BCECE Engineering BCECE Engineering Solved Paper-2007

  • question_answer
    Cathode rays of velocity \[{{10}^{6}}\,m/s\] describe an approximate circular path of radius 1 m in an electric field 300 V/cm. If the velocity of the cathode rays are doubled. The value of electric field so that the rays describe the same circular path, will be

    A)  2400 V/cm                         

    B)  600 V/cm           

    C)         1200 V/cm         

    D)         12000 V/cm

    Correct Answer: C

    Solution :

    Key Idea: The electric field acting on the particles provides the necessary centripetal force. Cathode rays are composed of electrons, when they move in electric field a force \[F=eE\]                                               ?(i) acts on them, this provides the necessary centripetal force to the particles.                 \[\therefore \]  \[F=\frac{m{{v}^{2}}}{r}\]                                            ?(ii) From Eqs. (i) and (ii), we get                                 \[eE=\frac{m{{v}^{2}}}{r}\]                 \[\Rightarrow \]               \[r=\frac{m{{v}^{2}}}{eE}=\frac{m{{({{10}^{6}})}^{2}}}{e(300)}\]                                ?(iii) When velocity is doubled same circular path is followed, hence radius is same                 \[r=\frac{m{{(2\times {{10}^{6}})}^{2}}}{eE}\]                                    ?(iv) Equating Eqs. (iii) and (iv), we get \[\frac{m\times {{({{10}^{6}})}^{2}}}{300\,e}=\frac{m\times {{(2\times {{10}^{6}})}^{2}}}{eE}\] \[\Rightarrow \]               \[E=300\times 4=1200\,V/cm\] Note: Since, \[E\propto {{u}^{2}}\]and \[v\]is doubled much larger field will be required.


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