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

  • question_answer
    A 30 cm long bar magnet is placed in the magnetic meridian with its north pole pointing south. The neutral point is obtained at a distance of 40cm from the center of the magnet. Find the pole strength of the magnet. The horizontal component of earth's magnetic field is 0.34 gauss.

    A) 25.7 Am

    B) 23.7 Am 

    C) 28.7 Am

    D) 26.7 Am 

    Correct Answer: D

    Solution :

    [d] \[\therefore \,\,\,\,\,\,M=\frac{4\pi }{{{\mu }_{0}}}\frac{{{B}_{H}}{{({{r}^{2}}-{{\ell }^{2}})}^{2}}}{2r}\] \[={{10}^{7}}\times \frac{(0.34\times {{10}^{-4}}).{{[{{(0.40)}^{2}}-{{(0.15)}^{2}}]}^{2}}}{2\times 0.40}\] \[=8.0\,A{{m}^{2}}\] The pole strength of the magnet is, \[m=\frac{M}{2\ell }=\frac{8.0}{0.30}=26.7\,Am\]


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