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JCECE Engineering JCECE Engineering Solved Paper-2011

  • question_answer
    A short bar magnet placed with its axis at \[{{30}^{o}}\] with a uniform external magnetic field of \[0.16\,\,T\] experience a torque of magnitude\[0.032\,\,J\]. 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 \[M=\frac{T}{B\sin \theta }=\frac{0.032}{0.16\times \sin {{30}^{o}}}\]                 \[=0.40\,\,J/T\]


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