4.19 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  is transverse to its initial velocity

Answers (1)
S Sayak

 (a) 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}

 Since the electron initially has velocity perpendicular to the magnetic field it will move in a circular path.

The magnetic field acts as a centripetal force. Therefore,

\\\frac{mv^{2}}{r}=evB\\ r=\frac{mv}{eB}\\ r=\frac{9.1\times 10^{-31}\times 2.67\times 10^{7}}{1.6\times 10^{-19}\times 0.15}=1.01mm 

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