Refer the attached picture.
Answers
SOLUTION
Energy of an electron beam, E= 18keV
=) 18× 10^3 eV
Change on an electron, e= 1.6×10^-19 C
E= 18× 10^3 × 1.6× 10^-19J
Magnetic field, B= 0.04G
Mass of an electron, me=9.11× 10^-19kg
Distance up to which the electron beam travels, d= 30cm= 0.3m
We can write the kinetic energy of the electron beam as:
E= 1/2mv^2
=) v= √2E/m
=) √2×18× 10^3× 1.6 × 10^-19× 10^-15/9.11× 10^-31
=) 0.795× 10^8 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× 10^-31× 0.795× 10^8/0.4× 10^-4× 1.6× 10^-19
=) 11.3metres
Let the up and down deflection of the electron beam be
x= r(1-cos theta)
where,
theta = angle of declination
sin theta= d/r
=) 0.3/ 11.3
theta= sin^-1. 0.3/11.3= 1.521°
& x= 11.3( 1-cos 1.521°)
=) 0.0039m= 3.9mm
Therefore, the up & down deflection of the beam is 3.9mm.