question_answer

An electron emitted by a heated cathode and accelerated through a potential difference of 2 kV, enters a region with a uniform magnetic field of 0.15 T. Determine the trajectory of the electrons if the magnetic field

(i) is transverse to its initial velocity.

(ii) makes an angle \[30{}^\circ \] with the initial velocity.

 

Answer:

            Velocity of electron accelerated through a potential difference V is given by \[\frac{1}{2}m{{v}^{2}}=eV\] \[\Rightarrow v=\sqrt{\frac{2eV}{m}}\] Given, \[V=2kV=2\times {{10}^{3}}V,\] \[e=1.6\times {{10}^{-19}}C,~\] mass of electron,\[m=9\times {{10}^{-31}}\,kg\] \[\therefore \]      \[v=\sqrt{\frac{2\times 1.6\times {{10}^{-19}}\times 2\times {{10}^{3}}}{9\times {{10}^{-31}}}}\] \[=\frac{8}{3}\times {{10}^{7}}\,m/s\] (i) When an electron enters the transverse magnetic field, its path is a circle of radius r, given by             \[\frac{m{{v}^{2}}}{r}=evB\] or \[r=\frac{mv}{eB}\] Substituting given values,             \[r=\frac{(9\times {{10}^{-31}})\times \left( \frac{8}{3}\times {{10}^{7}} \right)}{(1.6\times {{10}^{-19}})\times (0.15)}={{10}^{-3}}m\]             = 1 mm (ii) When electron enters at an angle \[30{}^\circ \] with magnetic fields, its path become a helix of radius,             \[r=\frac{mv\sin 30{}^\circ }{eB}\]             \[=\left( \frac{mv}{eB} \right)\times \sin 30{}^\circ \] Velocity component along* the magnetic field, \[{{v}_{II}}=v\cos 30{}^\circ \]             \[=\left( \frac{8}{3}{{10}^{7}}m/s \right)\times \frac{\sqrt{3}}{2}\]             \[=2.3\times {{10}^{7}}m/s\]