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JCECE Medical JCECE Medical Solved Paper-2015

  • question_answer
    A long insulated copper wire is closely wound as a spiral of \[{{N}_{0}}\]turn. The spiral lies in they-z plane and a steady current \[{{I}_{0}}\]flows through the wire. The X-component of the magnetic field at the centre of the spiral is (assume  inner radius as \[{{R}_{1}}\]and outer radius as \[{{R}_{2}}\]).

    A) \[\frac{{{\mu }_{0}}{{N}_{0}}{{l}_{0}}}{4({{R}_{2}}-{{R}_{1}})}\,\,\ln \,({{R}_{2}}/{{R}_{1}})\]

    B)  \[\frac{2{{\mu }_{0}}{{N}_{0}}{{l}_{0}}}{{{R}_{2}}-{{R}_{1}}}\,\,\ln \,({{R}_{2}}/{{R}_{1}})\]

    C)  \[\frac{{{\mu }_{0}}{{N}_{0}}{{l}_{0}}}{2({{R}_{2}}-{{R}_{1}})}\,\,\ln \,{{({{R}_{2}}/{{R}_{1}})}^{2}}\]

    D)  \[\frac{{{\mu }_{0}}{{N}_{0}}{{l}_{0}}}{2({{R}_{2}}-{{R}_{1}})}\,\,\ln \,({{R}_{2}}/{{R}_{1}})\]

    Correct Answer: D

    Solution :

     Consider a small strip of thickness di at distance \[r\] from the centre, then number of turns in this strip would be \[dN=\left( \frac{{{N}_{0}}}{{{R}_{2}}-{{R}_{1}}} \right)dr\] Magnetic field due to this element at the centre of the coil will be \[dB=\frac{{{\mu }_{0}}dN{{I}_{0}}}{2r}=\frac{{{\mu }_{0}}{{N}_{0}}{{I}_{0}}}{2({{R}_{2}}-{{R}_{1}})}\frac{dr}{r}\] Total magnetic field at the centre of the spiral \[B=\int_{r={{R}_{1}}}^{r={{R}_{2}}}{\frac{{{\mu }_{0}}{{N}_{0}}{{I}_{0}}}{2({{R}_{2}}-{{R}_{1}})}}\frac{dr}{r}=\frac{{{\mu }_{0}}{{N}_{0}}{{I}_{{}}}}{2({{R}_{2}}-{{R}_{1}})}\ln \,\left( \frac{{{R}_{2}}}{{{R}_{1}}} \right)\]


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