• # 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.

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, ${{v}_{II}}=v\cos 30{}^\circ$             $=\left( \frac{8}{3}{{10}^{7}}m/s \right)\times \frac{\sqrt{3}}{2}$             $=2.3\times {{10}^{7}}m/s$

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