5.22) 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_e = 9.11 \times 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.]

Answers (1)

The energy of electron beam   =   18 eV

                   We can write:-   

                                                        E\ =\ \frac{1}{2} mv^2

so                                                     v\ =\ \sqrt{\frac{2E}{M}}

We are given horizontal magentic field :          B  = 0.40 G

 Also,                                                Bev\ =\ \frac{mv^2}{r}   

                                     

We obtain,                                         r\ =\ \frac{1}{Be}\sqrt{2EM}

or                                                       r\ =\ 11.3\ m

Using geometry, we can write :- 

                                                       \sin \Theta =\ \frac{x}{r} =\ \frac{0.3}{11.3}

and                                                   y\ =\ r-r \cos\Theta

or                                                           =\ r(1- \sqrt{1- \sin^2\Theta })

or                                                     y\ \approx \ 4mm

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