MGIMS WARDHA MGIMS WARDHA Solved Paper-2015

  • 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\[\frac{R}{4}\]and\[\frac{R}{4}\]is

    A)  \[\frac{20\pi a{{R}^{4}}}{331}\]                

    B)  \[\frac{15\pi a{{R}^{4}}}{512}\]

    C)  \[\frac{\pi a{{R}^{4}}}{332}\]                     

    D)  \[\frac{3\pi a{{R}^{4}}}{864}\]

    Correct Answer: B

    Solution :

                    Given \[J=a{{r}^{2}}\] As       \[I=\int{J\,dA}\] Here,     \[dA=2\pi rdr\]                 \[[\because A=\pi {{r}^{2}}]\] So           \[I=\int_{R/4}^{R/2}{a{{r}^{2}}(2\pi r)dr}\]                 \[I=2\pi a\int\limits_{R/4}^{R/2}{{{r}^{3}}dr}=2\pi a\left[ \frac{{{r}^{4}}}{4} \right]_{R/4}^{R/2}\]                 \[=\frac{\pi a}{2}\left[ {{\left( \frac{R}{2} \right)}^{4}}-{{\left( \frac{R}{4} \right)}^{4}} \right]\]                 \[=\frac{\pi a{{R}^{4}}}{2}\left[ \frac{1}{16}-\frac{1}{256} \right]\]                 \[=\frac{\pi a{{R}^{4}}}{2}\left[ \frac{256-16}{4096} \right]\]                 \[=\frac{15\pi a{{R}^{2}}}{512}\]


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