CMC Medical CMC-Medical VELLORE Solved Paper-2009

  • question_answer
    A magnet 10 cm long and having a pole strength 2 Am is deflected through \[30{}^\circ \]from the magnetic meridian. The horizontal component   of   earths   induction is \[0.32\times {{10}^{-4}}T,\] then the value of  deflecting couple is

    A)  \[16\times {{10}^{-7}}Nm\]       

    B)  \[64\times {{10}^{-7}}Nm\]

    C)  \[48\times {{10}^{-7}}Nm\]       

    D)  \[32\times {{10}^{-7}}Nm\]

    Correct Answer: D

    Solution :

                     Deflecting couple is given by \[\tau =MH\,\sin \theta \]                           ?(i) where M = magnetic moment \[=\] pole strength \[\times \] effective length \[=2\times 10\times {{10}^{-2}}\] \[=2\times {{10}^{-1}}A{{m}^{2}}\] H = horizontal component of earths magnetic field \[=0.32\times {{10}^{-4}}T\] and        \[\theta =30{}^\circ \] From Eq. (i), we get \[\tau =2\times {{10}^{-1}}\times 0.32\times {{10}^{-4}}\times \sin 30{}^\circ \]    \[=0.64\times {{10}^{-5}}\times \frac{1}{2}\]    \[=0.32\times {{10}^{-5}}Nm\]    \[=32\times {{10}^{-7}}Nm\]


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