JEE Main & Advanced Sample Paper JEE Main - Mock Test - 23

  • question_answer
    Two long straight cylindrical conductors with resistivities \[{{\rho }_{1}}\] and \[{{\rho }_{2}}\] respectively are joined together as shown in figure. If current I flows through the conductors, the magnitude of the total free charge at the interface of the two conductors is

    A) zero                              

    B) \[\frac{({{\rho }_{1}}-{{\rho }_{2}})I{{\varepsilon }_{0}}}{2}\]

    C) \[{{\varepsilon }_{0}}I|{{\rho }_{1}}-{{\rho }_{2}}|\]   

    D)        \[{{\varepsilon }_{0}}I|{{\rho }_{1}}+{{\rho }_{2}}|\]

    Correct Answer: C

    Solution :

    [c] Apply Gauss's law: \[\frac{{{q}_{in}}}{{{\varepsilon }_{0}}}~(outgoing\text{ }flux-incoming\text{ }flux)=\Delta (EA)\] \[V=IR=\frac{I\rho \ell }{A}\Rightarrow \frac{V}{\ell }A=I\rho \Rightarrow EA=I\rho \] \[\Rightarrow \,\,\,\frac{{{q}_{in}}}{{{\varepsilon }_{0}}}=I|{{\rho }_{1}}-{{\rho }_{2}}|\,\,\,\,\Rightarrow \,\,\,\,{{q}_{in}}=I{{\varepsilon }_{0}}|{{\rho }_{1}}-{{\rho }_{2}}|\]

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