19. (b) An electron emitted by a heated cathode and accelerated through a potential difference of 2.0 kV, enters a region with uniform magnetic field of 0.15 T. Determine the trajectory of the electron if the field makes an angle of 30 \degreewith the initial velocity.

Answers (1)
S Sayak

The electron has been accelerated through a potential difference of 2.0 kV. 

Therefore K.E of electron = 1.6\times10-19\times2000=3.2\times10-16 J

\\\frac{1}{2}mv^{2}=3.2\times 10^{-16}\\ \\v=\sqrt{\frac{2\times 3.2\times 10^{-16}}{9.1\times 10^{-31}}}\\ v=2.67\times 10^{7}\ ms^{-1}

The component of velocity perpendicular to the magnetic field is

\\v_{p}=vsin30^{o}\\ v_{p}=1.33\times 10^{7}\ ms^{-1}

The electron will move in a helical path of radius r given by the relation,

\\\frac{mv^{2}_{p}}{r}=ev_{p}B\\ r=\frac{mv_{p}}{eB}\\ r=\frac{9.1\times 10^{-31}\times 1.33\times 10^{7}}{1.6\times 10^{-19}\times 0.15}

r=5m

r=5\times10-4 m

r=0.5 mm

The component of velocity along the magnetic field is

\\v_{t}=vcos30^{o}\\ v_{t}=2.31\times 10^{7}\ ms^{-1}

The electron will move in a helical path of pitch p given by the relation,

\\p=\frac{2\pi r}{v_{p}}\times v_{t}\\ p=\frac{2\pi \times 5\times 10^{-4}}{1.33\times 10^{7}}\times 2.31\times 10^{7}

p=5.45\times10^{-3} m

p=5.45 mm

The electron will, therefore, move in a helical path of radius 5 mm and pitch 5.45 mm.

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