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JEE Main & Advanced JEE Main Paper (Held on 12-4-2019 Morning)

  • question_answer
    A thin ring of 10 cm radius carries a uniformly distributed charge. The ring rotates at a constant angular speed of \[40\pi \,rad\,{{s}^{-1}}\]about its axis, perpendicular to its plane. If the magnetic field at its centre is \[3.8\times {{10}^{9}}T,\]then the charge carried by the ring is close to \[({{\mu }_{0}}=4\pi \times {{10}^{7}}\text{ }N/{{A}^{2}})\] : [JEE Main Held on 12-4-2019 Morning]

    A) \[2\times {{10}^{-6}}C\]                 

    B) \[3\times {{10}^{-5}}C\]

    C) \[4\times {{10}^{-5}}C\]                 

    D) \[7\times {{10}^{-6}}C\]

    Correct Answer: B

    Solution :

    \[B=\frac{{{\mu }_{0}}i}{2R}=\frac{{{\mu }_{0}}q\omega }{2R2\pi }\]\[\Rightarrow q=3\times {{10}^{-5}}C\]


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