JEE Main & Advanced Sample Paper JEE Main - Mock Test - 22

  • question_answer
    The current density varies with radial distance r as \[J=a{{r}^{2}}\], (where a is a constant) in a cylindrical wire of radius R. The current passing through the wire between radial distance R/3 and R/2 is,

    A) \[\frac{65\pi a{{R}^{4}}}{2592}\]        

    B) \[\frac{25\pi a{{R}^{4}}}{72}\]

    C) \[\frac{65\pi {{a}^{2}}{{R}^{4}}}{2938}\]    

    D) \[\frac{81\pi {{a}^{2}}{{R}^{4}}}{144}\]

    Correct Answer: A

    Solution :

    [a]: Current density \[J=a{{r}^{2}}\] Current \[I=\int_{{}}^{{}}{J.dA}=\int\limits_{R/3}^{R/2}{a{{r}^{2}}(2\pi rdr)}\] [where \[dA=2\pi dr\]] \[=2\pi a\left[ \frac{{{r}^{4}}}{4} \right]_{R/3}^{R/2}=\frac{65\pi a{{R}^{4}}}{2592}\]


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