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Manipal Engineering Manipal Engineering Solved Paper-2011

  • question_answer
    An electron moving with velocity \[2\times {{10}^{-7}}\]m/s, describes a circle in a magnetic field of strength \[2\times {{10}^{-2}}T\,.\] If\[\frac{e}{m}\] of electron is \[1.76\times {{10}^{11}}C/kg\]. Then the diameter of the circle will be

    A)  11 cm                                  

    B)  1.1 cm

    C)  1.1 mm               

    D)         1.1 m

    Correct Answer: B

    Solution :

    The force exerted on electron moving with velocity\[v\]is                 \[F=evB\sin \theta \] where\[\theta ={{90}^{o}}\] \[\therefore \]  \[{{F}_{B}}=evB\]                                             ... (i) Centripetal force is given by                 \[{{F}_{C}}=\frac{m{{v}^{2}}}{r}\]                                             ? (ii) Equating Eqs. (i) and (ii), we get                 \[r=\frac{m{{v}^{2}}}{eB}\] Diameter\[d=2r=\frac{2v}{(e/m)B}\] \[\Rightarrow \]               \[d=\frac{2v}{1.76\times {{10}^{11}}\times 2\times {{10}^{-2}}}\]                 \[=1.1\,\,cm\]


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