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JEE Main & Advanced Physics Magnetism Question Bank Earth's Magnetism

  • question_answer
    A short magnet of moment 6.75 Am2 produces a neutral point on its axis. If horizontal component of earth's magnetic field is \[5\times {{10}^{-5}}Wb/{{m}^{2}}\], then the distance of the neutral point should be                                     [SCRA 1994]

    A)            10 cm                                      

    B)            20 cm

    C)            30 cm                                      

    D)            40 cm

    Correct Answer: C

    Solution :

               At neutral point \[\left| \begin{matrix}   \,\text{Magnetic field due } \\   \text{to magnet} \\ \end{matrix} \right|=\,\left| \begin{matrix}   \text{Magnetic field due } \\   \text{to earth} \\ \end{matrix} \right|\]\[\frac{{{\mu }_{0}}}{4\pi }.\frac{2M}{{{d}^{3}}}=5\times {{10}^{-5}}\Rightarrow {{10}^{-7}}\times \frac{2\times 6.75}{{{d}^{3}}}=5\times {{10}^{-5}}\] \[\Rightarrow d=0.3\ m=30\ cm\]


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