JEE Main & Advanced JEE Main Paper (Held on 09-4-2019 Morning)

  • question_answer A rigid square loop of side 'a' and carrying current \[{{I}_{2}}\]is lying on a horizontal surface near a long current \[{{I}_{1}}\]carrying wire in the same plane as shown in figure. The net force on the loop due to wire will be : [JEE Main 9-4-2019 Morning]

    A) Attractive and equal to \[\frac{{{\mu }_{0}}{{I}_{1}}{{I}_{2}}}{3\pi }\]

    B) Repulsive and equal to \[\frac{{{\mu }_{0}}{{I}_{1}}{{I}_{2}}}{4\pi }\]

    C) Repulsive and equal to\[\frac{{{\mu }_{0}}{{I}_{1}}{{I}_{2}}}{2\pi }\]

    D) Zero

    Correct Answer: B

    Solution :

    \[{{F}_{3}}\And {{F}_{4}}\]cancel each other Force on PQ will be \[{{F}_{1}}={{I}_{2}}{{B}_{1}}a\]                                     \[={{I}_{2}}\frac{{{\mu }_{0}}{{I}_{1}}}{2\pi a}a\]                                     \[=\frac{{{\mu }_{0}}{{I}_{1}}{{I}_{2}}}{2\pi }\] Force on RS will be \[{{F}_{2}}={{I}_{2}}{{B}_{2}}a\]                                     \[={{I}_{2}}\frac{{{\mu }_{0}}{{I}_{1}}}{2\pi 2a}a\]                                     \[=\frac{{{\mu }_{0}}{{I}_{1}}{{I}_{2}}}{4\pi }\] Net force \[={{F}_{1}}-{{F}_{2}}=\frac{{{\mu }_{0}}{{I}_{1}}{{I}_{2}}}{4\pi }\]repulsion Option [b]


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