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JEE Main & Advanced Physics Magnetism Question Bank Self Evaluation Test - Magnetism and Matter

  • question_answer
    A bar magnet 8 cms long is placed in the magnetic merdian with the N-pole pointing towards geographical north. Two netural points separated by a distance of 6 cms are obtained on the equatorial axis of the magnet. If horizontal component of earth's field \[=3.2\times {{10}^{-5}}T\], then pole strength of magnet is    

    A) \[5\,ab-amp\times cm\]

    B) \[10\,ab-amp\times cm\]

    C) \[2.5\,ab-amp\times cm\]

    D) \[20\,ab-amp\times cm\]

    Correct Answer: A

    Solution :

    [a] Here, \[2\ell =8\,cm,\] \[\ell =4cm,\] \[d=\frac{6}{2}=3\,cm\]. At neutral point, \[H=B=\frac{{{\mu }_{0}}}{4\pi }\frac{M}{{{({{d}^{2}}+{{\ell }^{2}})}^{3/2}}}\] \[={{10}^{-7}}\frac{M}{{{(5\times {{10}^{-2}})}^{3}}}=\frac{M}{1250}\] \[\therefore \,\,\,M=1250H=1250\times 3.2\times {{10}^{-5}}\,A{{m}^{2}}\]              \[m=\frac{M}{2\ell }=\frac{1250\times 3.2\times {{10}^{-5}}}{8\times {{10}^{-2}}}Am.\,\,\,\,=0.5\]             \[Am=0.5\times \frac{1}{10}ab\,\] \[amp\times 100\,cm\]             = 5 ab-amp cm.


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