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Manipal Medical Manipal Medical Solved Paper-2005

  • question_answer
    The couple acting on a magnet of length 10 cm and pole strength 15 Am, kept in a field of \[B=2\times {{10}^{-5}}T,\]at an angle of\[30{}^\circ ,\]is:

    A)  \[1.5\times {{10}^{-5}}Nm\] 

    B)  \[1.5\times {{10}^{-3}}Nm\]

    C)  \[1.5\times {{10}^{-2}}Nm\]  

    D) \[1.5\times {{10}^{-6}}Nm\]

    Correct Answer: B

    Solution :

     \[C=MB\,sin\theta \] \[=(m\times 2l)\times 2\times {{10}^{-5}}\sin 30{}^\circ \] \[=15\times 10\times {{10}^{-2}}\times {{10}^{-5}}\times \frac{1}{2}\] \[=1.5\times {{10}^{-5}}Nm\]


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