west (c) What the magnitude of the deflection on the screen caused by the vertical component of the Earth's magnetic field, taken as 20.0 mu T down? (d) What is the direction of the defection on the screen caused by the vertical component of the Earth's magnetic field, taken as 20.0 mu T down? north south east west (e) Does an in this vertical magnetic field move as a projectile, perpendicular to a constant northward component of velocity? Yes No (f) Is a good approximation to assume it has this projectile motion? Yes No

Explanation / Answer

a) E = m v2 / 2, v = sqrt [2 E / m] = sqrt [2 * 2.42 * 103 * 1.6 * 10-19 / 9.11 * 10-31 ]

= 2.92 * 107 m/s

Deflection by earth's gravitational field = g t2 / 2 = g * (d/v)2 / 2 = [9.8 * (0.37 / (2.92 * 107))2]/2

= 7.87 * 10-16 m

b) Down

c) Deflection by earth's magnetic field = a t2 / 2 = (q v B / m) * (d/v)2 / 2 = q B d2 / 2 m v

= (1.6 * 10-19 * 20 * 10-6 * 0.372) / (2 * 9.11 * 10-31 * 2.92 * 107) = 8.2 mm

d) east

e) Yes

f) No

Because, For any charged particle moving in magnetic field perpendicular to magnetic field follow circular path

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