Question

A monoenergetic (18 keV) electron beam initially in the horizontal
direction is subjected to a horizontal magnetic field of 0.04 G normal
to the initial direction. Estimate the up or down deflection of the
beam over a distance of 30 cm (m. = 9.11 x 10-31 kg). (Note: Data in
this exercise are so chosen that the answer will give you an idea of
the effect of earth's magnetic field on the motion of the electron beam
from the electron gun to the screen in a TV set.)​

Answer 1

Answer:

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Explanation:

Velocity of the beam can be found from its kinetic energy.

=

2

1

mv

2

⟹v=

m

2E

=

9.1×10

−31

2×18×10

3

×1.6×10

−19

=0.795×10

8

m/s

The magnetic force acting on the beam tends it to follow a circular path. The magnetic force equals the centripetal force.

qvB=

r

mv

2

⟹r=11.3m

The up and down deflection of the electron beam is

x=r(1−cosθ)

sinθ=

r

d

=

11.3

0.3

⟹θ=1.521

∘

Therefore deflection x=11.3(1−cos1.521

∘

)=3.9mm