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Punjab Medical Punjab - MET Solved Paper-2010

  • question_answer
    A short bar magnet placed with its axis at\[{{30}^{o}}\]with a uniform external magnetic field of\[0.16\,T\]experiences a torque of magnitude\[0.032J\]. The magnetic moment of the bar magnet will be

    A) \[0.23\,\,J{{T}^{-1}}\]                    

    B) \[0.40\,\,J{{T}^{-1}}\]

    C) \[0.80\,\,J{{T}^{-1}}\]                    

    D)  zero

    Correct Answer: B

    Solution :

    Magnetic moment\[M=\frac{t}{B\sin \theta }=\frac{0.032}{0.16\times \sin {{30}^{o}}}=0.40\,\,J{{T}^{-1}}\]


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