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CET Karnataka Medical CET - Karnataka Medical Solved Paper-2015

  • question_answer
    A proton beam enters a magnetic field of \[{{10}^{-4}}\text{ }Wb\text{ }{{m}^{-2}}\]1normally. If the specific charge of the proton is \[{{10}^{11}}C\text{ }k{{g}^{-1}}\]and its velocity is \[{{10}^{9}}m\,{{s}^{-1}}\], then the radius of the circle described  will be

    A)  \[10m\]  

    B)  \[1\text{ }m~\]

    C)  \[0.1m\]

    D)  \[100m\]

    Correct Answer: D

    Solution :

    : When the proton beam enters the magnetic field B normally, it describes a circular path of radius r given by \[r=\frac{mv}{eB}=\frac{v}{\frac{e}{m}B}\] where \[\frac{e}{m}\] is the specific charge of the proton and v is its velocity. Here, \[v={{10}^{9}}\,m\,{{s}^{-1}},\frac{e}{m}={{10}^{11}}C\,k{{g}^{-1}}\] \[B={{10}^{-4}}\,Wb\,{{m}^{-2}}\] \[\therefore \]  \[r=\frac{{{10}^{9}}\,m\,{{s}^{-1}}}{({{10}^{11}}C\,k{{g}^{-1}})({{10}^{-4}}Wb\,{{m}^{-2}})}=100m\]


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