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Manipal Medical Manipal Medical Solved Paper-2005

  • question_answer
    The magnetic force on a point charge is \[\overrightarrow{F}=q(\overrightarrow{v}\times \overrightarrow{B})\] Here,\[q=\]electric charge \[\overrightarrow{v}=\]velocity of point charge \[\overrightarrow{B}=\]magnetic field The dimensions of B is:

    A)  \[[ML{{T}^{-1}}A]\]

    B)  \[[{{M}^{2}}L{{T}^{-2}}{{A}^{-1}}]\]

    C)  \[[M{{T}^{-2}}{{A}^{-1}}]\]

    D)  none of these

    Correct Answer: C

    Solution :

     \[\overrightarrow{F}=q\overrightarrow{v}\times \overrightarrow{g}\] or     \[F=q\upsilon B\sin \theta \] \[\therefore \] \[[B]=\left[ \frac{F}{q\upsilon } \right]=\frac{ML{{T}^{-2}}}{[AT][L{{T}^{-1}}]}\] \[=[M{{T}^{-2}}{{A}^{-1}}]\]


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