Uttarakhand PMT Uttarakhand PMT Solved Paper-2005

  • question_answer
    If a magnet of length 10 cm and pole strength 40 am is placed at an angle\[45{}^\circ \]in an uniform induction field intensity\[2\times {{10}^{-4}}T\]. The couple acting on it will be:

    A)  \[0.5656\times {{10}^{-3}}Nm\]

    B)  \[0.656\times {{10}^{-4}}N-m\]

    C)  \[0.5656\times {{10}^{-5}}Nm\]

    D) \[0.5656\times {{10}^{-4}}N-m\]

    Correct Answer: A

    Solution :

     Here: The length of the magnet\[=10\text{ }cm\] \[=0.1m\] Pole strength \[m=40\text{ }Am\] Angle         \[\theta =45{}^\circ \] Field intensity\[B=2\times {{10}^{-4}}\]tesla The couple is given by \[\overline{L}=mlB\text{ }sin\theta \] \[=40\times 0.1\times 2\times {{10}^{-4}}\sin 45{}^\circ \] \[=8\times {{10}^{-4}}\times 0.707\] \[=0.5656\times {{10}^{-3}}N-m\]


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