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BCECE Medical BCECE Medical Solved Papers-2005

  • question_answer
    An electron (mass \[=9.1\times {{10}^{-31}}kg\], charge \[1.16\times {{10}^{-19}}C\]) experiences no deflection, if subjected to an electric field of \[3.32\times {{10}^{5}}\] V/m, and a magnetic field of \[2.0\times {{10}^{-3}}Wb/{{m}^{2}}\] . Both the fields are normal to the path of electron and to each other. If the electric field is removed, then the electron will revolve in an orbit of radius :

    A)  45 m           

    B)  4.5 m

    C)  0.45 m         

    D)  0.045 m

    Correct Answer: C

    Solution :

    Since, electron has no deflection in electric and magnetic field, so magnetic force on electron = electric force on electron ie.,          \[Bev=eE\] or            \[v=\frac{E}{B}\] Given, \[E=3.2\times {{10}^{5}}V/m\],                 \[B=2.0\times {{10}^{-3}}Wb/{{m}^{2}}\] \[\therefore \] \[v=\frac{3.2\times {{10}^{5}}}{2.0\times {{10}^{-3}}}=1.6\times {{10}^{8}}m/s\] When electric field is switched off, then electron will move on circular path of radius                 \[r=\frac{mv}{eB}\]                 \[=\frac{9.1\times {{10}^{-31}}\times 1.6\times {{10}^{8}}}{1.6\times {{10}^{-19}}\times 2\times {{10}^{-3}}}\] = 0.45 m


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