# 19. (b) An electron emitted by a heated cathode and accelerated through a potential difference of 2.0 kV, enters a region with uniform magnetic field of 0.15 T. Determine the trajectory of the electron if the field makes an angle of with the initial velocity.

The electron has been accelerated through a potential difference of 2.0 kV.

Therefore K.E of electron = 1.610-192000=3.210-16 J

The component of velocity perpendicular to the magnetic field is

The electron will move in a helical path of radius r given by the relation,

r=5m

r=510-4 m

r=0.5 mm

The component of velocity along the magnetic field is

The electron will move in a helical path of pitch p given by the relation,

The electron will, therefore, move in a helical path of radius 5 mm and pitch 5.45 mm.

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