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JEE Main & Advanced JEE Main Paper (Held On 11 April 2015)

  • question_answer
    A short bar magnet is placed in the magnetic meridian of the earth with north pole pointing north. Neutral points are found at a distance of 30 cm from the magnet on the East - West line, drawn through the middle point of the magnet. The magnetic moment of the magnet in \[A{{m}^{2}}\]is close to : (Given \[\frac{{{\mu }_{0}}}{4\pi }={{10}^{-7}}\] in SI units and \[{{B}_{H}}=\]Horizontal component of earth's magnetic field \[=3.6\times {{10}^{-}}^{5}\] Tesla.) [JEE Main Online Paper (Held On 11 April 2015)]  

    A)  9.7

    B)  4.9

    C)  19.4

    D)  14.6

    Correct Answer: A

    Solution :

      At 30cm from the magnet on its equitorial plane \[{{\vec{B}}_{magnet}}=-\overrightarrow{{{B}_{M}}}\](newtral point) so by equating their magnitude \[\frac{{{\mu }_{0}}}{4\pi }\frac{M}{{{r}^{3}}}=3.6\times {{10}^{-5}}\]Tesla \[\frac{{{10}^{-7}}\times M}{{{(0.3)}^{3}}}=3.6\times {{10}^{-5}}\]Tesal \[M=3.6\times 0.027\times {{10}^{2}}=9.7A{{m}^{2}}\]


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