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BCECE Engineering BCECE Engineering Solved Paper-2005

  • question_answer
    An electron (mass \[9.1\times {{10}^{31}}\] kg, charge \[=1.6\times {{10}^{19}}C\]) experiences no deflection, if subjected to an electric field of \[3.2\times {{10}^{5}}V/m,\] 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 i.e.,             \[Bev=eE\] 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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