Question

A monoenergetic (18 keV) electron beam initially in the horizontaldirection is subjected to a horizontal magnetic field of 0.04 G normalto the initial direction. Estimate the up or down deflection of thebeam over a distance of 30 cm (me = 9.11 × 10–31 kg). [Note: Data inthis exercise are so chosen that the answer will give you an idea ofthe effect of earth’s magnetic field on the motion of the electron beamfrom the electron gun to the screen in a TV set.]

Answer 1

Given: the energy of electron beam is 18  keV, the strength of magnetic field is 0.04 G.

The energy of the electron beam is given as,

E= 1 2 m v 2 v= 2E m

Where, m is the mass of an electron and v is its velocity.

By substituting the given values in the above expression, we get

v= 2×18× 10 3 ×1.6× 10 −19 × 10 −15 9.11× 10 −31 =0.795× 10 8   ms −1

When the electron beam is subjected to a magnetic field normal to its direction, it gets deflected along a circular path.

The force on electron due to magnetic field is given as,

F=Bev

Where, e is the charge on electron and B is the applied magnetic field.

The force due to the magnetic field balances the centripetal force of the path. Hence,

Bev= m v 2 r r= mv Be

Where, r is the radius of circular path.

By substituting the given values in the above expression, we get

r= 9.11× 10 −31 ×0.795× 10 8 0.4× 10 −4 ×1.6× 10 −19 =11.3 m

The angle of declination is given as,

sinθ= d r

Where, d is the distance travelled by the electron beam.

By substituting the given values in the above expression, we get

sinθ= 0.3 11.3 θ= sin −1 ( 0.3 11.3 ) =1.521°

The deflection is given as,

x=r( 1−cosθ )

By substituting the given values in the above expression, we get

x=11.3( 1−cos( 1.521 ) ) =0.0039 m =3.9 mm

Thus, the up and down deflection of the beam is 3.9 mm.