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प्रश्न
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 (me = 9.11 × 10–31 kg).
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उत्तर
Energy of an electron beam, E = 18 keV = 18 × 103 eV
Charge on an electron, e = 1.6 × 10−19 C
E = 18 × 103 × 1.6 × 10−19 J
Magnetic field, B = 0.04 G
Mass of an electron, me = 9.11 × 10−19 kg
Distance up to which the electron beam travels, d = 30 cm = 0.3 m
We can write the kinetic energy of the electron beam as:
E = `1/2 "mv"^2`
v = `sqrt((2"E")/"m")`
= `sqrt((2 xx 18 xx 10^3 xx 1.6 xx 10^-19 xx 10^-15)/(9.11 xx 10^-31))`
= 0.795 × 108 m/s
The electron beam deflects along a circular path of radius, r.
The force due to the magnetic field balances the centripetal force of the path.
BeV = `"mv"^2/"r"`
∴ r = `"mv"/"Be"`
= `(9.11 xx 10^-31 xx 0.795 xx 10^8)/(0.4 xx 10^-4 xx 1.6 xx 10^-19)`
= 11.3 m
Let the up and down deflection of the electron beam be x = r(1 − cos θ)
Where,
θ = Angle of declination
sin θ = `"d"/"r"`
= `0.3/11.3`
θ = `sin^-1 0.3/11.3` = 1.521°
And x = 11.3(1 − cos 1.521°)
= 0.0039 m
= 3.9 mm
Therefore, the up and down deflection of the beam is 3.9 mm.
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