(a) What is the magnitude of the deflection on the screen caused by the Earth\'s

Question

(a) What is the magnitude of the deflection on the screen caused by the Earth's gravitational field?
m

(b) What is the direction of the deflection on the screen caused by the Earth's gravitational field?

updown     eastwest

(c) What is the magnitude of the deflection on the screen caused by the vertical component of the Earth's magnetic field, taken as 20.0 µT down?
mm

(d) What is the direction of the deflection on the screen caused by the vertical component of the Earth's magnetic field, taken as 20.0 µT down?

northsouth     eastwest

(e) Does an electron in this vertical magnetic field move as a projectile, with constant vector acceleration perpendicular to a constant northward component of velocity?

YesNo     

(f) Is it a good approximation to assume it has this projectile motion?

YesNo     

Explain.

Explanation / Answer

Kinetic energy of electron is eV, where is electron charge and v is accelerating potential.
a) Deflection due to garvitational force
  d = 1/2 g t2 = 1/2 g ( L/V)2  ( L is distance of screen and v is speed of electron)
   = 1/2 g L2 (m/2K)1/2 ( m is mass of electron and K is kinetic energy)
   = 1/2 g L2 (m/2ev)1/2
   = 1.71 x 10-8 m

b) deflection is in the downward direction.

c) deflection due to magnetic field, while moving in circular path is given by
   d' = R - (R2 - L2)
R is radius of circular path and given by ((2Km)1/2 /eB = (2eVm)1/2 /eB = ( 2vm/e)1/2 /B
R = 8.76 m
d' = 6.1x10-3 m

d) direction of deflection is given by VxB  and is towards east.

e) No it moves in circular path with magnitude of acceleration remaining constant but continously changing direction.

f) No, As deviation due to projectile motion is much smaller than deviation due to circular motion under magnetic force.

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